The gravitational force is given by

$$ F = \dfrac{-Gm_1m_2}{r^2} $$

But, since F = ma, then for an object placed at r distance away from the centre of the earth it would experience

$$ a = \dfrac{-Gm}{r^2} $$

where m is the mass of the earth. So if I wanted to find the relationship between the position and time of the object, I'd have to integrate acceleration once with respect to time for velocity, and again for the position. So I try to integrate:

$$ V = -Gm\int \dfrac{1}{r^2} dt $$

but since r itself is a function of time (which is what I want in the end), i'm not too sure how to integrate it?


$$ F = \dfrac{-Gm_1m_2}{r^2} $$ $$ a = \dfrac{-Gm}{r^2} $$ $$ V = -Gm\int \dfrac{1}{r^2} dt $$ $$ V^2 = -Gm\int \dfrac{1}{r^2} dt \dfrac{dr}{dt} $$ $$ V^2 = -Gm\int \dfrac{1}{r^2} dr $$ $$ V^2 = -Gm (-\dfrac{1}{r} + c) $$ assuming we are dropping a mass from a distance of s from the earths centre $$ 0 = -Gm (-\dfrac{1}{s} + c) $$ $$ c = 1/s $$ $$ V^2 = Gm(1/r-1/s) $$ $$ V = \pm \sqrt{Gm(1/r-1/s)} $$ assuming we are dropping a mass from a large r, and it is falling to earth, v is always negative $$ V = \sqrt{Gm(1/r-1/s)} $$

$$ \dfrac{dr}{dt} = \sqrt{Gm(1/r-1/s)} $$ $$ \dfrac{dt}{dr} = \dfrac{1}{\sqrt{Gm(1/r-1/s)}} $$ $$ t = \int \dfrac{1}{\sqrt{Gm(1/r-1/s)}} dr $$

$$ t = \dfrac{1}{\sqrt{Gm}} \int \dfrac{1}{\sqrt{1/r + 1/s}}dr $$

From here i think its just slogging it out

  • $\begingroup$ en.wikipedia.org/wiki/Classical_central-force_problem $\endgroup$ – Rahul Oct 11 '14 at 3:52
  • $\begingroup$ It should be 1 over $r^2$ $\endgroup$ – Jasser Oct 11 '14 at 3:57
  • $\begingroup$ Typically it would be easier to integrate the force with respect to position to obtain the work integral and work out what you want from there. $\endgroup$ – Eweler Oct 11 '14 at 4:00
  • $\begingroup$ If I do that I can find out the Gravitational Potential Energy at a particular point in space, but I'm trying to relate position to time, so I'm not sure if that helps $\endgroup$ – Joshua Lin Oct 11 '14 at 4:03
  • $\begingroup$ Line 4 is incorrect. You have to multiply line 2 by $dr/dt=v$ and then integrate. $\endgroup$ – Ted Shifrin Oct 11 '14 at 11:07

Hint: Multiply both sides of your equation by $v=dr/dt$ and then integrate.

  • $\begingroup$ Tried that, but I end up with an equation relating Velocity and displacement, and I end up with the same problem? v = -root(GM/r) $\endgroup$ – Joshua Lin Oct 11 '14 at 4:00
  • $\begingroup$ Integrate again by separating variables. Oh, and don't forget your constants of integration. :) $\endgroup$ – Ted Shifrin Oct 11 '14 at 4:19

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