Question

What happens to the force between two objects, if $\left( {\text{i}} \right)$ The mass of one object is doubled? $\left( {{\text{ii}}} \right)$ The distance between the objects is doubled and tripled? $\left( {{\text{iii}}} \right)$ The masses of both the object are doubled?

Answer 1

To solve this question, use Newton’s law of gravitation, according to which “every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.”

Formula used:
The magnitude of Gravitational force $F$ between two particles ${m_1}$ and ${m_2}$ placed at a distance $r$ is given by,
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Where $G$is the universal constant called the Gravitational constant.
$G = 6.67 \times {10^{ - 11}}{\text{ N - m/k}}{{\text{g}}^2}$

Complete step by step answer:
Let the mass of the two objects be ${m_1}$ and ${m_2}$ respectively and the distance between them be $r$. Thus, the force between these two objects will be,
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$

$\left( {\text{i}} \right)$ Given, the mass of one of the objects is doubled. Thus, the new mass of the object is $2{m_1}$. Since the mass of the object changes the force between them changes as it is dependent on mass. Thus, the force becomes,

\Rightarrow {F_1} = 2F$$ Hence, if mass of one of the objects is doubled force doubles $$\left( {{\text{ii}}} \right)$$ Given, the distance between the objects is doubled. Thus, the distance between the objects becomes $$2r$$. So, the force between the object becomes, $${F_2} = \dfrac{{G{m_1}{m_2}}}{{{{\left( {2r} \right)}^2}}} \\\ \Rightarrow {F_2} = \dfrac{{G{m_1}{m_2}}}{{4{r^2}}} \\\ \Rightarrow {F_2} = \dfrac{F}{4}$$ Therefore, if the distance is doubled the force between the objects becomes one-fourth. Given, now the distance is tripled. Thus, the distance between the objects becomes $$3r$$.So, the force between the object becomes, $${F_3} = \dfrac{{G{m_1}{m_2}}}{{{{\left( {3r} \right)}^2}}} \\\ \Rightarrow {F_3} = \dfrac{{G{m_1}{m_2}}}{{9{r^2}}} \\\ \Rightarrow {F_3}= \dfrac{F}{9}$$ Thus, if the distance is tripled the force between the objects becomes one-ninth. $$\left( {{\text{iii}}} \right)$$ Given, masses of both objects are doubled. The new masses are $$2{m_1}$$ and $$2{m_2}$$. So, the force between the object becomes, $${F_4} = \dfrac{{G2{m_1}2{m_2}}}{{{r^2}}} \\\ \Rightarrow {F_4} = \dfrac{{4G{m_1}{m_2}}}{{{r^2}}} \\\ \therefore {F_4} = 4F$$ **Thus, if the masses of the object are doubled the force between the object quadruples.** **Note:** Unlike the electrostatic force, Gravitational force is independent of the medium between the particles. It is conservative in nature. It expresses the force between two-point masses (of negligible volume). However, for external points of spherical bodies, the whole mass can be assumed to be concentrated at its center of mass.