The escape velocity on earth is given by Ve =Re2GMe
Escape velocity for the planet is
V′esc=4Re2G(9Me)...(i)
The escape velocity of earth is Ve=11.2×103ms−1
From equation (i)
V′esc=49(Re2GMe)=23×11.2×103ms−1⇒V′esc=16.8kms−1
A planet having mass 9Me and radius 4Re, where Me and Re are mass and radius of earth respectively, has escape velocity in kms−1 given by: (Given escape velocity on earth Ve=11.2×103ms−1)
Held on 13 Apr 2023 · Verified 6 Jul 2026.
67.2
16.8
11.2
33.6
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