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] $
BETA
