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)]}) #
