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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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An artificial satellite of mass m moves in an orbit whose radius is n times the radius of earth. Assuming the resistance to the motion is proportional to square of velocity ie $F= av^2$ where a is a constant. How long will the satellite take to fall to earth. M-mass of earth R- radius of earth

\[(a)\;\frac{m}{a} \bigg(\frac{R}{GM}\bigg)\sqrt {n-1} \quad (b)\;\frac{m}{a} \sqrt{\frac{R}{GM}}(\sqrt n -1) \quad (c)\;\frac{m}{a} \sqrt {\frac{RG}{M}}(\sqrt {n+1}) \quad (d)\;\frac{m}{a} \sqrt {\frac{RG}{M}}(\sqrt n-1) \]

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1 Answer

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Orbital velocity at distance r from center of earth
$v_0=\sqrt {\large\frac{GM}{r}}$
$KE= \large\frac{1}{2} $$ mv_0^2=\large\frac{GMm}{2r}$
$P.E= \large\frac{-GMm}{r}$
Total energy $E= KE+PE$
$E= \large\frac{-GMm}{2r}$
Now $\large\frac{dE}{dt}$$=F.v$
$\large\frac{GMm}{2r^2}\bigg(\large\frac{dr}{dt}\bigg)$$=-av^3$
$\qquad= -a \bigg(\sqrt {\large\frac{GM}{r}}\bigg)^3$
$\large\frac{GMm}{2r^2}\bigg(\large\frac{dr}{dt}\bigg)$$=-a \large\frac{GM}{r} \sqrt {\frac{GM}{r}}$
$\large\frac{a}{m}$$ \int \limits_0^t dt=-\large\frac{1}{2}\sqrt {\frac{1}{GM}}\int \limits_{r_i}^{r_f} $$r^{-1/2} dr$
$\large\frac{a}{m} $$\int \limits_0^t dt=-\large\frac{1}{2}\sqrt {\frac{1}{GM}}\int \limits_{nR}^{R} $$r^{-1/2} dr$
$t= \large\frac{m}{a} \sqrt {\frac{R}{GM} }$$ (\sqrt n -1)$
Hence b is the correct answer

 

answered Aug 29, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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