Understanding how the gravitational force of attraction between two masses changes with distance is fundamental in physics. This concept is explained by Newton's Law of Universal Gravitation.

Gravitational Force Explained

The gravitational force is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another. It is the force that gives weight to physical objects and causes them to fall to the ground when dropped.

Force of Universal Gravitation Explained

Sir Isaac Newton formulated the law which states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula for gravitational force ($F$) is:

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

Where:

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

Distance and Force Relationship

From the formula, we can see that the gravitational force ($F$) is inversely proportional to the square of the distance ($r$) between the two masses. This is often referred to as an "inverse-square law".

$$F \propto \frac{1}{r^2}$$

This means if the distance changes, the force changes by the square of the inverse of that change.

Force Change Calculation with Doubled Distance

Let's consider the initial scenario and then what happens when the distance is doubled.

• Initial State:

  Let the initial distance between the two masses be $r_1$.

  The initial gravitational force ($F_1$) is:

  $$F_1 = G \frac{m_1 m_2}{r_1^2}$$

• When Distance is Doubled:

  If the distance between the masses is doubled, the new distance ($r_2$) will be $2 \times r_1 = 2r_1$.

  Now, let's calculate the new gravitational force ($F_2$) using the new distance:

  $$F_2 = G \frac{m_1 m_2}{r_2^2} = G \frac{m_1 m_2}{(2r_1)^2}$$

  Simplifying the denominator:

  $$F_2 = G \frac{m_1 m_2}{4r_1^2}$$

  We can rewrite this expression by separating the constant factors:

  $$F_2 = \frac{1}{4} \left( G \frac{m_1 m_2}{r_1^2} \right)$$

  Since $F_1 = G \frac{m_1 m_2}{r_1^2}$, we can substitute $F_1$ into the equation for $F_2$:

  $$F_2 = \frac{1}{4} F_1$$

Gravitational Force Summary

This calculation shows that when the distance between the two masses is doubled, the new gravitational force of attraction becomes one-quarter of the original force. In simpler terms, the force is significantly reduced due to the inverse square relationship with distance.

Therefore, if the distance between the two masses is doubled, the gravitational force of attraction between them is reduced to a quarter.