First slide

Gravitaional potential energy

Question

The minimum and maximum distance of a satellite from the centre of the earth 2R and 4R respectively. Where R is the radius of the earth and M is the mass of earth. Its maximum speed is

Moderate

A
$\sqrt{\frac{G M}{6 R}}$
B

$\sqrt{\frac{2 G M}{3 R}}$
C

$\sqrt{\frac{G M}{3 R}}$
D

$\sqrt{\frac{2 G M}{5 R}}$

Solution

$\large v_1r_1=v_2r_2\;let\;v_1=v_{max}, v_2=v_{min}$
$\large i.e.\;v_1\times2R=v_2\times4R\Rightarrow v_2=\frac {v_1}{2}$
According to L.C.E
$\large \Rightarrow \frac{{ - GMm}}{{2R}} + \frac{1}{2}mv_1^2 = \frac{{ - GMm}}{{4R}} + \frac{1}{2}m{\left( {\frac{{{v_1}}}{2}} \right)^2}$
$\large \frac{1}{2}m\left( {v_1^2 - \frac{{v_1^2}}{4}} \right) = GMm\left[ {\frac{1}{{2R}} - \frac{1}{{4R}}} \right] = \frac{{GMm}}{{4R}}$
$\large \frac{1}{2}m\frac{{3v_1^2}}{4} = \frac{{GMm}}{{4R}} \Rightarrow {v_1} = \sqrt {\frac{{2GM}}{{3R}}}$

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