The gravitational force between two masses, P and Q, is 10N, find the new value of the force if both masses are doubled

Answer Details

Newton's law of universal gravitation states that the attractive force between two point masses is

\[F = \frac{G m_1 m_2}{r^{2}},\]

where \(G\) is the universal gravitational constant and \(r\) is the distance between their centres. The force is therefore directly proportional to the product of the two masses, and this question asks only how that product changes.

Doubling each mass replaces \(m_1 m_2\) by \((2m_1)(2m_2) = 4m_1 m_2\), while \(r\) is unchanged. Writing the new force as \(F_2\) and dividing one expression by the other lets \(G\) and \(r\) cancel:

\[\frac{F_2}{F_1} = \frac{(2m_1)(2m_2)}{m_1 m_2} = 4, \qquad F_2 = 4 \times 10 = 40\,\mathrm{N}.\]

The mistake to guard against is doubling the force to \(20\,\mathrm{N}\), which comes from doubling only one mass, or from treating the force as proportional to the sum of the masses rather than their product. A second useful habit for this formula is to keep the two dependences separate: the force scales with each mass to the first power but with distance to the power \(-2\). So if the masses were doubled and the separation also doubled, the factor would be \(4 \times \tfrac{1}{4} = 1\) and the force would stay at \(10\,\mathrm{N}\). Setting up the ratio \(F_2/F_1\) rather than trying to find \(G\) or the actual masses is always the fastest and safest method in these proportionality questions.