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# Does Force depends on Velocity? [closed]

A Engine $E$ is hinged on a Wooden block which is placed on a perfectly smooth surface:

Now the block is given an initial velocity $V$ (by some external agent) and the Engine is started.

The Engine has some sort of fuel and is capable of producing constant $10\ \mathrm W$ power.

The Engine applies some force on the Block (as a result of the generated power). We know , Power = Force x Velocity.

Again, the velocity of the block, engine $V$ system can't alter (because of absence of external forces).

So, can we say the force applied by the Engine $E$ on the block depends on the initial velocity $V$ of the block?

## closed as unclear what you're asking by John Rennie, sammy gerbil, Jon Custer, Kyle Kanos, GertJan 8 '17 at 18:53

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• can't alter velocity due to absence of external forces - then how do cars move? What exactly does the engine do in your system? Just sits on top of the block? – Gallifreyan Jan 6 '17 at 11:29
• Cars move due to Friction – Programmer Jan 6 '17 at 11:49
• I'm not saying there's change in momentum here, just a side note. What I don't understand is, what sort of force does the engine apply to the block? – Gallifreyan Jan 6 '17 at 12:37
• to answer generally the title: F=dp/dt, and as p=mv force is proportonal to the change in time of velocity times mass. I applies to all forces so you cannot maket the last statement, generally – anna v Jan 6 '17 at 13:11
• This really isn't such a bad question. It doesn't deserve all the down votes. It might be clearer if the engine was a rocket. The question might be rephrased like this. "A rocket burns fuel at a constant rate, and therefore has constant power. It applies a constant force to the block. Velocity continually increases. But $P = F \cdot V$. What gives?" – mmesser314 Jan 6 '17 at 14:37

Your setup makes no sense. It is unclear what happens with the power your engine generates. What is clear is that since there is no relative motion between the engine and the block (at least that's how I interpret what you are saying), that power is completely unrelated to either the force between the engine and the block or the velocity of the block. You can't just take an equation like $P=\mathbf F\cdot\mathbf V$ and plug in some random quantities. You would have to separate a (sub-)system that the power is being transferred to, and analyze forces and velocities on that subsystem.


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3 Tutor System
Starting just at 265/hour

# Show how you would connect three resistors, each of resistance 6 $$\Omega$$, so that the combination has a resistance of (i) 9 $$\Omega$$, (ii) 4 $$\Omega$$.

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors,

$$\Rightarrow$$ 6 $$\Omega$$ + 6 $$\Omega$$ + 6 $$\Omega$$ = 18 $$\Omega$$,

which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be $$\frac{6}{2}$$ = 3 $$\Omega$$ is also not desired.

Hence, we should either connect the two resistors in series or parallel.

(a) Two resistor in parallel

Two 6 $$\Omega$$ resistors are connected in parallel. Their equivalent resistance will be
$$\frac{1}{(\frac{1}{6} + \frac{1}{6} )}= \frac{6}{2} = 3 \Omega .$$

The third 6 $$\Omega$$ resistor is in series with 3 $$\Omega$$. Hence, the equivalent resistance of the circuit is 6 $$\Omega$$+ 3 $$\Omega$$ = 9 $$\Omega$$.

(b) Two resistor in parallel

Two 6 $$\Omega$$ resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 $$\Omega$$.
The third 6 $$\Omega$$ resistor is in parallel with 12 $$\Omega$$. Hence, equivalent resistance will be

$$12 x \frac{6}{12} + 6 = 4 \Omega .$$

Therefore, the total resistance is 4$$\Omega$$

format: plain


text:
Read the excerpt from chapter 8 of Obasan by Joy Kogawa. "Who are the letters from?” I ask loudly. She does not respond. Her face is expressionless. After a while she puts down the magnifying glass. "Everyone someday dies,” she says again. By repeating this so often, I suppose she is trying to make realizable what is real. Surely that is task enough for her today. But I ask once more, "Who are the letters from?” She stares steadily at the table. The greater my urgency to know, the thicker her si

Read the excerpt from chapter 8 of Obasan by Joy Kogawa. "Who are the letters from?” I ask loudly. She does not respond. Her face is expressionless. After a while she puts down the magnifying glass. "Everyone someday dies,” she says again. By repeating this so often, I suppose she is trying to m...

Does an object need a force to start moving

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Does The brake pedal and clutch pedal are the same????

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Which statement is correct? Group of answer choices none of these H2O is linear. The bonds in LiF have a more covalent character than those in F2. The diatomic molecule Cl2 is an example of a polar molecule. The molecule ClO2 cannot be accurately described by a Lewis structure consistent wi...

Write an equation that has a solution of 7, includes a variable, and uses multiplication. write a real-world problem thwt you could represent with youre equation. show how you know that 7 is the solution.

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HELPPP ITS QUESTION 17 BTW

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A bookstore has 45 copies of a bestseller available on Saturday morning. Twenty-one customers purchase the book that day. On Monday, a shipment of 25 books arrives and 14 copies are sold. Which statements are correct?

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4x+2y=-8 3x-2y=22x=y=

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Carries amino acids to the ribosome?

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Hey may someone please help me solve/graph and explain professionally (step by step) the following question? SOLVE the inequality. GRAPH the solution. Important* Use a good verbal explanation. 12b + 12 ≥ 60 I will mark as brainliest! Thank you!

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A realetor recieves 5% of the cost of every property he sells how much does the relator recieve for a property that sells for 50000

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Find the missing value. Hint: Use the number line to find the missing value. -(-7)=5 {  -15 → 15 -10 -5 0 5 10

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text:
# Why do satellites with greater altitude travel at a slower speed?

For example, a geostationary satellite travels slower than the ISS. What is going on?

Kepler's third law in this situation tells us that $$frac{R^3}{P^2} = frac{GM}{4pi^2},$$ where $R$ is the orbital radius (or semi-major axis for an elliptical orbit), $P$ is the orbital period and $M$ is the mass of the Earth (actually, it is the mas of the Earth plus the mass of the satellite, but we can neglect the latter). Thus the orbital period $P propto R^{3/2}$.

To calculate the orbital speed, we divide the orbital circumference by the orbital period $$v = frac{2pi R}{P}.$$

Then substituting for $P$ from Kepler's third law, we get $$v = sqrt{frac{GM}{R}}$$

Thus, the larger the orbital radius, the slower the orbital speed.

Or you can just use Newton's second law and equate the centripetal acceleration to the gravitational force: $$m frac{v^2}{R} = Gfrac{Mm}{R^2},$$ where $m$ is the mass of the satellite, which leads directly to the same result.

## Why a satellite in low earth orbit moves faster than a satellite in geostationary orbit?

The best answer is simply: That's the way gravity works.

If you take Newton's deceptively simple formula for the gravitational force

between two masses, and if you have enough geometry and calculus to

massage the formula around, look at it from different angles, and follow it

through to some of its implications, then some wonderful things happen.

The first result is that all of Kepler's laws of planetary motion fall out on the

table, and you no longer have to ask "Why do the planets do that ?" They

do everything they do because that's the way gravity works.

You also discover that neither the size of the orbit nor the speed in orbit depends

on the mass of the orbiting body. The speed depends only on the size of the orbit.

separating himself from his spacecraft but still staying in the same Earth orbit that

the spacecraft is in, so that he doesn't go sailing away from the spacecraft just

And you also discover the strange fact that the larger the orbit, the slower the

orbiting body moves. An astronaut on a space walk in low-earth-orbit is orbiting

## How Fast Do Satellites Travel?

A satellite requires a speed of 17,450 miles per hour in order to maintain a low Earth orbit. Satellites in higher orbits travel more slowly for example, a geostationary satellite only orbits at 6,858 miles per hour.

When satellites orbit, they are falling around the planet because the gravitational pull keeps the satellite in motion. The closer they are, the greater the pull is. Orbit around Earth was first thought of by Isaac Newton with his cannonball thought experiment. It stated that if a cannon on top of a mountain fired a ball, the faster it shot, the farther it would travel. He correctly reasoned that a speed could be reached that would allow an object to continue traveling around the planet.

## 24.4 Time in General Relativity

General relativity theory makes various predictions about the behavior of space and time. One of these predictions, put in everyday terms, is that the stronger the gravity, the slower the pace of time. Such a statement goes very much counter to our intuitive sense of time as a flow that we all share. Time has always seemed the most democratic of concepts: all of us, regardless of wealth or status, appear to move together from the cradle to the grave in the great current of time.

But Einstein argued that it only seems this way to us because all humans so far have lived and died in the gravitational environment of Earth. We have had no chance to test the idea that the pace of time might depend on the strength of gravity, because we have not experienced radically different gravities. Moreover, the differences in the flow of time are extremely small until truly large masses are involved. Nevertheless, Einstein’s prediction has now been tested, both on Earth and in space.

### The Tests of Time

An ingenious experiment in 1959 used the most accurate atomic clock known to compare time measurements on the ground floor and the top floor of the physics building at Harvard University. For a clock, the experimenters used the frequency (the number of cycles per second) of gamma rays emitted by radioactive cobalt. Einstein’s theory predicts that such a cobalt clock on the ground floor, being a bit closer to Earth’s center of gravity, should run very slightly slower than the same clock on the top floor. This is precisely what the experiments observed. Later, atomic clocks were taken up in high-flying aircraft and even on one of the Gemini space flights. In each case, the clocks farther from Earth ran a bit faster. While in 1959 it didn’t matter much if the clock at the top of the building ran faster than the clock in the basement, today that effect is highly relevant. Every smartphone or device that synchronizes with a GPS must correct for this (as we will see in the next section) since the clocks on satellites will run faster than clocks on Earth.

The effect is more pronounced if the gravity involved is the Sun’s and not Earth’s. If stronger gravity slows the pace of time, then it will take longer for a light or radio wave that passes very near the edge of the Sun to reach Earth than we would expect on the basis of Newton’s law of gravity. (It takes longer because spacetime is curved in the vicinity of the Sun.) The smaller the distance between the ray of light and the edge of the Sun at closest approach, the longer will be the delay in the arrival time.

In November 1976, when the two Viking spacecraft were operating on the surface of Mars, the planet went behind the Sun as seen from Earth (Figure 24.11). Scientists had preprogrammed Viking to send a radio wave toward Earth that would go extremely close to the outer regions of the Sun. According to general relativity, there would be a delay because the radio wave would be passing through a region where time ran more slowly. The experiment was able to confirm Einstein’s theory to within 0.1%.

### Gravitational Redshift

What does it mean to say that time runs more slowly? When light emerges from a region of strong gravity where time slows down, the light experiences a change in its frequency and wavelength. To understand what happens, let’s recall that a wave of light is a repeating phenomenon—crest follows crest with great regularity. In this sense, each light wave is a little clock, keeping time with its wave cycle. If stronger gravity slows down the pace of time (relative to an outside observer), then the rate at which crest follows crest must be correspondingly slower—that is, the waves become less frequent.

To maintain constant light speed (the key postulate in Einstein’s theories of special and general relativity), the lower frequency must be compensated by a longer wavelength. This kind of increase in wavelength (when caused by the motion of the source) is what we called a redshift in Radiation and Spectra. Here, because it is gravity and not motion that produces the longer wavelengths, we call the effect a gravitational redshift .

The advent of space-age technology made it possible to measure gravitational redshift with very high accuracy. In the mid-1970s, a hydrogen maser, a device akin to a laser that produces a microwave radio signal at a particular wavelength, was carried by a rocket to an altitude of 10,000 kilometers. Instruments on the ground were used to compare the frequency of the signal emitted by the rocket-borne maser with that from a similar maser on Earth. The experiment showed that the stronger gravitational field at Earth’s surface really did slow the flow of time relative to that measured by the maser in the rocket. The observed effect matched the predictions of general relativity to within a few parts in 100,000.

These are only a few examples of tests that have confirmed the predictions of general relativity. Today, general relativity is accepted as our best description of gravity and is used by astronomers and physicists to understand the behavior of the centers of galaxies, the beginning of the universe, and the subject with which we began this chapter—the death of truly massive stars.

### Relativity: A Practical Application

By now you may be asking: why should I be bothered with relativity? Can’t I live my life perfectly well without it? The answer is you can’t. Every time a pilot lands an airplane or you use a GPS to determine where you are on a drive or hike in the back country, you (or at least your GPS-enabled device) must take the effects of both general and special relativity into account.

GPS relies on an array of 24 satellites orbiting the Earth, and at least 4 of them are visible from any spot on Earth. Each satellite carries a precise atomic clock. Your GPS receiver detects the signals from those satellites that are overhead and calculates your position based on the time that it has taken those signals to reach you. Suppose you want to know where you are within 50 feet (GPS devices can actually do much better than this). Since it takes only 50 billionths of a second for light to travel 50 feet, the clocks on the satellites must be synchronized to at least this accuracy—and relativistic effects must therefore be taken into account.

The clocks on the satellites are orbiting Earth at a speed of 14,000 kilometers per hour and are moving much faster than clocks on the surface of Earth. According to Einstein’s theory of relativity, the clocks on the satellites are ticking more slowly than Earth-based clocks by about 7 millionths of a second per day. (We have not discussed the special theory of relativity, which deals with changes when objects move very fast, so you’ll have to take our word for this part.)

The orbits of the satellites are 20,000 kilometers above Earth, where gravity is about four times weaker than at Earth’s surface. General relativity says that the orbiting clocks should tick about 45 millionths of a second faster than they would on Earth. The net effect is that the time on a satellite clock advances by about 38 microseconds per day. If these relativistic effects were not taken into account, navigational errors would start to add up and positions would be off by about 7 miles in only a single day.

I quite agree that it is not intuitive. However, orbital mechanics are frequently not intuitive, probably because we don't get to experience an orbital environment on a regular basis (if ever).

Let's just assume we're talking about circular orbits for the remainder of my post, since you are a beginner in orbital mechanics.

There is only one speed that a given circular orbit of a certain altitude can go. Keep in mind that stable orbits do not require any force from an engine to keep going as they have been. Basically, in a circular orbit, the falling-toward-the-planet motion is matched exactly by the moving-forward motion.

Sir Issac Newton figured this out, and exemplified it with a thought experiment called Newton's Cannonball.

Note that if the orbital speed is too slow for that altitude, the cannonball crashed into the planet.

And if the orbital speed is too high for the altitude, the orbit will be an ellipse, rather than circular, or the cannonball may even escape Earth altogether!

Finally, if the cannonball is launched at the 'correct' orbital speed to be in a circular orbit at that altitude, it will neither crash, nor fly away, but will remain stable, traveling around earth at that particular velocity.

At different altitudes, this Goldilocks velocity is different. If the orbit is closer to the planet, the effect of gravity is higher, so the orbiting object must be moving faster to counteract the falling. When the orbiting object is further away, there is less falling force due to gravity (because gravitational force is based on distance), and so the object does not need to be moving as fast to counteract the falling force.

From Wikipedia's Geocentric Orbit article, we know that Low Earth Orbit could be, for example, an altitude of 160km. At this altitude, the Goldilocks velocity to keep a circular orbit is about 8000 m/s, and takes about 90 minutes.

Now what happens if we look at a slightly higher altitude? Well the velocity is lower, and the path the orbiting object travels gets bigger (the circle is bigger), so both of those factors make the orbit take longer. A slightly higher orbit might take 100 minutes instead of 90.

For a geosynchronous orbit, the orbit has to take 24 hours instead of 90 minutes, because the earth takes 24 hours to spin. This happens when the circle is expanded to an altitude of about 35000 km. The Goldilocks velocity at this altitude is about 3000 m/s.

This is all somewhat simplified, but the broad strokes are all there. As Organic Marble pointed out, you could try to force a craft to orbit at a different altitude in a 24 hour period, but it would not be a stable orbit, you would need engines to keep it going.

Simply put, for a circular orbit and a given central body, the orbital period is solely a function of the radius. A geosynchronous orbit is just the orbital radius at which the corresponding period is equal to the rotational period of the Earth.

You could fly around the Earth in 24 hours at any altitude, but not without propulsion.

Think of it this way. A circular orbit is characterized by the fact that the fictitious centrifugal force is exactly canceled out by the (centripetal) force of gravity. If that wasn't the case, if gravity was stronger, the satellite would begin to sink if gravity was weaker, it would begin to rise. In either case, it would no longer be in a circular orbit.

A geostationary orbit is characterized by its angular velocity (specifically, $2pi$ radians per day). The centrifugal force for circular motion at constant angular velocity is proportional to the radius. The gravitational force is proportional to the inverse square of the radius. So you have an equation in the (generic) form, $Ar = B/r^2$ where $A$ and $B$ are some numbers. This equation is not valid for arbitrary $r$ rather, you can calculate the value of $r$ by solving the equation for it.

When you plug in the numbers, this is exactly what happens. The centrifugal force for a mass $m$ is given by $F_c=mv^2/r = momega ^2r$ where $omega$ is the angular velocity. The gravitational force for a mass $m$ is $F_g = GMm/r^2$ where $G$ is Newton's constant of gravity and $M$ is the Earth's mass. When these two are equal, you have $momega^2 r = GMm/r^2$ or $r = sqrt[3]$. When you plug in the numbers, you get $r simeq 4.23 imes 10^7$ meters, or after subtracting the radius of the Earth, an altitude of approximately 36,000 km. This is the only value for which the two forces cancel at an angular velocity of one full revolution per day, so this is the geostationary altitude.

A satellite in a geosynchronous geostationary orbit is both at specific altitude (26199 miles high), specific direction (equatorial orbit going from west to east), and specific velocity (1.91 miles per second). The altitude implies the velocity because if the velocity were incorrect, the satellite would not stay in orbit.

This is my attempt at getting the value. It is off by a little bit but this may be due to accuracy of numbers used and considering the orbit perfectly circular.

Basically, in order for it to orbit correctly it must have the same angular velocity as earth (rotate at the same speed), which means having the same frequency or time period of rotation as the earth.

The weight of the object orbiting must then be equal to the centripetal force it has acting on it due to the circular motion. As others have said if these two forces are not equal then it will either crash into earth or fly off.

From this point onward it is just maths to calculate the actual value, remembering that this value of r gives the radius of orbit which is distance from the centre of the earth, so you must subtract R to get the height above earth.

From this you could calculate a velocity that the satellite is traveling at but in this area generally angular velocity is used more. Most people would not know what to do with this velocity either as it doesn't mean much and isn't useful.

What's so special about the magic number 22,000 that makes it possible to do a geosynchronous orbit at that altitude but not at any arbitrary altitude?

Lift an object to an orbital altitude of 1 meter. Let it go. What happens?

The centrifugal force of a geosynchronous orbit of 1 meter cannot support an object against gravity.

Then assume that Pluto is in a geosynchronous orbit. that is to say the dwarf planet needs to revolve around Earth in 24 hours. The speed it would need for that is approximately light speed. What happens?

Pluto will disappear out into the big black yonder, because Earth's gravity cannot possibly contain an object in a geosynchronous orbit of 7.5 billion kilometers.

Somewhere in between these two extremes is the altitude where gravity and the centrifugal force of a 24 hour orbit are equal and balance each other out.

That - special - altitude is 22 000 miles.

Move higher up and the centrifugal force of a 24 hours orbit is too strong. it will overcome gravity and result in an elliptic orbit, or make the object break away from Earth all-together. Move lower, and the centrifugal force is too weak to balance out gravity and the object will begin to lose altitude, again resulting in an eccentric orbit, or possibly even crash into the atmosphere.

You're falling around the earth at any altitude at any speed. Even if you throw a ball, it is falling around the earth. It just doesn't have enough velocity to keep from hitting it. So the sweet spot is for an orbit that you travel far enough that the curvature of the earth is equal to how far you fell. The closer you are the more gravity, the less distance you have to fall before you hit, the faster you have to go for the earth to curve away from / out of your fall. The higher you are the slower you can go as the earth curves out of your way — less gravity. This way you don't have to add any energy — you just keep falling. At a certain altitude, your speed exactly matches the rotation of the earth. This is great because we can point our satellite dish at it. If you want to be geosync at any other altitude, you can be — but you will need fuel/energy and a lot of it to do it and you won't be weightless. You are only weightless because you are falling. If there was a tower built up that high, you would stand on it with gravity just as you would down here. A little less gravity — but still gravity. Hence the falling. You are weightless when you fall down here too. You're just too worried about sticking the landing to notice.

There is no magical number 22,000.

If, as you say, you could achieve geostationary orbit at any altitude, then you could go to any location on Earth's equator, hold an object at arm's length, release it, and it expect it to remain in place, essentially hovering in the air. After all, you and the object are travelling some 1,000 miles per hour around Earth's axis. We all know the object would simply fall to ground.

We also know that objects in low-Earth orbit must travel at some 17,000 miles per hour to remain in orbit, taking some 90 minutes to complete one orbit. We also know that the Moon is in orbit around the Earth (strictly speaking, the Earth-Moon barycenter), is about 240,000 miles away, and completes one orbit in about 27 days, travelling something like 2,500 miles per hour. We also know that gravity follows the inverse-square law, diminishing in proportion to the square of the distance.

What does this tell us about orbits in general? For one thing, the closer an object to the body it is orbiting, the more it must oppose gravity, which it can only do by traveling faster, which requires greater acceleration to remain on the closed, curved path we call an orbit. Given the two examples of low Earth orbit and the Moon, there must be an infinite range of orbital distances, each of which has an associated velocity and period. There must therefore be an orbit where the period coincides with the rotation of the Earth, and it will have its own specific distance.

Given the above, knowing Earth's gravitational acceleration (

9.8 m/s/s at the surface), the radius of the Earth (the point at which gravity has that value), the inverse-square law, and the formula for circular motion relating radius and period to acceleration, we can calculate the distance at which an orbit will have a desired period. It turns out that the orbital distance at which the period coincides with the Earth's rotation occurs some 22,000 mile up.

## Changing orbits and changing speed

Long time reader, Fran, asked for a request and I can't turn it down. What happens when you have a spacecraft that wants to change orbital distances. Do you need to speed up or slow down? Let's begin.

So, I have some spacecraft orbiting a planet - say Earth in a perfectly circular orbit. What must be true? Well, I have done this one before, so let me cut to the chase. The space craft only has one force on it, the gravitational force. Also, it is accelerating because it is moving in a circle. Here is a diagram.

With just the gravitational force and with a circular acceleration, the following must be true in the direction of the center of the planet (and circle):

I could solve this for the velocity needed for an orbit with radius r - but I won't. Instead let me find the kinetic energy needed for an orbit. Multiplying both sides of that equation be r over 2, I get:

Now for energy. If I consider the satellite (or spacecraft) and the Earth as the system, then there is no external work on the system so:

The gravitational potential energy for the Earth-thing system is:

This gives a total energy for an object in orbit of:

Now let's pretend. Suppose we are in orbit at a distance r1 from the center of the Earth. This means that we would have to have an energy of:

With a kinetic energy and velocity:

Mission Command now wants the spacecraft at a lower orbit, say r2. Do I need to speed up or slow down? First, what happens to the total change in energy from E1 to E2?

Since r2 is smaller than r2, the change in energy for the system is negative. What does this mean? It means that the work on the system must be negative. Remember that the work-energy principle says:

The only way to make the work done negative would be to have a force (in this case from your rockets) in the opposite direction as the way you are going. Would your speed increase or decrease? Well, from the speed equation above, I can see that as r gets smaller then (oh, and forgive me for calling v the velocity when I am using it as a scalar):

So, the change in speed of the spacecraft is positive as it gets closer to the Earth (since r1 is greater than r2). This is kind of cool. You are doing negative work on the system, but the kinetic energy increases. It is not as odd as it might seem. Maybe it is just counter intuitive. But don't forget about the gravitational potential energy. As the spacecraft moves down, the potential energy decreases. It turns out the potential energy decreases more than energy needed to orbit. So, if you just "fell" to a lower orbit, you would be going too fast to be in a circular orbit. Maybe this energy graph will help.

I guess you just have to look at two things. The K curve gets bigger, but the total energy curve gets smaller. So, you speed up but you have to fire your rockets backwards.

## Einstein&rsquos Dilemma

Einstein was heavily influenced by the works of two great physicists. First, there were the laws of motion discovered by his idol, Newton, and second, were the laws of electromagnetism laid down by Maxwell. The two theories, however, were contradictory. Maxwell postulated that the speed of an electromagnetic wave, such as light, is fixed &mdash an exorbitant 186,000 miles per second. He claimed that this was a fundamental fact about the Universe.

Nothing travels faster than the speed of light. (Photo Credit: Pexels)

Whereas, Newton&rsquos law implied that velocities are always relative. The speed of a car traveling at 40 mph is 40 mph relative to a stationary observer, but only 20 mph relative to a car traveling adjacent to it at 20 mph. Or, 60 mph to the same car whizzing by in the opposite direction. This concept of relative velocity is incompatible with Maxwell&rsquos apparently fundamental fact when applied to the speed of light. This presented Einstein with a grievous dilemma.

The contradiction led Einstein to make a mind-boggling yet also one of the most groundbreaking claims in the history of physics &mdash a collocation of statements that is, of course, not surprising at all. To understand the contradiction and consequently why time slows down, consider another ingenious thought experiment, one of Einstein&rsquos absolute best. Einstein imagined a man on a station platform, on both sides of whom two lightning bolts strike. The man, standing right in the middle of these two points, observes the resulting beams of light from both sides at the same time.

However, things get peculiar when a fellow on a train views this scene while he moves past it at the speed of light. According to the laws of motion, light from the bolt closer to the train will reach the man earlier than the light from the bolt further from the train. The measurement of the speed of light made by both men will differ in their magnitude. But how is this possible when we recall that the speed of light, according to Maxwell, must be constant, regardless of the motion of an observer &ndash a so-called &ldquofundamental&rdquo fact of the Universe?

To compensate for this discrepancy, Einstein suggested that time itself slowed down such that the speed of light remained constant! Time for the man on the train passed slower relative to the time for the man on the platform. Einstein called this time dilation.

## Satellites

Weather Satellites are an important observational tool for all scales of NWS forecasting operations. Satellite data, having a global view, complements land-based systems such as radiosondes, weather radars, and surface observing systems.

There are two types of weather satellites: polar orbiting and geostationary. Both satellite systems have unique characteristics and produce very different products.The two polar orbiting satellites, in their north-south orbits, observe the same spot on the Earth twice daily, once during the daylight and once at night. Polar orbiting satellites provide imagery and atmospheric soundings of temperature and moisture data over the entire Earth. Geostationary satellites are in orbit 22,000 miles above the equator, spin at the same rate of the Earth and constantly focus on the same area. This enables the satellite to take a picture of the Earth, at the same location, every 30 minutes. Computer processing of this data creates &ldquomovie loops&rdquo of the data that forecasters use as their real-time &ldquobird&rsquos eye view&rdquo from space.

 The East-West orbit of GOES satellites depicted in the yellow circle. The North-South orbit of Polar orbiting satellites depicted in the yellow line.

The two US geostationary satellites provide imagery over North and South America and the Atlantic and Pacific oceans. During severe weather outbreaks, the geostationary satellites can be commanded to take images every 5- 15 minutes, and will focus in on smaller impacted area. On very special occasions the geostationary satellites can be commanded to take a picture every minute, but of a very small area like a severe thunderstorm. Geostationary satellites can also take atmospheric profiles of temperature and moisture, but at a reduced resolution compared to polar satellites and radiosonde soundings.

NOAA newest geostationary weather satellites, GOES-16 was successfully launched on November 19, 2016. When operational, GOES-16 will provide continuous imagery and atmospheric measurements of Earth&rsquos Western Hemisphere, total lightning data, and space weather monitoring to provide critical atmospheric, hydrologic, oceanic, climatic, solar and space data.

GOES-16&rsquos environmental data products, expected to be operational by late 2017, will support short-term 1-2 day weather forecasts and severe storm watches and warnings, maritime forecasts, seasonal predictions, drought outlooks and space weather predictions.

GOES-16 will offer 3x more types of imagery with 4x greater resolution, and available 5x faster than ever before.

GOES-16 can multi-task. The satellite will scan the Western Hemisphere every 15 minutes, the Continental U.S. every five minutes, and areas of severe weather every 30-60 seconds, all at the same time.

GOES-16 can provide images of severe weather as frequently as every 30 seconds!

The satellite&rsquos revolutionary Geostationary Lightning Mapper (GLM) will be the first-ever operational lightning mapper flown from geostationary orbit.

For a more detailed description of the polar and geostationary satellites visit:

To learn how Weather Satellites Imagery and Data products are used in NWS operations, visit:

## Mastering Physics Solutions Chapter 12 Gravity

Chapter 12 Gravity Q.1CQ
It is often said that astronauts in orbit experience weightlessness because they are beyond the pull of Earth’s gravity. Is this statement correct? Explain.
Solution:
No The force of Earth’s gravity is practically as strong in orbit as it is on the surface of Earth The astronauts experience weightlessness because they are in constant free fall.

Chapter 12 Gravity Q.1P
CE System A has masses m and m separated by a distance r system B has masses m and 2m separated by a distance 2r system C has masses 2m and 3m separated by a distance 2r, and system D has masses 4m and 5m separated by a distance 3r. Rank these systems in order of increasing gravitational force. Indicate ties where appropriate.
Solution:

Chapter 12 Gravity Q.2CQ
When a person passes you on the street, you do not feel a gravitational tug. Explain.
Solution:

Chapter 12 Gravity Q.2P
In each hand you hold a 0.16-kg apple. What is the gravitational force exerted by each apple on the other when their separation is (a) 0.25 m and (b) 0.50 m?
Solution:

Chapter 12 Gravity Q.3CQ
Two objects experience a gravitational attraction. Give a reason why the gravitational force between them does not depend on the sum of their masses.
Solution:
The force of gravity between two point masses m1 and m2, separated by a distance r, is attractive and of magnitude

where G is the universal gravitational constant.
Gravity exerts an action-reaction pair of forces on m1 and m2. That is, the force exerted by gravity on m1 is equal in magnitude but opposite in direction to the force exerted on m2. It is dependent on the product of masses. If the gravitational force depended on the sum of the two masses, it would predict a non-zero force even when one of the masses is zero. That is, there would be a gravitational force between a mass and a point in empty space, which is certainly not what we observed.

Chapter 12 Gravity Q.3P
A 6.1-kg bowling ball and a 7.2-kg bowling ball rest on a rack 0.75 m apart. (a) What is the force of gravity exerted on each of the balls by the other ball? (b) At what separation is the force of gravity between the balls equal to 2.0 × 10?9N?
Solution:

Chapter 12 Gravity Q.4CQ
Imagine bringing the tips of your index fingers together. Each finger contains a certain finite mass, and the distance between them goes to zero as they come into contact. From the force law F = Gm1m2/r2 one might conclude that the attractive force between the fingers is infinite, and, therefore, that your fingers must remain forever stuck together. What is wrong with this argument?
Solution:
As the tips of the fingers approach one another, we can think of them as being two small spheres that touch each other. Even though the two spheres touch each other, the distance between the centers is not zero. This is always a finite number. Therefore, the force between the spheres is always finite, even they touch each other. As such, the two fingers simply experience the finite force of two point masses separated by a finite distance.

Chapter 12 Gravity Q.4P
A communications satellite with a mass of 480 kg is in a circular orbit about the Earth. The radius of the orbit is 35,000 km as measured from the center of the Earth. Calculate (a) the weight of the satellite on the surface of the Earth and (b) the gravitational force exerted on the satellite by the Earth when it is in orbit.
Solution:

Chapter 12 Gravity Q.5CQ
Does the radius vector of Mars sweep out the same amount of area per time as that of the Earth? Why or why not?
Solution:
No. The amount of area swept out per time varies from planet to planet because the linear speeds of planets are different.

Chapter 12 Gravity Q.5P
The Attraction of Ceres Ceres, the largest asteroid known, has a mass of roughly 8.7 × 1020 kg. If Ceres passes within 14,000 km. of the spaceship in which you are traveling, what force does it exert on you? (Use an approximate value for your mass, and treat yourself and the asteroid as point objects.)
Solution:

Chapter 12 Gravity Q.6CQ
When a communications satellite is placed in a geosynchronous orbit above the equator. it remains fixed over a given point on the ground. Is it possible to put a satellite into an orbi t so that it remains fixed above the North Pole? Explain
Solution:
INot possiblel because a satellite will appear stationary only when it revolves in an orbit that is concentric and coplanar with the equatorial plane, has a period of revolution of 24 hours, and
has a sense of revolution from the west to the east of Earth. As the north pole is away from the equatorial plane. it will not be possible to put a geostationary satellite over the north pole.

Chapter 12 Gravity Q.6P
In one hand you hold a 0.11-kg apple, in the other hand a 0.24-kg orange. The apple and orange are separated by 0.85 m. What is the magnitude of the force of gravity that (a) the orange exerts on the apple and (b) the apple exerts on the orange?
Solution:

Chapter 12 Gravity Q.7CQ
The Mass of Pluto On June 22, 1978, James Christy made the first observation of a moon orbiting Pluto. Until that lime the mass of Pluto was not known, but with the discovery of its moon, Charon, its mass could be calculated with some accuracy. Explain.
Solution:

Chapter 12 Gravity Q.7P
IP A spaceship of mass m travels from the Earth to the Moon along a line that passes through the center of the Earth and the center of the Moon. (a) At what distance from the center of the Earth is the force due to the Earth twice the magnitude of the force due to the Moon? (b) How does your answer to part (a) depend on the mass of the spaceship? Explain.
Solution:

Chapter 12 Gravity Q.8CQ
Rockets arc launched into space from Cape Canaveral in an easterly direction Is there an advantage to launching to the east versus launching to the west? Explain
Solution:
Earth revolves from west to east (counterclockwise) about its polar axis. Therefore, all the particles on Earth have a velocity from west to east. This velocity is at a maximum along the
equatorial line, as y = Rw,where R is the radius of Earth and w is the angular velocity of Earth’s revolution about its polar axis.
ICape Canaveral is situated at the equator so when a rocket is launched from west to east in this place. the maximum linear velocity is added to the launching velocity of the rocket Because of this. launching becomes easied

Chapter 12 Gravity Q.8P

Solution:

Chapter 12 Gravity Q.9CQ
One day in the future you may take a pleasure cruise to the Moon While there you might climb a lunar mountain and throw a rock horizontally from its summit If. in principle, you could throw the rock fast enough, it might end up hitting you in the back Explain.
Solution:
Ion the Moon. where there is no atmosphere, a rock can orbit at any altitudel where it clears the mountains — as long as it has sufficient speed If we could give a rock enough speed. it would orbit the Moon and return to us from the other side (behind).

Chapter 12 Gravity Q.9P

Solution:

Chapter 12 Gravity Q.10CQ
Apollo astronauts orbiting the Moon at low altitude noticed occasional changes ¡n their orbit that they attributed to localized concentrations of mass below the lunar surface. Just what effect would such ‘Thascons” have on their orbit?
Solution:
As the astronauts approach a mass concentration, its increased gravitational attraction would increase the speed of the craft Similarly, as they pass the mass concentration, its ravitationaI attraction is in the backward direction, which decreases their speed I

Chapter 12 Gravity Q.10P

Solution:

Chapter 12 Gravity Q.11CQ
If you light a candle on the space shuttle—which would not be a good idea—would it burn the same as on the Earth? Explain
Solution:
No. In the weightless environment of the shuttle, there is no convection which is required to bring fresh oxygen to the flame. Without convection, a flame usually goes out very quickly. In carefully controlled experiments on the shuttle, however, small flames have been maintained for considerable times These “weightless” flames are spherical in shape. as opposed to the tear- shaped flames on Earth

Chapter 12 Gravity Q.11P
IP Three 6.75-kg masses are at the corners of an equilateral triangle and located in space far from any other masses. (a) If the sides of the triangle are 1.25 m long, find the magnitude of the net force exerted on each of the three masses. (b) How does your answer to part (a) change if the sides of the triangle are doubled in length?
Solution:

Chapter 12 Gravity Q.12CQ
The force exerted by the Sun on the Moon is more than twice the force exerted by the Earth on the Moon. Should the Moon be thought of as orbiting the Earth or the Sun? Explain.
Solution:

Chapter 12 Gravity Q.12

Solution:

Chapter 12 Gravity Q.13CQ

Solution:
The net force acting on the moon is always directed toward the Sun, never away from the Sun. Therefore, the Moon’s orbit must always curve toward the Sun. It curves sharply toward the Sun when Earth is between the Moon and the Sun, and curves only slightly toward the Sun when the Moon is between the Sun and Earth.

Chapter 12 Gravity Q.13P
Suppose that three astronomical objects (1, 2, and 3) are observed to lie on a line, and that the distance from object 1 to object 3 is D. Given that object 1. has four times the mass of object 3 and seven times the mass of object 2, find the distance between objects 1 and 2 for which the net force on object 2 is zero.
Solution:

Chapter 12 Gravity Q.14P
Find the acceleration due to gravity on the surface of (a) Mercury and (b) Venus.
Solution:

Chapter 12 Gravity Q.15P
At what altitude above the Earth’s surface is the acceleration due to gravity equal to g/2?
Solution:

Chapter 12 Gravity Q.16P
Two 6-7-kg bowling balls, each with a radius of 0.11 m, are in contact with one another. What is the gravitational attraction between the bowling balls?
Solution:

Chapter 12 Gravity Q.17P
What is the acceleration due to Earth’s gravity at a distance from the center of the Earth equal to the orbital radius of the Moon?
Solution:

Chapter 12 Gravity Q.18P
Gravity on Titan Titan is the larges t moon o f Saturn and the only moon in the solar system known to have a substantial atmosphere. Find the acceleration due to gravity on Titan’s surface, given that its mass is 1.35 × 1023 kg and its radius is 2570 km.
Solution:

Chapter 12 Gravity Q.19P
IP At a certain distance from the center of the Earth, a 4.6-kg object has a weight of 2.2 N. (a) Find this distance, (b) If the object is released at this location and allowed to falï toward the Earth, what is its initial acceleration? (c) If the object is now moved twice as far from the Earth, by what factor does its weight change? Explain, (d) By what factor does its initial acceleration change? Explain.
Solution:

Chapter 12 Gravity Q.20P
Tine acceleration due to gravity on the Moon’s surface is known to be about one-sixth the acceleration due to gravity on the Earth. Given that the radius of the Moon is roughly one-quarter that of the Earth, find the mass of the Moon in terms of the mass of the Earth.
Solution:

Chapter 12 Gravity Q.21P
IP An Extraterrestrial Volcano Several volcanoes have been observed erupting on the surface of Jupiter’s closest Galilean moon, lo. Suppose that material ejected from one of these volcanoes reaches a height of 5.00 km a fter being projected straight upward with an initial speed of 134 m/s. Given that the radius of lo is 1820 km, (a) outlinca strategy thatallows you to calculate the mass of To. (b) Use your strategy to calculate Io’s mass.
Solution:

Chapter 12 Gravity Q.22P
IP Verne’s Trip to the Moon In his novel From the Earth to the Moon, Jules Verne imagined that astronauts inside a spaceship would walk on the floor of the cabin when the force exerted on the ship by the Earth was greater than the force exerted by the Moon. When the force exerted by the Moon was greater, he thought the astronauts would walk on the ceiling of the cabin, (a) At what distance from the center of the Earth would the forces exerted on the spaceship by the Earth and the Moon be equal? (b) Explain why Verne’s description of gravitational effects is incorrect.
Solution:

Chapter 12 Gravity Q.23P
Consider an asteroid with a radius of 19 km and a mass of 3.35 X 1015 kg. Assume the asteroid is roughly spherical, (a) What is the acceleration due to gravity on the surface of the asteroid? (b) Suppose the asteroid spins about an axis through its center, like the Earth, with a rotational period T. What is the smallest value T can have before loose rocks on the asteroid’s equator begin to fly off the surface?
Solution:

Chapter 12 Gravity Q.24P
CE Predict/Explain The Speed of the Earth The orbital speed of the Earth is greatest around January 4 and least around July 4. (a) Is the distance from the Earth to the Sun on January 4 greater than, less than, or equal to its distance from the Sun on July 4? (b) Choose the best explanation from among the following:
I. The Earth’s orbit is circular, with equal distance from, the Sun at all times.
II. The Earth sweeps out equal area in equal time, thus it must be closer to the Sun when it is moving faster.
III. The greater the speed of the Earth, the greater its distance from the Sun.
Solution:
a) The distance from the Earth to the Sun on January 4, is less than the distance from the Sun on July 4.
b) The Earth sweeps out equal area in equal time, thus it must be closer to the sun when it is moving faster.

Chapter 12 Gravity Q.25P
C E A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine is fired in such a way that its speed increases rapidly by a small amount. As a result, do the (a) apogee distance and (b) perigee distance increase, decrease, or stay the same?
Solution:
Use the concept of orbital transfer to place the satellite into a new orbit.
(a)
The decelerating or accelerating rockets at some point in the circular orbit of the satellite would allow the satellite into a new orbit which is not a circle. The new orbit is an ellipse. The largest distance between the Earth and the satellite in an elliptical orbit is called the apogee distance. In the case of transfer of orbits, the apogee distance increases if the speed of the rocket increases a while in the original orbit.
(b)
The smallest distance between the Earth and the satellite in an elliptical orbit is nothing but the perigee distance. In case of transfer of orbits, the perigee distance doesn’t change and equal to the radius of the original circular orbit.

Chapter 12 Gravity Q.26P
g Repeat the previous problem., only this time with the rocket engine of the satellite fired in such a way as to slow the satellite.
Solution:
(A) The satellite drops into an elliptical orbit that brings it closer to Earth.
(B) The apogee distance remains unchanged.
(C) The perigee distance is reduced.

Chapter 12 Gravity Q.27P
CE Predict/Explain The Earth-Moon Distance Is Increasing Laser reflectors left on the surface of the Moon by the Apollo astronauts show that the average distance from the Earth to the Moon is increasing at the rate of 3.8 cm per year. (a) As a result, will the length of the month increase, decrease, or remain the same? (b) Choose the best expianation from among the following: I. The greater the radius of an orbit, the greater the period,
which implies a longer month.
II. The length of the month will remain the same due to conservation of angular momentum,
III. The speed of the Moon is greater with increasing radius therefore, the length of the month will be less.
Solution:
a) If the average distance increases, then the length of the month also increases.
b) The period depends upon the radius. Greater the radius, greater will be the period. Option (1) is correct.

Chapter 12 Gravity Q.28P
Apollo Missions On Apollq missions to the Moon, the command module orbited at an altitude of 110 km above the lunar surface. How long did it take for the command module to complete one orbit?
Solution:

Chapter 12 Gravity Q.29P
Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 X 107 m above the surface of the Earth.
Solution:

Chapter 12 Gravity Q.30P
An Extrasolar Planet In July of 1999 a planet was reported to be orbiting the Sun-like star Iota Horologii with a period of 320 days. Find the radius of the planet’s orbi t, assuming that iota Horologii has the same mass as the Sun. (This planet is presumably similar to Jupiter, but it may have large, rocky moons that enjoy a relatively pleasant climate.)
Solution:

Chapter 12 Gravity Q.31P
Phobos, one of the moons of Mars, orbits at a distance of 9378 km from the center of the red planet. What is the orbital period of Phobos?
Solution:

Chapter 12 Gravity Q.32P
· The largest moon in the solar system is Ganymede, a moon of Jupiter. Ganymede orbits at a distance of 1.07 X 109 m from the center of Jupiter with an orbital period of about 6.18 X 10′ s. Using this information, find the mass of Jupiter.
Solution:

Chapter 12 Gravity Q.33P
IP Am Asteroid with Its Own Moon The asteroid 243 Ida has its own small moon, Dactyl. (See the photo on p. 390) (a) Outline a strategy to find the mass of 243 Ida, given that the orbital radius of Dactyl is 89 km arid its period is 19 hr. (b) Use your strategy to calculate the mass of 243 Ida.
Solution:

Chapter 12 Gravity Q.34P
GPS Satellites GPS (Global Positioning System) satellites orbit at an altitude of 2.0 x 107 m. Find (a) the orbital period, and (b) the orbital speed of such a satellite.
Solution:

Chapter 12 Gravity Q.35P
IP Two satellites orbit the Earth, with satellite 1 at a greater altitude than satellite 2. (a) Which satellite has the greater orbital speed? Explain, (b) Calculate the orbital speed of a satellite that orbits at an altitude of one Earth radius above the surface of the Earth, (c) Calculate the orbital speed of a satellite that orbits at an altitude of two Earth radii above the surface of the Earth.
Solution:

Chapter 12 Gravity Q.36P
IP Calculate the orbital periods of satellites that orbit (a) one Earth radius above the surface of the Earth and (b) two Earth radii above the surface of the Earth, (c) How do your answers to parts (a) and (b) depend on the mass of the satellites? Explain, (d) How do your answers to parts (a) and (b) depend on the mass of the Earth? Explain.
Solution:

Chapter 12 Gravity Q.37P
S P The Martian moon Deimos has an orbital period that is greater than the other Martian moon, Phobos. Both moons have approximately circular orbits, (a) Is Deimos closer to or farther from Mars than Phobos? Explain, (b) Calculate the distance from the center of Mars to Deimos given that its orbital period is 1.10 × 105 s.
Solution:

Chapter 12 Gravity Q.38P
Binary Stars Centauri A and Centauri B are binary stars with a separation of 3.45 × 1012 m and an orbital period of 2.52 × 109 s. Assuming the two stars are equally massive (which is approximately the case), determine their mass.
Solution:

Chapter 12 Gravity Q.39P
Find the speed of Centauri A and Centauri B, using the information given in the previous problem.
Solution:

Chapter 12 Gravity Q.40P
Sputnik The first artificial satellite to orbit the Earth was Sputnik I, Saunched October 4,1957. The mass of Sputnik 1 was 83.5 kg, and its distances from the center of the Earth at apogee and perigee were 7330 km-and 6610 km, respectively. Find the difference in gravitational potential energy for Sputnik I as it moved from apogee to perigee.
Solution:

Chapter 12 Gravity Q.41P
CE Predict/Explain (a) Is the amount of energy required to get a spacecraft from the Earth to the Moon greater than, less than, or equal to the energy required to get the same spacecraft from the Moon to the Earth? (b) Choose the best explanation from among the following:
I. The escape speed of the Moon is less than that of the Earth therefore, less energy is required to leave the Moon.
II. The situation is symmetric, and hence the same amount of energy is required to travel in either direction.
III. It takes more energy to go from the Moon to the Earth because the Moon is orbiting the Earth.
Solution:
Use the concept of escape speed of the planet. The escape speed of the planet is the minimum speed at which the object frees from the gravitational attraction of the planet.
(a)
The escape speed of an object launched from the planet depends only on the mass and size of the planet, but not on the mass of the object. The escape speed of the Earth is much greater than that of the Moon. Since the kinetic energy is directly proportional to the square of the velocity, the more energy is required to launch the spacecraft from the Earth to the Moon than that required to launch the spacecraft from the Moon to the Earth.
(b)
The option (I) is correct.

Chapter 12 Gravity Q.42P

Solution:

Chapter 12 Gravity Q.43P
Calculate the gravitational potential energy of a 8.8-kg mass (a) on the surface of the Earth and (b) at an altitude of 350 km. (c) Take the difference between the results for parts (b) and (a), and compare with nigh, where h = 350 km.
Solution:

Chapter 12 Gravity Q.44P
Two 0.59-kg basketballs, each with a radius of 12 cm, are just touching. How much energy is required to change the separation between the centers of the basketballs to (a) 1.0 m and (b) 10.0 m? (Ignore any other gravitational interactions.)
Solution:

Chapter 12 Gravity Q.45P
Find the minimum kinetic energy needed for a 39,000-kg rocket to escape (a) the Moon or (b) the Earth.
Solution:

Chapter 12 Gravity Q.46P
CE Predict/Explain Suppose the Earth were to suddenly shrink to half its current diameter, with its mass remaining constant, (a) Would the escape speed of the Earth increase, decrease, or stay the same? (b) Choose the best explanation from among the following:
I. Since the radius of the Earth would be smaller, the escape speed would also be smaller.
II. The Earth would have the same amount of mass, and hence its escape speed would be unchanged.
III. The force of gravity would be much stronger on the surface of the compressed Earth, leading to a greater escape speed.
Solution:
a) The escape speed of the earth increases.
b) The force of gravity would be much stronger on the surface of compressed Earth, leading to a greater escape speed. Option (III) is correct.

Chapter 12 Gravity Q.47P
CE Is the energy required to launch a rocket vertically to a height h greater than, less than, or equal to the energy required to prit the same rocket into orbit at the height hi Explain.
Solution:
The energy required to launch a rocket vertically to a height h is equal to the potential energy of the rocket at that height. However, for a rocket to be put into orbit at a height h, both kinetic energy and potential energy are required. So the energy required for the rocket to be put into orbit is greater than the energy required to launch a rocket vertically to the same height.

Chapter 12 Gravity Q.48P
Suppose one of the Global Positioning System satellites has a speed of 4.46 km/s at perigee and a speed of 3.64 km/s at apogee. If the distance from the center of the Earth to the satellite at perigee is 2.00 × 104 lem, what is the corresponding distance at apogee?
Solution:

Chapter 12 Gravity Q.49P

Solution:

Chapter 12 Gravity Q.50P
· Referring to Example 12-1, if the Millennium Eagle is at rest at point A, what is its speed at point B?
Solution:

Chapter 12 Gravity Q.51P
What is the launch speed of a projectile that rises vertically above the Earth to an altitude equal to one Earth radius before coming to rest momentarily?
Solution:

Chapter 12 Gravity Q.52P
A projectile launched vertically from the surface of the Moon? rises to an altitude of 365 km. What was the projectile’s initial speed?
Solution:

Chapter 12 Gravity Q.53P
Find the escape velocity for (a) Mercury and (b) Ventis.
Solution:

Chapter 12 Gravity Q.54P
IP Halley’s Comet Halley’s comet, which passes around the Sun every 76 years, has an elliptical orbit. When closest to the Sun (perihelion) it is at a distance of 8.823 x 1010 m and moves with a speed of 54.6 km/s. The greatest distance between Halley’s comet and the Sun (aphelion) is 6.152 x 1012 m. (a) Is the speed of Halley’s comet greater than or less than 54.6 km/s when it is at aphelion? Explain, (b) Calculate its speed ai aphelion.
Solution:

Chapter 12 Gravity Q.55P
The End of the Lunar Module On Apollo Moon missions, the lunar module would blast off from the Moon’s surface and dock with the command module in lunar orbit. After docking, the lunar module would be jettisoned and allowed to crash back onto the lunar surface. Seismometers placed on the Moon’s surface by the astronauts would then pick up the resulting seismic waves. Find the impact speed of the lunar module, given that it is jettisoned from an orbit 110 km above the lunar surface moving with a speed of 1630 m/s.
Solution:

Chapter 12 Gravity Q.56P
If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how high above the Earth’s surface is it when its speed is half the escape speed?
Solution:

Chapter 12 Gravity Q.57P
Suppose a planet is discovered orbiting a distant star. If the mass of the planet is 10 times the mass of the Earth, and its radius is one-tenth the Earth’s radius, how does the escape speed of this planet compare with that of the Earth?
Solution:

Chapter 12 Gravity Q.58P
A projectile is launched vertically from the surface of the Moon with an initiaL speed of 1050 m/s. At what altitude is the projectile’s speed one-half its initial value?
Solution:

Chapter 12 Gravity Q.59P
To what radius would the Sun have to be contracted for its escape speed to equal the speed of light? (Black holes have escape speeds greater than the speed of light hence we see no light from them.)
Solution:

Chapter 12 Gravity Q.60P
IP Two baseballs, each with a mass of 0.148 kg, are separated by a distance of 395 m in outer space, far from any other objects. (a) If the balls are released from rest, what speed do they have when their separation has decreased to 145 m? (b) Suppose the mass of the balls is doubled. Would the speed found in part (a) increase, decrease, or stay the same? Explain.
Solution:

Chapter 12 Gravity Q.61P
On Earth, a person can jump vertically and rise to a height h. What is the radius of the largest spherical asteroid from which this person could escape by jumping straight upward? Assume that each cubic meter of the asteroid has a mass of 3500 kg.
Solution:

Chapter 12 Gravity Q.62P
As will be shown in Problem 63, the magnitude of the tidal force exerted on an object of mass m and length a is approximately 4GmMa/r3. In this expression, M is the mass of the body causing the tidal force and r is the distance from the center of m to the center of M. Suppose you are 1 million miles away from a black hole whose mass is a million times that of the Sun. (a) Estimate the tidal force exerted on your body by the black hole. (b) At what distance will the tidal force be approximately 10 times greater than your weight?
Solution:

Chapter 12 Gravity Q.63P

Solution:

Chapter 12 Gravity Q.64P

Solution:

Chapter 12 Gravity Q.65GP
CE You weigh yourself on a scale inside an airplane flying due east above the equator. If the airplane now turns around and heads due west with the same speed, will the reading on the scale increase, decrease, or stay the same? Explain.
Solution:
SOLUTION:
The reading on the scale is due to the force of gravity between the person on the plane and the Earth.
F = Gm1m2 / R2
Where R is the difference between the passenger on the plane and the center of the Earth. As the plane switches direction from East to West, the R value remains unchanged. Since the mass of the person and the mass of the Earth are both the same, the magnitude of gravitational force will be the same.

Chapter 12 Gravity Q.66GP

Solution:

Chapter 12 Gravity Q.67GP

Solution:

Thus the increasing order of gravitational force is given by
object C >object A >object B

Chapter 12 Gravity Q.68GP

Solution:

Chapter 12 Gravity Q.69GP
CE A satellite goes through one complete orbit of the Earth. (a) Is the net work done on it by the Earth’s gravitational force positive, negative, or zero? Explain, (b) Does your answer to part (a) depend on whether the orbit is circular or elliptical?
Solution:
(A) When a satellite goes through one complete orbit, this means the satellite returns to
the initial point at which it started. The resulting net displacement is zero. So the net work done on it by Earth’s gravitational force is zero.
(B) No, the answer to part (A) is independent of the shape of the orbit (i.e., whether the orbit is circular or elliptical). It is dependent on the displacement by the satellite.

Chapter 12 Gravity Q.70GP
CE The Crash of Skylab Skylab, the largest spacecraft ever to fall back to the Earth, met its fiery end on July 11,1979, after flying directly over Everett, WA, on its last orbit. On the CBS Evening News the night before the crash, anchorman Walter Cronkite, in his rich baritone voice, made the following statement: “NASA says there is a little chance that Skylab will land in a populated area.” After the commercial, he immediately corrected himself by saying,”I meant to say ‘there is little chance’ Skylab will hita populated area.” In fact, it landed primarily in the Indian Ocean off the west coast of Australia, though several pieces were recovered near the town of Espérance, Australia, which later sent the U.S. State Department a \$400 bill for littering. The cause of Skylab’s crash was the friction it experienced in the upper reaches of the Earth’s atmosphere. As the radius of Skylab’s orbit decreased, did its speed increase, decrease, or stay the same? Explain.
Solution:
The speed of the Skylab increases with decreasing radius. We might think that friction would slow Skylab just like other objects are slowed by friction – but by dropping Skylab to a lower orbit, friction is ultimately responsible for an increase in speed.

Chapter 12 Gravity Q.71GP
Consider a system consisting of three masses on the x axis. Mass m1 = 1.00 kg is at x = 1.00 m mass m2 = 2.00 kg is at x = 2.00 m and mass m3 = 3.00 kg is at x = 3.00 m. What is the total gravitational potential energy of this system?
Solution:

Chapter 12 Gravity Q.72GP
An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 3860 km. When the astronaut jumps upward with an initial speed of 3.10 m/s, she rises to a height of 0.580 m. What is the mass of the planet?
Solution:

Chapter 12 Gravity Q.73GP
IP When the Moon is in its third-quarter phase, the Earth, Moon, and Sun form a right triangle, as shown in Figure 12-22. Calculate the magnitude of the force exerted on the Moon by (a) the Earth and (b) the Sun. (c) Does it make more sense to think of the Moon as orbiting the Sun, with a small effect due to the Earth, or as orbiting the Earth, with a small effect due to the Sun?
Solution:

Chapter 12 Gravity Q.74GP

Solution:

Chapter 12 Gravity Q.75GP

Solution:

Chapter 12 Gravity Q.76GP
A Near Miss! In the early morning hours of June 14, 2002, the Earth had a remarkably close encounter with an asteroid the size of a small city. The previously unknown asteroid, now designated 2002 MN, remained undetected until three days after it had passed the Earth. At its closest approach, the asteroid was 73,600 miles from the center of the Earth?about a third of the distance to the Moon. (a) Find the speed of the asteroid at closest approach, assuming its speed at infinite distance to be zero and considering only its interaction with the Barth. (b) Observations indica te the asteroid to have a diameter of about 2.0 km. Estimate the kinetic energy of the asteroid at closest approach, assuming it has an average density of 3.33 g/cm3 (For comparison, a 1-megaton nuclear weapon releases about 5.6 × 1015J of energy.)
Solution:

Chapter 12 Gravity Q.77GP
IP Suppose a planet is discovered that has the same amount of mass in a given volume as the Earth, but has half its radius. (a) Is the acceleration due to gravity on this planet more than, less than, or the same as the acceleration due to gravity on the Earth? Explain. (b) Calculate the acceleration due to gravity on this planet.
Solution:

Chapter 12 Gravity Q.78GP
IP Suppose a planet is discovered that has the same total mass as the Earth, but half its radius. (a) Is the acceleration due to gravity on this planet more than, less than, or the same as the acceleration due to gravity on the Earth? Explain. (b) Calculate the acceleration due to gravity on this planet.
Solution:

Chapter 12 Gravity Q.79GP

Solution:

Chapter 12 Gravity Q.80GP

Solution:

Chapter 12 Gravity Q.81GP

Solution:

Chapter 12 Gravity Q.82GP
Using the results from Problem 54. find the angular momentum of Halley’s comet (a) at perihelion and (b) at aphelion (Take the mass of Halley’s comet to be 9.8 x 1014 kg.)
Solution:

Chapter 12 Gravity Q.83GP
Exploring Mars Inthe not-too-distant future astronauts will travel to Mars to carry out scientific explorations. As part of their mission, it is likely that a “geosynchronous” satellite will be placed above a given point on the Martian equator to facilitate communications. At what altitude above the surface of Mars should such a satellite orbit? (Note: The Martian “day” is 24.6229 hours, Other relevant information can be found in Appendix C.)
Solution:

Chapter 12 Gravity Q.84GP
IP A satellite is placed in Earth orbit 1000 miles higher than the altitude of a geosynchronous satellite. Referringto Active Example 12-1, we see that the altitude of the satellite is 23,300 mi. (a) Is the period of this satellite greater than or less than 24 hours? (b) As viewed from the surface of the Earth, does the satellite move eastward or westward? Explain. (c) Find the orbital period of this satellite.
Solution:

Chapter 12 Gravity Q.85GP
Find the speed of the Millennium Eagle at point A in Example 12-1 if its speed at point B is 0.905 m/s.
Solution:

Chapter 12 Gravity Q.86GP
Show that the force of gravity between the Moon and the Sun is always greater than the force of gravity between the Moon and the Earth.
Solution:

Chapter 12 Gravity Q.87GP

Solution:

Chapter 12 Gravity Q.88GP
(a) Find the kinetic energy of a 1720-kg satellite in a circular orbit about the Earth, given that the radius of the orbit is 12,600 miles. (b) How much energy is required to move this satellite to a circular orbit with a radius of 25,200 miles?
Solution:

Chapter 12 Gravity Q.89GP
IP Space Shuttle Orbit On a typical mission, the space shuttle (m = 2.00 × 106 kg) orbits at an altitude of 250 km above the Earth’s surface. (a) Does the orbital speed of the shuttle depend on its mass? Explain. (b) Find the speed of the shuttle in its orbit. (c) How long does it take for the shuttle to complete one orbit of the Earth?
Solution:

Chapter 12 Gravity Q.90GP
IP Consider an object of mass m orbiting the Earth at a radius r. (a) Find the speed of the object. (b) Show that the total mechanical energy of this object is equal to (?1) times its kinetic energy. (c) Does the result of part (b) apply to an object orbiting the Sun? Explain.
Solution:

Chapter 12 Gravity Q.91GP
In a binary star system two stars orbit about their common center of mass. Find the orbital period of such a system, given that the stars are separated by a distance d and have masses m and 2m.
Solution:

Chapter 12 Gravity Q.92GP

Solution:

Chapter 12 Gravity Q.93GP
Find an expression for the kinetic energy of a satellite of mass m in an orbit of radius r about a planet of mass M.
Solution:

Chapter 12 Gravity Q.94GP
Referring to Example 12-1, find the x component of the net force acting on the Millennium Eagle as a function of x. Plot your result, showing both negative and positive values of x.
Solution:

Chapter 12 Gravity Q.95GP
A satellite orbits the Earth in an elliptical orbit. At perigee its distance from the center of the Earth is 22,500 km and its speed is 4280 m/s. At apogee its distance from the center of the Earth is 24,100 km and its speed is 3990 m/s. Using this information, calculate the mass of the Earth.
Solution:

Chapter 12 Gravity Q.96PP

Solution:

Chapter 12 Gravity Q.97PP

Solution:

Chapter 12 Gravity Q.98PP

Solution:

Chapter 12 Gravity Q.99PP

Solution:

Chapter 12 Gravity Q.100IP
Find the orbital radius that corresponds to a “year” of 150 days.
Solution:

Chapter 12 Gravity Q.101IP
Suppose the mass of the Sun is suddenly doubled, but the Earth’s orbital radius remains the same. (a) Would the length of an Earth year increase, decrease, or stay the same? (b) Find the length of a year for the case of a Sun with twice the mass. (c) Suppose the Sun retains its present mass, but the mass of the Earth is doubled instead. Would the length of the year increase, decrease, or stay the same?
Solution:

Chapter 12 Gravity Q.102IP
(a) If the mass of the Earth were doubled, would the escape speed of a rocket increase, decrease, or stay the same? (b) Calculate the escape speed of a rocket for the case of an Earth with twice its present mass. (c) If the mass of the Earth retains its present value, but the mass of the rocket is doubled, does the escape speed increase, decrease, or stay the same?
Solution:

The escape speed depends on the mass of Earth, radius of Earth, and the universal gravitational constant. However, it does not depend on the mass of the rocket.

Chapter 12 Gravity Q.103IP
Suppose the Earth is suddenly shrunk to half its present radius without losing any of its mass. (a) Would the escape speed of a rocket increase, decrease, or stay the same? (b) Find the escape speed for an Earth with half its present radius.
Solution:

## Can I be taught general and special Relativity on this forum

So, now you claim that GPS' uses general relativity, and not Special Relativity?

Lets us continue with exactly what we can measure with SR and GR.
gentlemen, If you dont know, say so.
However, show me on scientists very own Thought Experiment where I am wrong.

Both. Besides, SR is built in GR.

Because you are unwilling to learn, and want to stay with your incorrect premises. At least that's what it looks like for me, an external observer. Especially when you write something like this:

So, now you claim that GPS' uses general relativity, and not Special Relativity?

Nice of you to investigate the sources I quote.
Do you know how many people would not even have bothered?

The article reads: "GPS satellites travel at approximately 8,700 mph (14,000 km/h) with respect to Earth. This means time runs 7,200 nanoseconds per day slower for a satellite relative to us on Earth as described by Special Relativity.

However, using General Relativity it is possible to calculate that time goes faster for a GPS satellite by 45,900 nanoseconds per day, due to the satellite being 19,000km above the Earth (therefore in weaker gravity). This means overall time runs 38,700 (45,900 – 7,200) nanoseconds faster per day for a GPS satellite relative to us stationary on Earth."

1. As Special relativity describes, time runs slower for the Satellite, than what Time runs on the earth.
2. But according to General Relativity, Time on the Satellite runs Faster.

We will come back to this statement.

Nice of you to investigate the sources I quote.

Do you know how many people would not even have bothered?

No point. You just misunderstood what was written. Please note that, as has been pointed out several times, the sooner you accept that the problems are in your understanding of relativity, not in relativity, the sooner you will start to learn.

Relativity might not be a perfectly accurate model of reality, but it is not inconsistent with itself. Any logical problems you find are in your conception of relativity, not in relativity itself.

No. It is an apparent paradox that arises from an incomplete understanding of relativity. There are others, e.g. the “barn and pole paradox”.

In my experience, in nine out of ten cases, the resolution lies in an understanding of the relativity of simultaneity.

Really? Ever thought of doing a search for experimental evidence for SR?

I will pick one at random:
Time dilation is confirmed in heavy ion storage rings, such as the TSR at the MPIK, by observation of the Doppler effect of lithium, and those experiments are valid in the electron, proton, and photon sector.

You have a direct answer to your direct question. I have also explained the why of both SR and GR, but at present your math is not up to understanding them. I have carefully considered your situation and gave a list of reading material so you can understand it. It is now in your court.

If you insist on this line now you have what you directly asked for, then if this thread is worthwhile continuing will be looked at by the mentors - and yes I am one. Consider this a friendly bit of advice on posting etiquette on this forum.

There are a thousand and one threads on the twin paradox in this forum. Have a look at some of them. It's clearly not an actual paradox in relativity or we wouldn't be able to resolve it.

It is a paradox in a lot of common misunderstandings of relativity, which is why it's often used as a teaching tool - to force you to confront a possible misunderstanding.

Explain what - his experience? In other words you want all the situations that led him to this view listed. That is simply unrealistic - as is very obvious.

This thread is very fast devolving into being valueless.

Please, all those involved can we ensure it gets back on track?

I am pretty sure I explained around here we do not spoon feed - you must do some work yourself.

Its simple to look up the things you are challenging us with.

Anyhow, I still want to know what this mysterious measurement of Special Relativity, Time dilation, Length contraction and mass increase is.

Simple, what are they measuring?

Time dilation is the observation (measurement) of different rates of the passage of time and therefore different amounts of elapsed time between clocks.

Simply put if you have a local repeater for a distant clock, it may not be synchronized with your clock.

Here is what it said -
Each satellite carries with it an atomic clock that "ticks" with a nominal accuracy of 1 nanosecond (1 billionth of a second). A GPS receiver in an airplane determines its current position and course by comparing the time signals it receives from the currently visible GPS satellites (usually 6 to 12) and trilaterating on the known positions of each satellite. The precision achieved is remarkable: even a simple hand-held GPS receiver can determine your absolute position on the surface of the Earth to within 5 to 10 meters in only a few seconds. A GPS receiver in a car can give accurate readings of position, speed, and course in real-time!

Its utterly clear what is being measured - the time difference between atomic clocks on the satellites. It uses that to calculate its position - but the effects of both SR and GR must be taken into account.

I will speak plainly - if you insist doing this the tread will be shut down. You must think a bit and stop asking things that are utterly obvious.

You will have to think a LOT harder if you want to understand SR and GR, which is what your question asked - can I learn it from people here. You can, but you will need to do research yourself - we can guide you - but you must do it. Above all you must THINK - which so far you have not shown you really want to do. I am pretty sure you are much more intelligent than this. If you are merely trying to have fun at our expense then you will find it will not last long. You are dealing with people with a very low tolerance to that sort of thing.


text:
Volume 18, issue 1 (2014)

 Recent Issues
 The Journal About the Journal Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Subscriptions Author Index To Appear Contacts ISSN (electronic): 1364-0380 ISSN (print): 1465-3060
Contact manifolds with symplectomorphic symplectizations

Sylvain Courte

Geometry & Topology 18 (2014) 1–15
Abstract

We provide examples of contact manifolds of any odd dimension greater than or equal to $5$ which are not diffeomorphic but have exact symplectomorphic symplectizations.

Keywords
symplectization, Mazur trick, Weinstein cobordism, Whitehead torsion
Primary: 53D10


text:
# Convert InputField to a number for ListLinePlot

I have a list of InputFields that I want the user to enter. This list should then be plotted. When I try to plot the list it does not show up because the items in the list have the Head InputField and are not a number. I have tried using all the conversion tricks I can.

HATableData :=
Select[InsertColumn[Table[With[{i = i, j = j},
InputField[Dynamic[HA[[i, j]]],Number]],{i, 1, Length[HA]},
{j, 2, Length[HA[[1]]]}], HTNames, 1], #[[1]] == Truck &]

{{CAT 793D, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897, 0.897}}

Head[HATableData[[1, 2]]]

InputField

ListLinePlot[HATableData[[1, 2 ;;]], PlotMarkers -> Automatic]

The Plot returns a graph with nothing on it. I have even tried to convert it to a string first but all I get is this

ToString[HATableData[[1, 2]]]

"InputField[Dynamic[HA[[1,2]]], Number, FieldSize -> {4, 1}, Appearance -> Frameless, Alignment -> Right]"

In other similar questions it looks like you can use InputField data inside a function and then plot the function using Plot but in this case I want to simply plot the numbers themselves on a ListLinePlot and not using a function inside a Plot.

-
It looks like ListLinePlot[HA[[1, 1 ;;]], PlotMarkers -> Automatic], or simply ListLinePlot[HA[[1]], PlotMarkers -> Automatic], would work, but I can't check without working example code. Perhaps the index 1 should be changed to the first Truck. HATableData appears to be the input GUI, and HA appears to be the data it works on. You might look up Setting, but working with the data seems better. –  Michael E2 Jun 24 '13 at 12:41
add comment


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For instance, when radio stations stream their content onto the internet some ads must be stripped out for legal reasons. Advertisements featuring radio personalities often have to be removed because there is no arrangement to pay for the rights of on-air talent, writers and producers. This commonly used advertising type, featuring a trusted personality endorsing a product of service, will not be heard by the listeners who choose to listen on internet radio. Thus, advertisers have to consider the benefits of producing these types of ads in the new environment.

RightSpot and similar agencies are working to turn a nascent field into a viable business that benefits advertisers of any size. They expect the percentage of internet radio listeners to increase from 24 percent of listeners in 2006 to 50 percent in 2010. This increase will further add to the necessity for radio advertisers to change methodologies. Those that embrace the changes will certainly see positive results.

The Primary purpose of the article is to

A) Demonstrate why radio ads featuring an on-air personality are no longer effective.
D) Describe how radio advertising has remained relatively stable for the past 30 years.
E) Describe the ways in which Right Spot and similar agencies are turning a nascent field into a viable business.
[Reveal] Spoiler:
C

The author implies that all of the following are changes internet radio is causing in the industry EXCEPT

A) Advertisements featuring radio personalities must are always removed because there is no arrangement to pay for the rights of on-air talent.
B) Radio ads from 30 years ago are similar to those aired today
D) The percentage of internet radio listeners is expected to more than double from 2006 to 2010

[Reveal] Spoiler:
A

According to the article, RightSpot believes internet radio provides new opportunities for
A) On-Air personalities

[Reveal] Spoiler:
C

The tone of the author towards advertising opportunities in internet radio can best be described as
A) Distrustful
B) Frustrated
C) Enthusiastic
D) Disinterested
E) Fervent
[Reveal] Spoiler:
C

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I appreciate your effort for posting this RC.

Nice and Interesting stuff.

easy to undersatnd.

questions were preety straightforward.

reading time : 0348 overall time :0739

expecting some more RCs from You(fresh meat)

regards
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C, A, C, C.

Hope to see the OA.
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Can you please explain why A for q2 ... it is clearly mentioned as a change here For instance, when radio stations stream their content onto the internet some ads must be stripped out for legal reasons. Advertisements featuring radio personalities often have to be removed because there is no arrangement to pay for the rights of on-air talent, writers and producers.
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All the others one were pretty easy, apart from q2. q2 says
The author implies that all of the following are changes internet radio is causing in the industry EXCEPT
Option B is not caused by internet radio.
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roopika2990 wrote:
Can you please explain why A for q2 ... it is clearly mentioned as a change here For instance, when radio stations stream their content onto the internet some ads must be stripped out for legal reasons. Advertisements featuring radio personalities often have to be removed because there is no arrangement to pay for the rights of on-air talent, writers and producers.

Because "often have to be done" is not the same with "MUST be done". The fist means in most cases, while the second one means always. This kind of trap is quite often used in GMAT.
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Dmitriy wrote:
roopika2990 wrote:
Can you please explain why A for q2 ... it is clearly mentioned as a change here For instance, when radio stations stream their content onto the internet some ads must be stripped out for legal reasons. Advertisements featuring radio personalities often have to be removed because there is no arrangement to pay for the rights of on-air talent, writers and producers.

Because "often have to be done" is not the same with "MUST be done". The fist means in most cases, while the second one means always. This kind of trap is quite often used in GMAT.

Exactly - watch for the difference in language. The passage says that it is often removed but the answer choice says it must be removed. The adding of extreme language is a trap.
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aryabhatta wrote:
All the others one were pretty easy, apart from q2. q2 says
The author implies that all of the following are changes internet radio is causing in the industry EXCEPT
Option B is not caused by internet radio.

As I understood the question we should eliminate all reasons that changed the industry. Fot instance, lets look on D. The growing number of users is not caused by the internet radio, but obviously it is the reason of changes. B plays the same role. The industry had not been changed and was outdated; therefore, it had to be changed for needs of the internet marketing.

In contrast, "A" gives us the reason not mentioned in the article. So A.
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Dmitriy wrote:
aryabhatta wrote:
All the others one were pretty easy, apart from q2. q2 says
The author implies that all of the following are changes internet radio is causing in the industry EXCEPT
Option B is not caused by internet radio.

As I understood the question we should eliminate all reasons that changed the industry. Fot instance, lets look on D. The growing number of users is not caused by the internet radio, but obviously it is the reason of changes. B plays the same role. The industry had not been changed and was outdated; therefore, it had to be changed for needs of the internet marketing.

In contrast, "A" gives us the reason not mentioned in the article. So A.

But doesnt the language mean that ...the changes internet radio are causing....
A may b extreme but is still relevant
B is something IR has no effect on
D may be thought of as ...the features are causing the viwershi to increase as in ..its the effect of the good things abt IR

B still seem the right answer may b i understood something incorectly lease explain
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therock99 wrote:
Dmitriy wrote:
aryabhatta wrote:
All the others one were pretty easy, apart from q2. q2 says
The author implies that all of the following are changes internet radio is causing in the industry EXCEPT
Option B is not caused by internet radio.

As I understood the question we should eliminate all reasons that changed the industry. Fot instance, lets look on D. The growing number of users is not caused by the internet radio, but obviously it is the reason of changes. B plays the same role. The industry had not been changed and was outdated; therefore, it had to be changed for needs of the internet marketing.

In contrast, "A" gives us the reason not mentioned in the article. So A.

But doesnt the language mean that ...the changes internet radio are causing....
A may b extreme but is still relevant
B is something IR has no effect on
D may be thought of as ...the features are causing the viwershi to increase as in ..its the effect of the good things abt IR

B still seem the right answer may b i understood something incorectly lease explain

I second this as well... B is the lesser of two evils but i do believe the chances of finding such ambiguity in the answer choices on GMAT is very unlikely.
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But doesnt the language mean that ...the changes internet radio are causing....
A may b extreme but is still relevant
B is something IR has no effect on
D may be thought of as ...the features are causing the viwershi to increase as in ..its the effect of the good things abt IR

B still seem the right answer may b i understood something incorectly lease explain
[/quote]

I second this as well... B is the lesser of two evils but i do believe the chances of finding such ambiguity in the answer choices on GMAT is very unlikely.[/quote]

I have similar feeling. i went with B and it seems like the right answer
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11:32 min.
All correct. Passage was quite straightforward, but I spent much time on grasping every word.
It can be solved within 6-9min.
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Hi

1.C
2.A
3.C
4.C

Easy passage enjoyed doing the questions.
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Concentration: Strategy, Healthcare
Schools: Sloan '18 (A)
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