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An artifical satellite (mass \mathrm{m} ) of a planet (mass \mathrm{M} ) revolves in a circular orbit whose radius is \mathrm{n} times the radius \mathrm{R} of the planet. In the process of motion, the satellite experiences a slight resistance due to cosmic dust. Assuming resistance force on satellite to depend on velocity, as \mathrm{\mathrm{F}=a v^2} where \mathrm{' a ' }is a constant, how long the satellite will stay in the orbit before it falls onto the planet's surface.
 

Option: 1

(\sqrt{\mathrm{n}}-1) \frac{\mathrm{m}}{\mathrm{a} \sqrt{\mathrm{gR}}}

 


Option: 2

(\sqrt{2 \mathrm{n}}-1) \frac{\mathrm{m}}{\mathrm{a} \sqrt{\mathrm{gR}}}
 


Option: 3

\quad(\sqrt{\mathrm{n}}-1) \frac{2 \mathrm{~m}}{\mathrm{a} \sqrt{\mathrm{gR}}}
 


Option: 4

\sqrt{\mathrm{n}}-1) \frac{\mathrm{m}}{\mathrm{a} \sqrt{2 \mathrm{gR}}}


Answers (1)

best_answer

Air resistance \mathrm{F=-a v^2\: where \: v=\sqrt{\frac{G M}{r}} ; r=} the distance of the satellite from earth's center.

\mathrm{ \Rightarrow \mathrm{F}=-\frac{\mathrm{GMa}}{\mathrm{r}} }

The work done by the resistance force

\mathrm{ =d W=F \cdot d x=[F \vee d t] }

\mathrm{ =\left[\frac{G M a}{r} \sqrt{\frac{G M}{r}} d t\right] }

\mathrm{ =\left[\frac{(\mathrm{GM})^{3 / 2} \mathrm{a}}{\mathrm{r}^{3 / 2} \mathrm{dt}}\right] }          .................(1)

\RightarrowThe loss of energy of the satellite =\mathrm{dE}

\mathrm{\therefore \frac{\mathrm{dE}}{\mathrm{dr}}=\frac{\mathrm{d}}{\mathrm{dr}}\left[-\frac{\mathrm{GM} \mathrm{m}}{2 \mathrm{r}}\right]=\frac{\mathrm{GMm}}{2 \mathrm{r}^2}}

\mathrm{\Rightarrow \mathrm{dE}=\frac{\mathrm{GMm}}{2 \mathrm{r}^2} \mathrm{dr}}               ............(2)

Since \mathrm{d E=d W} (work energy theorem)

\mathrm{\frac{G M m}{2 r^2} d r=\frac{(G M)^{3 / 2} a}{r^{3 / 2}} d t}

\mathrm{\Rightarrow \mathrm{t}=\frac{\mathrm{m}}{2 \mathrm{a} \sqrt{\mathrm{GM}}} \int_{\mathrm{nR}}^{\mathrm{R}} \frac{\mathrm{dr}}{\sqrt{\mathrm{r}}}}

\mathrm{\Rightarrow t=\frac{m \sqrt{R}(\sqrt{n}-1)}{a \sqrt{G M}}=(\sqrt{n}-1) \frac{m}{a \sqrt{g R}}}

Hence option 1 is correct.




 

Posted by

Gautam harsolia

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