Question

The earth $(mass=6\times {10}^{24}kg)$ revolves around 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

• A.
• B. Zero
• C.
• D.

Answer 1

Answer: (D)

Step 1: Given that:

Mass of earth = $6\times {10}^{24}kg$

Angular velocity(ω) of the earth around the sun = $2\times {10}^{-7}rad{s}^{-1}$

Radius(r) of circular orbit = $1.8\times {10}^{8}km=1.8\times {10}^{11}m$

Step 2: Calculation of the force exerted by the sun on the earth:

The force exerted by the sun on the earth is given by;

$F=m{\omega }^{2}R$
By substituting the value we can get,

$F=6\times {10}^{24}\times (2\times {10}^{-7}{)}^2\times 1.5\times {10}^{11}$

$F=36\times {10}^{21}N$

Thus,

The force exerted by the sun on the earth is $36\times {10}^{21}N$

Hence,

Option D) $36\times {10}^{21}N$ is the correct option.