In Newton's law of universal gravitation, how does the magnitude of the gravitational force between two point masses change when the distance between them is doubled (masses unchanged)?

• A. It remains the same because distance does not appear in $F=GmM$.
• B. It becomes one-half as large because $F\propto \frac{1}{r}$.
• C. It becomes twice as large because $F\propto r$.
• D. It becomes one-fourth as large because $F\propto \frac{1}{r^2}$.

Answer: (D)

Verified step by step guidance

1. Recall Newton's law of universal gravitation, which states that the magnitude of the gravitational force \(F\) between two point masses \(m\) and \(M\) separated by a distance \(r\) is given by the formula: \[F = G \frac{mM}{r^2}\] where \(G\) is the gravitational constant.
2. Notice that the force \(F\) is inversely proportional to the square of the distance \(r\) between the two masses. This means if the distance changes, the force changes according to the factor \(\frac{1}{r^2}\).
3. If the distance \(r\) is doubled, then the new distance becomes \$2r\(. Substitute this into the formula to find the new force \)F_{new}$: \[F_{new} = G \frac{mM}{(2r)^2}\]
4. Simplify the denominator: \[(2r)^2 = 4r^2\] So the new force becomes: \[F_{new} = G \frac{mM}{4r^2} = \frac{1}{4} \times G \frac{mM}{r^2} = \frac{1}{4} F\]
5. Therefore, when the distance between the two masses is doubled, the gravitational force becomes one-fourth as large, illustrating the inverse square relationship between force and distance.