Time for particle to fall to Earth

Question

I have seen the time quoted for a particle to fall to Earth from a height h (variations of gravity are taken into account) as

$\sqrt{\frac{2h}{g}}\left(1+ \frac{5h}{6R}\right)$

where R is the radius of the Earth. Having set up the equation of motion with g factored for a height of h, this solution still evades me. Can any one shed some light on this result ?

Answer 1

Just to expand on @YvesDaoust's answer (including the label change from $h$ to $h_0$), the substitution $x:=\sqrt{\frac{R+h}{h_0-h}}$ and abbreviation $y:=\frac{h_0}{R}$ gives the time as $$t=\frac{(R+h_0)^{3/2}}{R\sqrt{2g}}\int_{R/h_0}^\infty\frac{\sqrt{x}dx}{(x+1)^2}=\sqrt{\frac{h_0}{2g}}(1+y)^{3/2}\left[\frac{1}{1+y}+\frac{1}{\sqrt{y}}\arctan\sqrt{y}\right].$$Since $y$ is small, $$t=\sqrt{\frac{h_0}{2g}}\left(1+\frac32 y+o(y)\right)\left(1-y+1-\frac13 y+o(y)\right)=\sqrt{\frac{2h_0}{g}}\left(1+\frac56 y\right).$$

Answer 2

Hint:

The gravitational force is expressed by

$$f=-\frac{mgR^2}{(R+h)^2}$$ so that

$$\ddot h=-\frac{gR^2}{(R+h)^2}.$$

From this,

$$\dot h\ddot h=-\frac{gR^2}{(R+h)^2}\dot h$$ and by integration, assuming a null initial speed,

$$\dot h^2=2gR^2\left(\frac1{R+h}-\frac 1{R+h_0}\right)=2gR^2\frac{h_0-h}{(R+h_0)(R+h)}.$$

Then

$$dt=\sqrt{\frac{R+h_0}{2gR^2}}\sqrt{\frac{R+h}{h_0-h}}dh,$$

that you integrate from $h_0$ to $0$.