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

Hint: 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,
\[{F_1} = \dfrac{{G2{m_1}{m_2}}}{{{r^2}}} \\
\Rightarrow {F_1} = 2F\]
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.