Question:
The ratio of the energy required to raise a satellite upto a height \(h\) above the earth to the kinetic energy of the satellite into the orbit is
The ratio of the energy required to raise a satellite upto a height \(h\) above the earth to the kinetic energy of the satellite into the orbit is
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The PE change to lift a satellite to height $h$ is $mgh/(1+h/R)$, and the orbital KE is $mgR/[2(1+h/R)]$. Their ratio simplifies to $2h/R$.
Updated On: Apr 20, 2026
- \(h:R\)
- \(R:2h\)
- \(2h:R\)
- \(R:h\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Energy to raise satellite (against gravity) vs kinetic energy in orbit at height \(h\).
Step 2: Detailed Explanation:
Energy to raise: \(E_1 = \Delta U = \dfrac{mgh}{1+h/R}\). Kinetic energy in orbit: \(E_2 = \dfrac{1}{2}mv_0^2 = \dfrac{GMm}{2(R+h)} = \dfrac{mgR^2}{2R(1+h/R)} = \dfrac{mgR}{2(1+h/R)}\). \[ \frac{E_1}{E_2} = \frac{mgh/(1+h/R)}{mgR/[2(1+h/R)]} = \frac{2h}{R} \]
Step 3: Final Answer:
\(E_1 : E_2 = 2h : R\).
Energy to raise satellite (against gravity) vs kinetic energy in orbit at height \(h\).
Step 2: Detailed Explanation:
Energy to raise: \(E_1 = \Delta U = \dfrac{mgh}{1+h/R}\). Kinetic energy in orbit: \(E_2 = \dfrac{1}{2}mv_0^2 = \dfrac{GMm}{2(R+h)} = \dfrac{mgR^2}{2R(1+h/R)} = \dfrac{mgR}{2(1+h/R)}\). \[ \frac{E_1}{E_2} = \frac{mgh/(1+h/R)}{mgR/[2(1+h/R)]} = \frac{2h}{R} \]
Step 3: Final Answer:
\(E_1 : E_2 = 2h : R\).
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