Question:
Work required to shift an artificial satellite from an orbit of radius \(r\) to an orbit of radius \(2r\) is
Work required to shift an artificial satellite from an orbit of radius \(r\) to an orbit of radius \(2r\) is
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Remember:
\[
E=-\frac{GMm}{2R}
\]
Higher orbit means higher (less negative) total energy.
Updated On: Jun 17, 2026
- \( \dfrac{GMm}{2r} \)
- \( \dfrac{GMm}{4r} \)
- \( \dfrac{GMm}{8r} \)
- Zero
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The Correct Option is B
Solution and Explanation
Concept:
Total energy of a satellite in circular orbit:
\[
E=-\frac{GMm}{2R}
\]
Work done in shifting orbit equals change in total energy.
Step 1: Find initial total energy. Initial radius: \[ R=r \] So, \[ E_i=-\frac{GMm}{2r} \]
Step 2: Find final total energy. Final radius: \[ R=2r \] Therefore, \[ E_f=-\frac{GMm}{2(2r)} \] \[ E_f=-\frac{GMm}{4r} \]
Step 3: Compute work required. \[ W=E_f-E_i \] \[ W=-\frac{GMm}{4r}+\frac{GMm}{2r} \] Taking LCM: \[ W=\frac{-GMm+2GMm}{4r} \] \[ W=\frac{GMm}{4r} \] Hence, \[ \boxed{\frac{GMm}{4r}} \]
Step 1: Find initial total energy. Initial radius: \[ R=r \] So, \[ E_i=-\frac{GMm}{2r} \]
Step 2: Find final total energy. Final radius: \[ R=2r \] Therefore, \[ E_f=-\frac{GMm}{2(2r)} \] \[ E_f=-\frac{GMm}{4r} \]
Step 3: Compute work required. \[ W=E_f-E_i \] \[ W=-\frac{GMm}{4r}+\frac{GMm}{2r} \] Taking LCM: \[ W=\frac{-GMm+2GMm}{4r} \] \[ W=\frac{GMm}{4r} \] Hence, \[ \boxed{\frac{GMm}{4r}} \]
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