Gravitational Force Calculation

Newton's Law of Universal Gravitation states that the force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The formula is given by:

$F = G \frac{m_1 m_2}{r^2}$

Where:

• $F$ is the gravitational force.
• $G$ is the gravitational constant.
• $m_1$ and $m_2$ are the masses of the two bodies.
• $r$ is the distance between the centers of the two bodies.

Solving Force Change with Doubled Distance

Let the initial force between the Earth and the Moon be $F_1$, the mass of the Earth be $M_E$, the mass of the Moon be $M_M$, and the initial distance between them be $r_1$.

According to the question, $F_1 = F$. So, we have:

$F = G \frac{M_E M_M}{r_1^2} \quad (1)$

Now, consider the case where the distance between the Earth and the Moon is doubled. The new distance, $r_2$, will be $2r_1$. All other parameters (masses and G) remain the same.

The new force, $F_2$, will be:

$F_2 = G \frac{M_E M_M}{r_2^2}$

Substitute $r_2 = 2r_1$ into the equation:

$F_2 = G \frac{M_E M_M}{(2r_1)^2}$

$F_2 = G \frac{M_E M_M}{4r_1^2}$

We can rewrite this expression by factoring out $\frac{1}{4}$:

$F_2 = \frac{1}{4} \left( G \frac{M_E M_M}{r_1^2} \right)$

From equation (1), we know that $G \frac{M_E M_M}{r_1^2} = F$. Substituting this back:

$F_2 = \frac{1}{4} F$

Therefore, if the distance between the Earth and the Moon is doubled, the force of gravitation between them will be $F/4$.