Question

If the distance between two objects is increased two times, then by how many times will the mass of one of the objects be changed to maintain the same gravitational force?

Answer 1

Hint: We should first write the expression for the gravitational force at a certain distance r then at two times this distance keep the mass of one of the objects as variable and other fixed. Then by equating the forces in the two cases, we can get the required value of increase in mass required to keep the force constant.

Complete step by step answer:
Consider two masses ${m_1}$ and ${m_2}$ which are at a distance r from each other initially. Then the gravitational force between them is given as
$F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$ …(i)
Now the distance between these two objects is increased two times but we need to keep the magnitude of force between the objects same as before. Let us keep the mass of the first object constant and we will vary the mass of the second object. Let the mass of the second object be ${m_2}'$. Then we can write the gravitational force as follows:
$F = G\dfrac{{{m_1}{m_2}'}}{{{{\left( {2r} \right)}^2}}} = G\dfrac{{{m_1}{m_2}'}}{{4{r^2}}}$ …(ii)
Now we can equate the equations (i) and (ii), doing so, we get
$
  G\dfrac{{{m_1}{m_2}}}{{{r^2}}} = G\dfrac{{{m_1}{m_2}'}}{{4{r^2}}} \\
  {m_2}' = 4{m_2} \\
 $
Therefore, the mass of one of the objects should be increased four times to increase the distance between the objects two times in order to keep the value of gravitational force the same.

Additional information:
Newton’s law of gravitation says that two bodies which have mass exert an attractive force on each other. This force is called gravitational force or gravity. According the Newton, mathematically this force is
Directly proportional to the masses of the two bodies under consideration
$F \propto {m_1}{m_2}{\text{ }}...{\text{(i)}}$
where ${m_1}$and ${m_2}$ are the masses of the two bodies.
Inversely proportional to the square of the distance by which they are separated from each other
$F \propto \dfrac{1}{{{r^2}}}{\text{ }}...{\text{(ii)}}$
where r signifies the separation distance between the two bodies.
This means that greater the mass of the bodies, greater is the gravitational force between the two bodies and if we increase the distance between the two bodies, then the gravitational force between them will also decrease and vice-versa.
Combining equation (i) and (ii), we get
$
  F \propto \dfrac{{{m_1}{m_2}}}{{{r^2}}}{\text{ }} \\
   \Rightarrow F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}{\text{ }} \\
 $
where G is the constant of proportionality called the universal gravitational constant. Its value is given as
$G = 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$

Note:
It should be noted that the $\dfrac{1}{{{r^2}}}$ dependence in the expression for gravitational force is known as the inverse square dependence. The consequence of this dependence is that with increase in distance between two objects, the gravitational force between them decreases non-linearly as the square of the distance between them.