# Is the gravitational force exerted by the Earth on the Moon equal to the centripetal force acting on the Moon?

The gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force. An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon’s orbit as circular (it is not, but this is a good approximation for our purposes).

The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:

ag=G(MEarth+MMoon)/r2

Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:

acentr=(4 pi2 r)/T2

Where T is the period. Since the two accelerations have to be equal, we obtain:

(4 pi2 r) /T2=G(MEarth+MMoon)/r2

Which implies:

r3/T2=G(MEarth+MMoon)/4 pi2=const.

This is the so-called third Kepler law, that states that the cube of the radius of the orbit is propprtional to the square of the period. This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet’s period you can determine its orbit radis. Edited from Cornell

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### 3 Responses to Is the gravitational force exerted by the Earth on the Moon equal to the centripetal force acting on the Moon?

1. alfy says:

I don’t think it answers the question, which is: What is the value of the force which pushes the moon outward as a result of its orbital speed around the earth? I have read articles which question the validity of concepts like “centrifugal force” and “centripetal force”. This passage begins by assuming the two forces are equal, which I am sure they are, but that misses the point. The idea is to establish the physics of the moon’s orbital motion, and calculate the value of the outward force. We can then see if the two are equal. If they aren’t there must be a mistake in the sums or the assumptions about the outward force.
As I tried to explain in the pub, my best understanding of the physics is that the circular motion of the moon creates an acceleration towards the orbital centre, and this produces an outward force in reaction. I have yet to find a formula that says, F = something involving mass, orbital speed, distance etc. All the equations seem to be about something slightly different. I want a four door saloon, but the salesman keeps showing me 4wds. sports cars, or vans.

• Deskarati says:

I think the reason that this doesn’t answer the question is because the question assumes an outward force that is not actually there. This is explained in the following presentation of Dynamics of Uniform Circular Motion (see page 9).

2. alfy says:

I have had a quick look at the article, and need to read it carefully, but at first glance it seems likely to answer the question. Thanks for continuing the quest.