How to integrate acceleration due to gravity with height? If I project a body at a very large velo

Jamir Melendez

Jamir Melendez

Answered question

2022-04-12

How to integrate acceleration due to gravity with height?
If I project a body at a very large velocity such that it reaches a height h which is comparable to Earth's radius, then how to derive an equation for 'h' in terms of initial velocity 'u' and other constants.
This is where I ended up with this problem
S = u²/2g
But g is variable with height as h approaches R
So dg = GM/(R+dr)² --- this is where I think I might be wrong
Hence dS = u²(R+dr)²/GM
So how do I go ahead?

Answer & Explanation

Madelyn Lynch

Madelyn Lynch

Beginner2022-04-13Added 15 answers

As I cannot fully understand your question, I will somewhat provide you with the tools so that you can answer your own question.
g = G M R 2
The above mentioned equation gives you the acceleration due to the gravitational at some point point that is R units of distance away from the center of any body of mass M. This can also be called as the gravitational field.
For earth, M = 6 10 24 (approximately), and for any point on the surface of the earth R will be approximately 6400 k m
Now, only using the variables..
Let g be the acceleration due to gravity on the earth's surface. Therefore,
g = G M R 2
, where R = radius of earth.
Let g be the acceleration due to gravity at a height h from the earth's surface. Now,
g = G M ( R + h ) 2
With the ratio g g , you can compare the two fields. For h << R, use binomial expansion for negative index and approximate. (Higher powers of h / R are negligible).
If you want to find the potential energy, integrate with limits from infinity to the desired distance ( R + h) the following expression.
W = F d x = ( R + h ) G m 1 m 2 x 2 d x
Note: Gravitational potential energy at a point is defined as the work done to bring a mass from infinity to that point.
Hope I am right and hope this helps you.

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