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This may be a repeated question.

Using Newton's law of gravitation F= GMm/r^2 , if I substitute the values of mass of earth and moon, distance between earth and moon, I get an value of around 1.9679×10^15 N (given below).

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If I use force law, F=ma and divide F value by their masses, I get two acceleration values 2.684×10^-8 m/s² (divided by moon's mass, m), 3.29521×10^-10m/s² (divided by earth's mass, M).

  1. What did these two different acceleration values represent?

So, using this value, I can understand (roughly) that the acceleration of moon towards earth, which is very very small compared to object inside the atmosphere; so that moon stays in the orbit. I also read before that,moon's linear velocity balances this gravitation force to maintain its constant path- 'Newton's orbital cannon'. But, moon is external body right, not projected from earth. So, I assume its orbit might determined by external balancing (moon come inside earth's gravitational field, but does not enters further due to its linear velocity action) of these two forces. Is this correct? If so,

  1. can I take that above gravitational force (GMm/r^2) and centripetal force (mv^2 /r) equals each other for the constant revolution of moon in its orbit? If yes, can anyone provide the calculation.
Rajesh R
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2 Answers2

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If you consider the problem in a non-rotating inertial reference frame, there is no balance of forces. The gravitational force causes the Moon to accelerate toward the Earth. This centripetal acceleration, combined with the Moon's velocity, causes the body to follow (approximately) a circular trajectory (or orbit).

You could also consider a frame of reference centered at Earth which rotates at the same rate as the Moon orbits Earth. From this perspective the Moon is stationary and gravitational force is balanced by the centrifugal pseudo-force.

FTT
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  1. What did these two different acceleration values represent?

You can calculate the gravitational attraction between the Earth and the Moon in the exact way that you did, using Newton’s gravitational formula. Considering Newton’s third law, this is a force that both objects are experiencing. Dividing this force by the objects respective masses simply gives you the acceleration that the Earth and Moon respectively are under - just like the Moon is orbiting Earth, Earth is also being attracted by the Moon. Specifically, both the Earth and the Moon orbit the centre of mass of the Earth-Moon system, which is inside the Earth. This is also often called the “Barycenter”[1]

  1. can I take that above gravitational force (GMm/r^2) and centripetal force (mv^2 /r) equals each other for the constant revolution of moon in its orbit? If yes, can anyone provide the calculation.

This is simple cancellation and rearrangement of variables, and for a circular orbit you get the simple answer that v=sqrt(GM/r).

I am not sure what you mean by “external balancing” in this part, Earth’s gravity simply provides the necessary centripetal acceleration for the object to remain in circular motion at said distance.

[1]: https://en.wikipedia.org/wiki/Barycenter_(astronomy)#:~:text=In%20astronomy%2C%20the%20barycenter%20(or,point%2C%20not%20a%20physical%20object.

skytak picus
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