Question
Physics Question on Gravitation
What is the escape velocity for a body on the surface of a planet on which the acceleration due to gravity is (3.1)2ms−2 and whose radius is 8100 km?
A
2790km−s−1
B
27.9km−s−1
C
527.9km−1
D
27.95km−s−1
Answer
527.9km−1
Explanation
Solution
Escape velocity ve=2gR
Given, g=(3.1)2ms−2,
R=8100km=8100x103m.
= ve=2×(3.1)2×8100×103
ve=12.5×103ms−1
⇒ve=12.5kms−1
≈527.9kms−1