Question

The earth (mass = \[6\times {{10}^{24}}\] kg) revolves around the sun with an angular velocity of \[2\times {{10}^{-7}}\]rad/s in a circular orbit of radius \[1.5\times {{10}^{9}}\]km. The force exerted by the sun on the earth, in newtons, is:
A. \[36\times {{10}^{21}}\] N
B. \[16\times {{10}^{24}}\] N
C. \[25\times {{10}^{16}}\] N
D. zero

Answer 1

Hint: In this problem, we know the movement of planets around the sun happens under the influence of mutual gravitational forces. So, we can make use of gravitational force to solve this problem. But here the mass of the sun is not mentioned, so we can make use of angular velocity to find out the force.

Complete step by step answer:
We know force when the body is moving in a circular motion, is always the centripetal force. Here, the cause of centripetal force is the gravitational force of attraction between the earth and the sun. We know centripetal force on a body of mass m, moving in a circular orbit of radius r, with linear velocity v, is given by: \[F=\dfrac{m{{v}^{2}}}{r}\].
But we don’t have the linear velocity, so we make use of relation between linear velocity and angular velocity: \[v=r\omega \]
Thus, centripetal force becomes: \[F=mr{{\omega }^{2}}\]
now mass=\[6\times {{10}^{24}}\] kg
radius= \[1.5\times {{10}^{9}}\]km= \[1.5\times {{10}^{12}}\] m
angular velocity= \[2\times {{10}^{-7}}\]
$ F=mr{{\omega }^{2}} \\
\Rightarrow F=6\times {{10}^{24}}\times 1.5\times {{10}^{12}}\times {{(2\times {{10}^{-7}})}^{2}} \\
\therefore F=36\times {{10}^{21}}N$
So, the force comes out to be \[36\times {{10}^{21}}N\].

Hence, the correct option is A.

Note:Whenever the body moves in a circular motion, it has linear components of velocity and acceleration and also, angular velocity and angular acceleration. We need to keep in mind that if a body is acted upon by a force and then that body starts moving in a circular motion, then that force must be equal to the centripetal force and in this problem it is provided by the gravitational force.