AP EAMCET · PHYSICS · Gravitation
The potential energy of a satellite of mass ' \(m\) ' revolving around the earth at a height of \(R_e\) from the surface of the earth is
( \(R_e\) - radius of earth; \(g\) - acceleration due to gravity)
- A \(-0.5 \mathrm{mgR}_{\mathrm{e}}\)
- B \(-m g R_e\)
- C \(-2 m g R_e\)
- D \(-4 m g R_e\)
Answer & Solution
Correct Answer
(A) \(-0.5 \mathrm{mgR}_{\mathrm{e}}\)
Step-by-step Solution
Detailed explanation
Potential energy \(U = -\frac{GMm}{r}\) Distance from Earth's center \(r = R_e + h = R_e + R_e = 2R_e\) \(U = -\frac{GMm}{2R_e}\) Substitute \(GM = gR_e^2\): \(U = -\frac{(gR_e^2)m}{2R_e}\) \(U = -\frac{mgR_e}{2} = -0.5 mgR_e\)
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