Q. The earth (mass = $6 \times 10^{24}$kg) revolves around the sun with angular velocity $2 \times 10^{- 7}\ rads^{- 1}$ in a circular orbit of radius $1.5 \times 10^{8}$km. The force exerted by the sun on the earth in newton is
A) $Zero$
B) $18 \times 10^{25}$
C) $27 \times 10^{39}$
D) $36 \times 10^{21}$
Explanation
(d) $F = mR\ \omega^{2}$
$= 6 \times 10^{24} \times \left( 1.5 \times 10^{11} \right)\left( 2 \times 10^{- 7} \right)^{2}$
$= 36 \times 10^{21}N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.