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)]}) #