question_answer

The earth of 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 \[1.5\times {{10}^{8}}\,km,\] the force exerted by the sun, on the earth is:

A)  \[\text{27 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{39}}}\text{N}\]                      

B)  \[\text{36 }\!\!\times\!\!\text{ 1}{{\text{0}}^{21}}\text{N}\]

C)  \[\text{18 }\!\!\times\!\!\text{ 1}{{\text{0}}^{25}}\text{N}\]                    

D)         \[\text{6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{19}}\text{N}\]

Correct Answer: B

Solution :

Key Idea: Earth is acted upon by a centripetal force directed towards the sun. On the basis of Keplers law, Newton drew the following conclusion that planet is acted upon by a centripetal force directed towards the sun.                 Force is given by \[F=\frac{m{{v}^{2}}}{r}\] where v is velocity and r is the radius. Also,      \[v=r\omega \] \[\therefore \]  \[F=mr{{\omega }^{2}}\] Putting the numerical values, we have \[F=(6\times {{10}^{24}})\times (1.5\times {{10}^{11}})\times {{(2\times {{10}^{-7}})}^{2}}\] \[=36\times {{10}^{21}}\,N\] Note: This force is attractive in nature.