# Force to use in harmonic oscillation through the inside of a planet

I am to find an equation for the time it takes when one falls through a planet to the other side and returns to the starting point. I have seven different sets of values - mass of object falling, mass of planet, radius of the planet, and time. I'm not including friction in the calculations.

I think this qualifies as a harmonic oscillator, and thus I work with the formula

$$T = 2\pi \sqrt{\frac{m}{k}}$$

To find the spring constant $k$ I need force $F$, and this is where I get uncertain. Should I work with the gravitational force between the object and the planet when the fall begins? In other words

$$F = G\times\frac{m \times M}{R^2}$$

When I try this I find that

$$F = kx \iff k = \frac{F}{x}$$

$$\iff k = \frac{G\times\frac{m \times M}{R^2}}{2R} = \frac{G \times m \times M}{2R^3}$$

$$\Rightarrow T = 2\pi \sqrt{\frac{m}{\frac{G \times m \times M}{2R^3}}} \iff T = 2\pi \sqrt{\frac{2R^3}{G \times M}}$$

Using this equation for the values I have, however, I get the wrong results - $T = 7148$ instead of $T = 5055$. What am I doing wrong?

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Just as an aside, you could have edited the earlier question to make it clear that it was not a duplicate then flagged for moderator attention. We try to be responsive to such things. – dmckee♦ Sep 15 '12 at 2:13
The first one really was a duplicate. Finding that original question helped me start to visualise the solution I later needed help finishing (in this question). Thank you though. – Quispiam Sep 15 '12 at 14:34

## 2 Answers

The key to this problem is the fact that the planet's mass $M$ as it appears in Newton's law of gravitation, $$F=\frac{GMm}{r^2},$$ is not actually constant. This is because the layers of the planet that are above you cause zero net force: if you are inside of a hollow spherical shell of mass then diametrically opposite elements of solid angle exert equal forces in opposite directions.

Thus, the effective mass of the planet in this problem is only that of a sphere of radius $r$ and density $3M_0/4\pi R^3$, i.e. $M(r)=\frac{r^3}{R^3}M_0$. The force is then $$F=\frac{GM_0m}{R^3}r$$ and it of course causes harmonic motion, with "spring constant" $k=GM_0m/R^3$.

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Wow, this solution gives me exactly the right answer. Thank you. I'm still not sure I grasp what exactly the solution does, but I'm getting there. It's just about mentally visualising it now... :) – Quispiam Sep 15 '12 at 14:29
@Quispiam If you still are having trouble with concepts, then we have failed. The purpose of this site is not to do your homework for you, but to help you understand the physics behind it (which is not a bad philosophy for all your studies). If you have any further, specific conceptual issues, you should bring them up, since walking away with the right answer but no deeper understanding is the same as walking away with nothing. – Chris White Sep 15 '12 at 16:24
I understand mathematically how you get $M_r = \frac{r^3}{R^3}M_0$, but I can't grasp why this is done. Why is the density of the sphere multiplied by the volume of the planet, and not the volume of the sphere itself? – Quispiam Sep 15 '12 at 16:27
@ChrisWhite I know. I want to understand the concepts. If I was happy with the solution I would have written down what Emilio wrote and been happy with it. I'm not happy until I understand what is happening. The mathematical solution is just a step closer to understanding the concept. – Quispiam Sep 15 '12 at 16:29
The important concept is that the mass above you does not add to the gravitational attraction, so that you are effectively on the surface of a smaller planet of radius $r$ which has constant density $\rho=M_0/\frac{4\pi}{3}R^3$ for $R$ the planet's radius. Thus the mass of the "effective planet" is $\frac{4\pi}{3}r^3\rho$. In your terms, the density of the planet is multiplied by the volume of the sphere of radius $r$. – Emilio Pisanty Sep 15 '12 at 16:56
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The period is indeed $T = 2\pi \sqrt{m/k}$, when $k$ is the proportionality constant between displacement and force, as in your third equation. So far so good. Now, why did you replace $x$ with $2R$? $x$ is the displacement from equilibrium at which you evaluated $F$.

There are two ways to go. Either say leave $x$ unknown and evaluate $F$ in terms of it, or choose a value for $x$ and find the force in that particular case. Both methods should agree. In the former, you'll know you have a simple harmonic oscillator if the $x$-dependence drops out when you find $k = F/x$. If you choose the latter, remember what $x$ is: displacement from equilibrium. Where is the equilibrium position of your intra-planet traveler, and how far away from that point are you when you evaluate the force on said traveler?

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To the downvoter: it would be great if you could leave a comment suggesting improvements or saying how this answer is bad. – Chris White Sep 15 '12 at 4:14
The answer misses the point of the problem, which is the breakdown of the simple inverse square formula when inside the planet. The harmonic motion is around the centre of the planet and therefore the displacement from equilibrium $x$ coincides with the radial coordinate $r$ up to a sign, but taking $k=F/x=F/r$ while keeping $F\propto1/r^2$ will always fail - motion outside the planet is not harmonic! – Emilio Pisanty Sep 15 '12 at 11:58
Just to be clear, neither I nor the OP ever said $F \propto 1/r^2$, nor was motion outside the planet brought up. – Chris White Sep 15 '12 at 16:19
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