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?
HANG ON I THINK I GOT 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
Hint: Multiply both sides of your equation by $v=dr/dt$ and then integrate.