alleryIf the radius of a planet shrinks to half while its mass is unchanged, the escape velocity from its surface will:
ABecome half its old value
BStay exactly the same
CBecome √2 times its value
DBecome twice its old value
Answer & Solution
Correct answer: C. Become √2 times its value
1. v_e = √(2 G M / R).
2. Mass M is constant; R → R/2, so v_e' = √(2 G M / (R/2)) = √(4 G M / R) = 2 · √(G M / R).
3. Wait, recompute: v_e' = √(2 · 2 G M / R) = √2 · √(2 G M / R) = √2 · v_e.
4. Hence v_e becomes √2 times the original — about 41 % larger.
5. "Double" would require R → R/4, not R/2.
_Source: syllabus Physics Ch 7 "Gravitation", §7.8 (escape-speed scaling)_
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