An artificial satellite revolves around earth in a circular orbit of radius 'r' with time period 'T' .The satellite is made to stopo in the orbit which makes it fall onto earth.Time taken by the satellite to fall onto earth is

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Solution

$Suppose the satellite stops due to the loss in energy and suppose the rate of losing energy is C . Let the velocity of the satellite is v_{1} . \frac{m_{s} {v_{1}}^{2}}{r} = \frac{m_{s} m_{e}}{r^{2}} G where {v_{1}}^{2} = \frac{m_{e} G}{r} [Because centripetal force on the satellite = Force of attraction due to the earth] Similarly v_{2} is the velocity of satellite when its orbit touches the earth . \frac{m_{s} {v_{2}}^{2}}{r_{e}} = \frac{m_{s} m_{e}}{{r_{e}}^{2}} G where, {v_{2}}^{2} = \frac{m_{e} G}{r_{e}} Kinetic energy of the satellite in orbit of radius r is \frac{1}{2} m_{s} {v_{1}}^{2} i . e \frac{1}{2} m_{s} \frac{m_{e} G}{r} Kinetic energy of the satellite in orbit of radius r_{e} when it touches the earth is \frac{1}{2} m_{s} {v_{2}}^{2} i . e \frac{1}{2} m_{s} \frac{m_{e} G}{r_{e}} Potential energy of the satellite when at distance r from the centre of the earth = - m_{s} \frac{m_{e} G}{r} Potential energy of the satellite when at distance r_{e} from the centre of the earth = - m_{s} \frac{m_{e} G}{r_{e}} Total energy of the satellite when in orbit of radius r = - m_{s} \frac{m_{e} G}{r} + \frac{1}{2} m_{s} \frac{m_{e} G}{r} = - \frac{1}{2} m_{s} \frac{m_{e} G}{r} Total energy of the satellite when on the surface of the earth = - m_{s} \frac{m_{e} G}{r_{e}} + \frac{1}{2} m_{s} \frac{m_{e} G}{r_{e}} = - - \frac{1}{2} m_{s} \frac{m_{e} G}{r_{e}} Thus, loss of energy in falling to the earth = - \frac{1}{2} m_{s} \frac{m_{e} G}{r} - [- \frac{1}{2} m_{s} \frac{m_{e} G}{r_{e}}] = \frac{1}{2} m_{s} m_{e} G [\frac{1}{r_{e}} - \frac{1}{r}] Suppose if the satellite takes time' t' to fall to the earth, then the energy lost = Ct Thus, Ct = \frac{1}{2} m_{s} m_{e} G [\frac{1}{r_{e}} - \frac{1}{r}] or t = \frac{1}{2 C} m_{s} m_{e} G [\frac{1}{r_{e}} - \frac{1}{r}]$

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