A) \[\frac{1}{\alpha }\frac{\left[ \sqrt{\eta R}-\sqrt{R} \right]}{\sqrt{GM}}\]
B) \[\frac{1}{\alpha }\frac{\left[ \sqrt{\eta R}-\sqrt{R} \right]}{M}\]
C) \[\frac{m}{\alpha }\frac{\left[ \sqrt{\eta R}-\sqrt{R} \right]}{\sqrt{GM}}\]
D) \[\frac{m}{\alpha }\frac{\left[ \sqrt{\eta R}-\sqrt{R} \right]}{\sqrt{GM}}\]
Correct Answer: C
Solution :
| The energy of the satellite in a orbit of radius r,\[E=\frac{-GMm}{2r}.\] |
| Here m is the mass of satellite and M is the mass of the moon. If \[dE/dt\]is the instantaneous rate of decrease of energy of the satellite, then we have\[\left( \frac{dE}{dt} \right)=Fv\] |
| Given F=\[\alpha {{v}^{2}}\] |
| \[\therefore \frac{d\left[ \frac{-GMm}{2r} \right]}{dt}=(\alpha {{v}^{2}})v\] |
| Or \[\frac{GMm}{2{{r}^{2}}}dr=-\alpha {{v}^{3}}dt\] |
| Orbital velocity of satellite,\[v=\sqrt{\frac{CM}{r}}.\] |
| \[\therefore \frac{GMm}{2{{r}^{2}}}dr=-\alpha {{\left[ \frac{GM}{r} \right]}^{3/2}}dt\] |
| \[\frac{m{{r}^{1/2}}}{2}dr=-\alpha {{(GM)}^{1/2}}dt\] |
| Integrating above expression, we get \[\frac{m}{2}\int\limits_{\eta R}^{R}{{{r}^{-1/2}}dr=-\alpha {{(GM)}^{1/2}}\int\limits_{0}^{t}{dt}}\] |
| \[m\left| \sqrt{r} \right|_{\eta R}^{R}=-\alpha \sqrt{GM}t\] |
| Or \[m\left[ \sqrt{\eta R}-\sqrt{R} \right]=\alpha \sqrt{Gm}\,t\] |
| \[\therefore t=\frac{m}{\alpha }\frac{\sqrt{\eta R}-\sqrt{R}}{\sqrt{GM}}\] |
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