JEE Mains · Physics · STD 11 - 7. gravitation
A body of mass \(m\) is taken from the surface of earth to a height equal to twice the radius of earth (\(R_e\)). The increase in potential energy will be _______.
(\(g\) is acceleration due to gravity at the surface of earth)
- A \(\dfrac{1}{2}mgR_e\)
- B \(\dfrac{3}{4}mgR_e\)
- C \(\dfrac{1}{4}mgR_e\)
- D \(\dfrac{2}{3}mgR_e\)
Answer & Solution
Correct Answer
(D) \(\dfrac{2}{3}mgR_e\)
Step-by-step Solution
Detailed explanation
Initial potential energy of the body at the surface of the earth is given by: \(U_i = -\dfrac{GMm}{R_e}\) Final potential energy of the body at a height \(h = 2R_e\) from the surface of the earth is: \(U_f = -\dfrac{GMm}{R_e + h} = -\dfrac{GMm}{R_e + 2R_e} = -\dfrac{GMm}{3R_e}\)…
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