Question

what happens to the gravitation force between two objects that are 15 m apart, when one of them moves 3 m closer? (1 point)
it decreases by a factor of ( 1 \frac { 1 } { 4 } ).
it increases by a factor of ( 1 \frac { 1 } { 4 } ).
it decreases by a factor of ( 1 \frac { 9 } { 16 } ).
it increases by a factor of ( 1 \frac { 9 } { 16 } ).

Explanation:

Step1: Recall the gravitational force formula

The gravitational force \(F = \frac{Gm_1m_2}{r^{2}}\), where \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the two objects, and \(r\) is the distance between them.

Step2: Calculate the initial and final distances

The initial distance \(r_1=15\space m\). The final distance \(r_2 = 15 - 3=12\space m\).

Step3: Find the ratio of the forces

Let \(F_1=\frac{Gm_1m_2}{r_1^{2}}\) and \(F_2=\frac{Gm_1m_2}{r_2^{2}}\). Then \(\frac{F_2}{F_1}=\frac{r_1^{2}}{r_2^{2}}\). Substitute \(r_1 = 15\) and \(r_2=12\): \(\frac{F_2}{F_1}=\frac{15^{2}}{12^{2}}=\frac{225}{144}=\frac{25}{16}=1\frac{9}{16}\). Since \(F_2>F_1\) (as the distance decreases, the force increases), the force increases by a factor of \(1\frac{9}{16}\).

Answer:

It increases by a factor of \(1\frac{9}{16}\).