Question

A planet of mass $3 \times {10^{29}}gm$ moves around a star with a constant speed of $2 \times {10^8}m{s^{ - 1}}$ in a circle of radii$1.5 \times {10^{12}}m$. The gravitational force exerted on the planet by the star is
A) $6.67 \times {10^{22}}dyne$
B) $6.67 \times {10^{26}}N$
C) $8 \times {10^{26}}N$
D) $8 \times {10^{27}}dyne$

Answer 1

Here the planet and the Sun attract each other and the planet is being pulled by the sun at its center and the force exerted by the Sun is known as centripetal force. Here, we have to apply the formula for centripetal force which is the gravitational force.

Formula used:
${F_c} = \dfrac{{m{v^2}}}{r}$ ;
${F_c}$= Centripetal Force.
m = mass.
v = velocity
r = distance

Complete step by step solution:
Find the gravitational force by applying the formula for centripetal force.
${F_c} = \dfrac{{m{v^2}}}{r}$;
Put the value in the above equation and solve,
${F_c} = \dfrac{{(3 \times {{10}^{29}} \times {{10}^{ - 3}})(2 \times {{10}^{16}})}}{{1.5 \times {{10}^{12}}}}$;
Here, the unit is in dyne, make sure to convert it into Newton.
${F_c} =$ $8 \times {10^{31}}dyne = 8 \times {10^{26}}N$; $1N = {10^5}dyne$

Final Answer: Gravitational force exerted on the planet by the star is $8 \times {10^{26}}N$.

Additional Information: The centripetal force is not the fundamental force in nature; it is a force which allows an object to move in circular orbit or rotation. Here the centripetal force is the external force that keeps an object in circular motion. The direction of the centripetal force is towards the center of the body which is applying the centripetal force.

Note: Here, the general formula for gravitational force will not work as we have not been given the mass of the Sun in accordance to that we have been given the velocity of the planet. So, here the formula applicable is of the centripetal force which acts as gravitational force.