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?

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/63590/2451 and links therein. $\endgroup$ – Qmechanic Jun 1 '19 at 16:56
  • $\begingroup$ Your expression for $dg$ doesn't make sense, you have to apply the chain rule there on the right hand side $\endgroup$ – Triatticus Jun 1 '19 at 20:05

As I cannot fully understand your question, I will somewhat provide you with the tools so that you can answer your own question. $$ g = \frac{GM}{R^2} $$ 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 $6400$ $km$.

Now, only using the variables..

Let $g$ be the acceleration due to gravity on the earth's surface. Therefore, $$g = \frac{GM}{R^2}$$, where R = radius of earth.

Let $g*$ be the acceleration due to gravity at a height h from the earth's surface. Now, $$g* = \frac{GM}{(R + h)^2}$$

With the ratio $\frac{g*}{g}$, 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. $$ W = \int Fdx = \int_\infty^{(R+h)} \frac{Gm_1m_2}{x^2}dx$$

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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  • $\begingroup$ Thank you for the response, I have updated my question to make it a bit more clear, really appreciate you taking time. $\endgroup$ – Fardeen Jun 1 '19 at 16:12

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