Understanding Gravitational Force and Distance

The question asks how the gravitational force between the Earth and the Moon changes if the distance between them is reduced to half of its original value. To answer this, we need to use Newton's Law of Universal Gravitation.

Newton's Law of Universal Gravitation

Newton's law 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 ($F$) between two objects with masses $m_1$ and $m_2$ separated by a distance $r$ is given by:

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

Where:

• \( F \) is the gravitational force
• \( G \) is the gravitational constant
• \( m_1 \) is the mass of the first object (e.g., Earth)
• \( m_2 \) is the mass of the second object (e.g., Moon)
• \( r \) is the distance between the centers of the two objects

Analyzing the Effect of Changing Distance

In this problem, the masses of the Earth ($m_1$) and the Moon ($m_2$) remain constant, and the gravitational constant ($G$) is also constant. The only thing changing is the distance between them. The original distance is \( r \). The new distance is half of the original distance, which we can call \( r' \).

So, \( r' = \frac{r}{2} \).

Let the original gravitational force be \( F_{original} \). Using the formula:

\[ F_{original} = G \frac{m_1 m_2}{r^2} \]

Now, let the new gravitational force when the distance is halved be \( F_{new} \). We replace \( r \) with \( r' = \frac{r}{2} \) in the formula:

\[ F_{new} = G \frac{m_1 m_2}{(r')^2} \]

Substitute \( r' = \frac{r}{2} \) into the equation for \( F_{new} \):

\[ F_{new} = G \frac{m_1 m_2}{\left(\frac{r}{2}\right)^2} \]

Simplify the denominator:

\[ \left(\frac{r}{2}\right)^2 = \frac{r^2}{2^2} = \frac{r^2}{4} \]

So, the expression for \( F_{new} \) becomes:

\[ F_{new} = G \frac{m_1 m_2}{\frac{r^2}{4}} \]

To divide by a fraction, we multiply by its reciprocal:

\[ F_{new} = G \frac{m_1 m_2}{1} \times \frac{4}{r^2} \]

Rearranging the terms:

\[ F_{new} = 4 \times G \frac{m_1 m_2}{r^2} \]

Notice that the term \( G \frac{m_1 m_2}{r^2} \) is the original gravitational force, \( F_{original} \). Therefore:

\[ F_{new} = 4 \times F_{original} \]

This shows that when the distance between the Earth and the Moon is halved, the gravitational force of attraction between them increases to four times its original value.

Comparing with Options

Let's look at the options again:

• reduce to half of its original value. (Incorrect, the force increases)
• increase to two times of its original value. (Incorrect, it increases by a factor of 4)
• remain the same as the original value. (Incorrect, changing distance changes the force)
• increase to four times of its original value. (Correct, as calculated above)

The calculation clearly shows that the gravitational force increases to four times its original value.

Revision Table: Gravitational Force vs. Distance

Additional Information: Inverse Square Law

The relationship between gravitational force and distance is an example of an inverse square law. This means that the magnitude of a physical quantity (like force or intensity) is inversely proportional to the square of the distance from the source of that quantity.

Other examples of inverse square laws in physics include:

• The intensity of light from a point source.
• The intensity of sound from a point source.
• The electric field strength around a point charge (Coulomb's Law).

In the case of gravity, doubling the distance reduces the force to one-fourth ($1/2^2$). Tripling the distance reduces the force to one-ninth ($1/3^2$). Conversely, halving the distance increases the force by a factor of four ($1/(1/2)^2 = 4$). This inverse square relationship is fundamental in understanding many physical phenomena that spread out from a source in three dimensions.