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 - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

The centripetal force is provided by gravity: \frac{m v ^{2}}{R + x} = \frac{GM m}{( R + x ) ^{2}} ⟹ v = \frac{GM}{R + x} We know g = \frac{GM}{R ^{2}} ⟹ GM = g R^{2} .Substituting GM : v = \frac{g R ^{2}}{R + x} = (\frac{g R ^{2}}{R + x})^{1/2}

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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:

gx

\frac{g R}{R - x}

\frac{R g}{R + x}

(\frac{g R ^{2}}{R + x})^{1/2}

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