A satellite of mass m is revolving round the earth at a height R above the surface of earth. If g is gravitational acceleration at surface of earth and R is radius of earth then magnitude of momentum of satellite is

\sqrt{\frac{m^{2} R g}{2}}

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m \sqrt{R g}

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\sqrt{\frac{m^{2} R g}{4}}

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\sqrt{2 m^{2} R g}

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Solution

The correct option is A $\sqrt{\frac{m^{2} R g}{2}}$
$\frac{G M m}{(2 R)^{2}} = \frac{m v^{2}}{2 R} \Rightarrow V = \sqrt{\frac{G M}{2 R}} = \sqrt{\frac{G M R}{2 R^{2}}} = \sqrt{\frac{g R}{2}}$

momentum P=mV= $m \sqrt{\frac{g R}{2}} = \sqrt{\frac{m^{2} R g}{2}}$

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A satellite of mass m is revolving round the earth at a height R above the surface of earth. If g is gravitational acceleration at surface of earth and R is radius of earth then magnitude of momentum of satellite is

The correct option is A √m2Rg2GMm(2R)2=mv22R⇒V=√GM2R=√GMR2R2=√gR2 momentum P=mV=m√gR2=√m2Rg2

The correct option is A $\sqrt{\frac{m^{2} R g}{2}}$
$\frac{G M m}{(2 R)^{2}} = \frac{m v^{2}}{2 R} \Rightarrow V = \sqrt{\frac{G M}{2 R}} = \sqrt{\frac{G M R}{2 R^{2}}} = \sqrt{\frac{g R}{2}}$

momentum P=mV= $m \sqrt{\frac{g R}{2}} = \sqrt{\frac{m^{2} R g}{2}}$