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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)2ms2(3.1)^2 ms^{-2} and whose radius is 8100 km?

A

2790kms12790\, km-s^{-1}

B

27.9kms127.9\,km-s^{-1}

C

27.95km1\frac{27.9}{\sqrt5}\,km^{-1}

D

27.95kms127.9\sqrt5 \,km-s^{-1}

Answer

27.95km1\frac{27.9}{\sqrt5}\,km^{-1}

Explanation

Solution

Escape velocity ve=2gRv_e=\sqrt{2gR}
Given, g=(3.1)2ms2,g=(3.1)^2ms^{-2},
R=8100km=8100x103m.R = 8100km = 8 1 00 x 10^3 m.
= ve=2×(3.1)2×8100×103v_e=\sqrt{2\times(3.1)^2\times8100\times10^3}
ve=12.5×103ms1v_e=12.5\times10^3ms^{-1}
ve=12.5kms1\Rightarrow v_e=12.5 kms^{-1}
27.95kms1\approx\frac{27.9}{\sqrt5}kms^{-1}