# Question c397a

Nov 21, 2017

You are almost there.... but I doubt if the integral has a closed form solution.
$T \left(H\right) = \setminus {\int}_{{R}_{\setminus \oplus}}^{{R}_{\setminus \oplus} + H} \frac{\mathrm{dr}}{\setminus \sqrt{2 G {M}_{\setminus \oplus} \left[\frac{1}{r} + \frac{1}{{R}_{\setminus \oplus} + H}\right]}}$

#### Explanation:

Consider an object of mass $m$ dropped from a height $H$ from the Earth's surface. The time taken for the object to impact Earth's surface is $T$. We are seeking to find an expression for this time in terms of the height $H$.

Measuring the position of the object from Earth's centre, this is an 1D problem in which the object moves along a straight line from an initial point ${r}_{i} = {R}_{\setminus \oplus} + H$ to a final point ${r}_{f} = {R}_{\setminus \oplus}$, where ${R}_{\setminus \oplus}$ is the radius of the Earth.

The kinematic quantities, velocity and acceleration are functions of position ($r$).

Time to impact:
$v \left(r\right) = \setminus \frac{\mathrm{dr}}{\mathrm{dt}} \setminus q \quad \setminus \rightarrow \mathrm{dt} = \frac{\mathrm{dr}}{v \left(r\right)} \setminus q \quad \setminus \rightarrow \setminus {\int}_{0}^{T} \mathrm{dt} = \setminus {\int}_{{r}_{i}}^{{r}_{f}} \frac{\mathrm{dr}}{v \left(r\right)}$
$T \left(H\right) = \setminus {\int}_{{R}_{\setminus \oplus} + H}^{{R}_{\setminus \oplus}} \frac{\mathrm{dr}}{v \left(r\right)}$ ...... (1)

To proceed further, we need to find an expression for the velocity as a function of position. For this we employ the mechanical energy conservation condition. Calculate the change in potential and kinetic energies of the object as it falls from a initial position ${r}_{0} = {R}_{\setminus \oplus} + H$ to an arbitrary point at a distance $r$ from the centre.
Potential Energy: U(r) = -(GM_{\oplus}m)/r;
$\setminus \Delta U = {U}_{f} \left(r\right) - {U}_{i} \left({r}_{0}\right) = - G {M}_{\setminus \oplus} m \left[\frac{1}{r} - \frac{1}{r} _ 0\right]$

Kinetic Energy: $K = \frac{1}{2} m {v}^{2}$
$\setminus \Delta K = {K}_{f} \left(v\right) - {K}_{i} \left(0\right) = \frac{1}{2} m {v}^{2} \left(r\right) - 0 = \frac{1}{2} m {v}^{2} \left(r\right)$

Mechanical Energy Conservation: $\setminus \Delta E = \setminus \Delta K + \setminus \Delta U = 0$
\Delta K = - \Delta U; \qquad 1/2mv^2(r) = GM_{\oplus}m[1/r-1/(R_{\oplus}+H)]#
$v \left(r\right) = - \setminus \sqrt{2 G {M}_{\setminus \oplus} \left[\frac{1}{r} - \frac{1}{{R}_{\setminus \oplus} + H}\right]}$ ...... (2)

The negative sign indicates that the velocity vector points downward, towards the coordinate origin

Substituting (2) in (1)

$T \left(H\right) = - \setminus {\int}_{{R}_{\setminus \oplus} + H}^{{R}_{\setminus \oplus}} \frac{\mathrm{dr}}{\setminus \sqrt{2 G {M}_{\setminus \oplus} \left[\frac{1}{r} + \frac{1}{{R}_{\setminus \oplus} + H}\right]}}$

$T \left(H\right) = \setminus {\int}_{{R}_{\setminus \oplus}}^{{R}_{\setminus \oplus} + H} \frac{\mathrm{dr}}{\setminus \sqrt{2 G {M}_{\setminus \oplus} \left[\frac{1}{r} + \frac{1}{{R}_{\setminus \oplus} + H}\right]}}$