Question

The distance of Pluto from the sun is $40$ times the distance of earth, if the masses of earth and Pluto are equal, what will be the ratio of gravitational forces of sun on these planets?

Answer 1

Hint: To solve this question, we need to use the expression for Newton's law of gravitation for the earth and Pluto. On dividing the two expressions, we will get the required ratio.
Formula used: The formula used to solve this question is given by
${F_G} = \dfrac{{GMm}}{{{r^2}}}$, here ${F_G}$ is the gravitational force exerted by a body having a mass of $M$ on another body of mass $m$, separated by a distance of $r$ from it.

Complete step-by-step solution:
Let the mass of the sun be equal to ${M_S}$.
According to the question, the masses of earth and Pluto are equal. So $m$ be the mass of each of them.
Also, it is given that the distance of Pluto from the sun is $40$ times the distance of earth. So if the distance of the earth from the sun is equal to $d$, then the distance of Pluto from the sun becomes equal to $40d$.
From the Newton’s law of gravitation, the gravitational force is given by
${F_G} = \dfrac{{GMm}}{{{r^2}}}$....................(1)
For the force exerted on the earth due to the Sun, we substitute $M = {M_S}$ and $r = d$ in (1) to get
${F_E} = \dfrac{{G{M_S}m}}{{{d^2}}}$....................(2)
Also, for the force exerted on Pluto due to the Sun, we substitute $M = {M_S}$ and \[r = 40d\] in (1) to get
${F_P} = \dfrac{{G{M_S}m}}{{{{\left( {40d} \right)}^2}}}$
 \[ \Rightarrow {F_P} = \dfrac{{G{M_S}m}}{{1600{d^2}}}\] (3)
Dividing (2) by (3) we get
$\dfrac{{{F_E}}}{{{F_P}}} = \dfrac{{\dfrac{{G{M_S}m}}{{{d^2}}}}}{{\dfrac{{G{M_S}m}}{{1600{d^2}}}}}$
On simplifying, we get
$\dfrac{{{F_E}}}{{{F_P}}} = 1600$

Hence, the ratio of the gravitational force exerted on the earth to that on Pluto is equal to $1600:1$.

Note: Since the universal constant of gravity, the mass of the sun, and the mass of earth and Pluto are the same, we could also use the proportionality of the distance on the gravitational force due to the sun. Using this, we need not write the gravitational force on each planet and obtain the ratio directly.