Sometimes questions go unasked. The subject of this article is
one. It appears that no one has questioned the amount of energy
required to leave the moon’s orbit. It has been calculated and
successfully used; yet, the reason for the apparently large amount
is not questioned.
It may be that NASA and other space exploration organizations
are satisfied with knowing that it takes less energy to leave the
moon than it does to leave the earth. This is important for
determining how much fuel was required for the Apollo missions to
return home. It also was instrumental in determining the shape of
the craft used as a lander. The lower than earth delta-v, the term
used to identify the energy requirement, also has implications for
any future space exploration missions.
First of all, let’s get acquainted with the term “delta-v”. Wikipedia’s
definition is located at
http://en.wikipedia.org/wiki/Delta-v . However, basically, it
is the amount of energy used measured in velocity and has nothing
to do with fuel or power plant requirements. The unit of measure
is in kilometers per second. The amount to leave earth’s orbit is
the tremendous figure of 11.2 km/sec. The moon’s requirement is a
much less 2.38 km/sec. This is where questions should begin to
The delta-v required to leave earth’s orbit is said to be
influenced by gravity, atmospheric resistance, and the shape of
the craft. Therefore, the same influences must also apply to the
moon. These influential elements should be looked at.
Anyone who has seen the Apollo rockets and images of the lunar
landers will immediately notice a difference in the shape of the
crafts. The Apollo rockets are streamlined and smooth. They look
like they could slip through most anything. Giant needles, if you
The lunar landers; however, look as aerodynamic as a brick.
They have hard edges with all manner of protruding objects. Look
at one on the ground and have someone tell you it flies. You would
laugh in their face. This is addressed because it will be
presented by many as a possible answer for the main question.
Despite the doubts that exist.
Delta-v is used as the unit of measurement because it is a
reference to velocity. The use of velocity allows for the
differing sizes and shapes. Once the velocity has been achieved,
it is possible to leave orbit, regardless of size or shape. If you
get your pickup truck going fast enough it can go into outer
At this point, you need to look at the main influence on the
amount of energy we need to leave orbit on either body. That
influence is the force of gravity. Earth has a gravity six times
that of the moon. Conversely, the moon’s gravity is only 1/6th the
earth’s. It should, therefore, stand to reason that the amount of
delta-v required to leave the moon’s orbit should not exceed 1/6th
that needed to leave the earth’s.
This is not how the math works out. In fact, it is reported
that 2.38 km/sec for the moon is actually 1.4.71th that of the
earth. Simple logic states that if the earth’s delta-v is 11.2
km/sec, then the moon’s should be no more than 1.87 km/sec. There
is a major difference in the amount of energy required and that
which basic math says it should be. The difference could then be
attributed to the shape of the vehicle. Right?
Not so fast!! Are we not told that the moon has no atmosphere?
That a total vacuum exists on the surface of the moon? So, why
would aerodynamics even be an issue?
Unless, ten pounds of clay molded into a bust of some famous
person has a different weight that a ten pound block of clay, the
ratio of gravity should be the main influencing factor on the
amount of delta-v. Remember, as stated earlier, delta-v is a
measurement of velocity. As such, we should be able to understand
the possibility for our pickup trucks leaving earth orbit if the
proper delta-v is achieved.
The shape of an object is not important until you consider the
resistance caused by the atmosphere. All of the substances between
the surface and the unseen border to outer space cause the
resistance. Oxygen, nitrogen, water vapor, CO2, and a great number
of other things act as a barrier by creating friction against our
space craft. This is the majority factor in determining the amount
of energy. These items add to the overall total delta-v that must
With the lack of these elements and substances at the moon, it
would stand to reason that the delta-v for the moon would be even
less than 1/6th that of earth as calculated based on gravitational
difference. Referring to the above delta-v figures for the moon,
it is easy to see that simple logic does not follow. Why?
Unless the gravity of the moon is stronger than 1/6th that of
the earth, there has to be a yet unidentified and unseen force on
the moon. What this force is and how it is created is a good
question to investigate. It is probably worth the time and effort.
Our understanding of this force could be our key to better deal
with future space exploration; especially if we ever conduct a
manned mission to Mars. It would be a crying shame to get there
and be unable to get off due to a miscalculation of the proper
One possible answer to the question is that there is no
unidentified force. NASA may have just added a major fudge-factor
to the delta-v to ensure that they did not underestimate. If this
is what happened, it should be admitted. Guessing is o.k. if it is
an educated guess that takes the unforeseen into account.
Copyright Arthur Ryan 2010
Presented with permission of the author
Articles by Art Ryan
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About the Author
Arthur “Art” Ryan is an independent researcher and
author, who is a full-time employee supporting a major railroad
and a small business owner of a process engineering consulting
company, that enjoys all manner of things science. He is a
baseball and football fan, who values his privacy and therefore
writes under a pseudonym.
His book concerning Earth Changes is called “Are We
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Questions and comments can be addressed to Art through his
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