In Newton's law of universal gravitation, if the distance between two point masses is doubled while the masses remain the same, how does the gravitational force change?

• A. It becomes one-half as large: $F^{\prime }=\frac{F}{2}$.
• B. It becomes twice as large: $F^{\prime }=2F$.
• C. It becomes four times as large: $F^{\prime }=4F$.
• D. It becomes one-fourth as large: $F^{\prime }=\frac{F}{4}$.

Answer: (D)

Verified step by step guidance

1. Recall Newton's law of universal gravitation, which states that the gravitational force \(F\) between two point masses \(m_1\) and \(m_2\) separated by a distance \(r\) is given by the formula: \[F = G \frac{m_1 m_2}{r^2}\] where \(G\) is the gravitational constant.
2. Identify the initial distance between the masses as \(r\) and the initial force as \(F\). When the distance is doubled, the new distance becomes \(r' = 2r\).
3. Substitute the new distance into the gravitational force formula to find the new force \(F'\): \[F' = G \frac{m_1 m_2}{(2r)^2}\]
4. Simplify the denominator: \[(2r)^2 = 4r^2\] So the new force becomes: \[F' = G \frac{m_1 m_2}{4r^2}\]
5. Compare the new force \(F'\) to the original force \(F\): \[F' = \frac{1}{4} \times G \frac{m_1 m_2}{r^2} = \frac{1}{4} F\] This shows that doubling the distance reduces the gravitational force to one-fourth of its original value.