Question

What happens to the gravitational force between two objects, if
(i) The mass of one object is doubled?
(ii) The distance between the objects is doubled and tripled?
(iii) The masses of both objects are doubled?

Answer 1

Gravitational force can be defined as an attractive force that occurs between all objects that have mass, an object which has mass attracts another object with mass, the amount of force is directly proportional to the masses of the two objects and inversely proportional to the square of the distance between them. Hence by using the mathematical equation If we can double the mass and the distance according to the conditions given in the question we can find the solution.

Complete answer:
The Gravitational force is given by the relation
$F = \dfrac{{G{m_1}{m_2}}}{{{d^2}}}$Where G is the universal gravitational constant $G = 6.67 \times {10^{ - 11}}N{m^2}/k{g^2}$ (which means it has the same value throughout the universe), ${m_{1\;}}$and ${m_2}\;$are the masses of the objects in kgs, and $d$ is the distance between them in $m$
For the first case let us consider the mass of the object$\;1$ is doubled ${m_1} = 2{m_1}$
Which gives us ${F^1} = \dfrac{{G \times 2{m_1} \times {m_2}}}{{{d^2}}}$
Therefore ${F^1} = 2\dfrac{{G{m_1}{m_2}}}{{{d^2}}}$ which implies ${F^1} = 2F$
Thus gravitational force will also get doubled.
For the second case let us consider the distance between the objects is doubled and tripled
If the distance between the objects is doubled i.e. $d\prime = 2d$
Therefore ${F^1} = \dfrac{{G \times {m_1} \times {m_2}}}{{4{d^2}}}$ which implies ${F^1} = \dfrac{1}{4}F$
Thus gravitational force will be reduced by $4$ time.
If the distance between the objects is tripled i.e. $d\prime = 3d$
${F^1} = \dfrac{{G \times {m_1} \times {m_2}}}{{9{d^2}}}$ Which implies ${F^1} = \dfrac{1}{9}F$
Thus gravitational force will be reduced by $9$ time.
For the third case let us consider that both the masses are doubled
If the mass of both the objects are doubled i.e. ${m_1} = 2{m_1}$​ And ${m_2} = 2{m_2}$
Therefore we get ${F^1} = \dfrac{{G \times 2{m_1} \times 2{m_2}}}{{{d^2}}} = \dfrac{{4G{m_1}{m_2}}}{{{d^2}}}$ which implies ${F^1} = 4F$
Thus gravitational force will become $4$ time.

Note:
The scientist Cavendish found the value of the universal gravitation constant, which helps us to know the weight of the Earth. As we evaluate this equation we can see that as the masses increase, the gravitational force also increases. But as the distance between them increases the Force decreases. Because the force is directly proportional to $\dfrac{1}{{{d^2}}}$. Another thing to note is the distances are dependent on the center of gravity and so even if we are standing on the earth, we are from a distance of $R = 6400{\text{ }}km$or $3980{\text{ }}m\;$ from the earth’s center.