A satellite of mass m revolves around the Ear | vedclass

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(C) The orbital speed $v_0$ of a satellite at a distance $r = R + x$ from the center of the Earth is given by $v_0 = \sqrt{\frac{GM}{r}} = \sqrt{\frac

Vedclass · Accepted Answer

${\left( {\frac{{gR^2}}{{R + x}}} \right)^{1/2}}$

Vedclass · Answer

(C) The orbital speed $v_0$ of a satellite at a distance $r = R + x$ from the center of the Earth is given by $v_0 = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{R + x}}$.We know that the acceleration due to gravity on the surface of the Earth is $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.Substituting $GM = gR^2$ into the expression for $v_0$,we get:$v_0 = \sqrt{\frac{gR^2}{R + x}} = \left( \frac{gR^2}{R + x} \right)^{1/2}$.Thus,the correct option is $C$.

$A$ satellite of mass $m$ revolves around the Earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the Earth,the orbital speed of the satellite is:

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