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Answer: 2. \(\frac{5GmM}{6R}\)

Key idea: The minimum energy is the difference between the total energy in the orbit and the total energy at the surface.

The orbit radius is \(r=R+2R=3R\).

  • Energy at the surface: \(-\frac{GMm}{R}\).
  • Energy in the orbit: \(-\frac{GMm}{2r}=-\frac{GMm}{6R}\).

\(\Delta E=-\frac{GMm}{6R}+\frac{GMm}{R}=\frac{5GMm}{6R}\).

Option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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