Question

Newton's law of universal gravitation describes the attractive force F LaTeX: F\:=\frac{Gm_1m_2}{r^2}F = G m 1 m 2 r 2 between point masses m1 and m2 separated by distance r. Note that due to Newton's 3rd law, both masses experience this force. What would the value of the gravitational force be in terms of F if the mass of m1 were doubled?

Answer 1

Answer:

2F

Explanation:

Given;

$F=\frac{Gm_1m_2}{r^2}................(1)$

G is is the universal gravitational constant. Since G is a constant, we could rewrite equation (1) as follows;

$\frac{Fr^2}{m_1m_2}=constant (G)................(2)$

When one of the masses is altered, the force changes to a new value, say $F_2$.

In this case it is said that $m_1$ is doubled.

Equation (2) therefore implies that we can write the following;

$\frac{Fr^2}{m_1m_2}=\frac{F_2r^2}{2m_1m_2}...........(3)$

We are assuming that all other parameters a parts from $m_1$ remain the same.

By appropriately cancelling out parameters from both side of the equation (3), we obtain the following;

$F=\frac{F_2}{2}$

The new gravitational force in terms of F is therefore'

$F_2=2F$