The escape speed from the surface of a planet of mass M and radius R is given by:
Av_e = √(G M / R)
Bv_e = √(2 G M / R)
Cv_e = √(G M / R²)
Dv_e = √(2 G M / R²)
Answer & Solution
Correct answer: B. v_e = √(2 G M / R)
1. To just escape, total mechanical energy at the surface = 0 (KE + PE = 0 at infinity).
2. ½ m v_e² + (−G M m / R) = 0.
3. Solve: v_e² = 2 G M / R ⇒ v_e = √(2 G M / R).
4. Using g = G M / R², an equivalent form is v_e = √(2 g R).
5. The distractor √(GM/R) is the ORBITAL speed at the surface — easy confusion.
_Source: NCERT Class 11 Physics Ch 7 "Gravitation", §7.8 Escape Speed, p. 136_
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