Gravitation Question 283

Question: The earth (mass $ [=6\times 10^{24}kg)] $ ) revolves round the sun with angular velocity $ [2\times {{10}^{-7}}rad/s] $ in a circular orbit of radius $ [1.5\times 10^{8}km] $ . The force exerted by the sun on the earth in newtons, is [CBSE PMT 1995; AFMC 1999; Pb. PMT 2003]

Options:

A)$ [18\times 10^{25}] $

B) Zero

C) $ [27\times 10^{39}] $

D) $ [36\times 10^{21}] $

Show Answer

Answer:

Correct Answer: D

Solution:

$ [m=6\times 10^{24}kg,] $

$ [\omega =2\times {{10}^{-7}}rad/s,] $

$ [R=1.5\times 10^{11}m] $

The force exerted by the sun on the earth $ [F=m{{\omega }^{2}}R] $

By substituting the value we can get, $ [F=36\times 10^{21}N] $



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