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_{\oplus} + H to a final point r_f=R_{\oplus}, where R_{\oplus} is the radius of the Earth.
The kinematic quantities, velocity and acceleration are functions of position (r).
Time to impact:
v(r) = \frac{dr}{dt} \qquad \rightarrow dt = (dr)/(v(r)) \qquad \rightarrow \int_0^Tdt = \int_{r_i}^{r_f}(dr)/(v(r))
T(H) = \int_{R_{\oplus}+H}^{R_{\oplus}} (dr)/(v(r)) ...... (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_{\oplus} + H to an arbitrary point at a distance r from the centre.
Potential Energy: U(r) = -(GM_{\oplus}m)/r;
\DeltaU = U_{f}(r) - U_i(r_0) = -GM_{\oplus}m[1/r - 1/r_0]
Kinetic Energy: K = 1/2 mv^2
\DeltaK = K_{f}(v) - K_i(0) = 1/2mv^2(r) - 0 = 1/2mv^2(r)
Mechanical Energy Conservation: \Delta E = \DeltaK + \DeltaU = 0
\Delta K = - \Delta U; \qquad 1/2mv^2(r) = GM_{\oplus}m[1/r-1/(R_{\oplus}+H)]
v(r) = -\sqrt{2GM_{\oplus}[1/r-1/(R_{\oplus}+H)]} ...... (2)
The negative sign indicates that the velocity vector points downward, towards the coordinate origin
Substituting (2) in (1)
T(H) = -\int_{R_{\oplus}+H}^{R_{\oplus}} (dr)/(\sqrt{2GM_{\oplus}[1/r+1/(R_{\oplus}+H)]})
T(H) = \int_{R_{\oplus}}^{R_{\oplus}+H} (dr)/(\sqrt{2GM_{\oplus}[1/r+1/(R_{\oplus}+H)]})
