Answered

If the distance between two objects is cut in half, what happens to the
gravitational force between them?
A. It decreases to 1/2 its original magnitude.
B. It increases to 2 times its original magnitude.
C. It increases to 4 times its original magnitude.
D. It decreases to 1/4 its original magnitude.
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Verified Answer :

GinnyAnswer

According to Newton's law of universal gravitation, the gravitational force \(\( F \)\) between two masses \(\( m_1 \)\) and \(\( m_2 \)\) is inversely proportional to the square of the distance \(\( R \)\) between them. This relationship is given by the formula:

\(\[ F = G \frac{{m_1 \cdot m_2}}{{R^2}} \]\)

where \(\( G \)\) is the gravitational constant. Now let's analyze what happens when the distance \(\( R \)\) is cut in half.

1. Original Distance \(\( R \)\):

\(\[ F = G \frac{{m_1 \cdot m_2}}{{R^2}} \]\)

2. New Distance \(\( R/2 \)\):

When we halve the distance, the new distance becomes \(\( R' = R/2 \)\). We substitute \(\( R/2 \)\) into the gravitational force formula:

\(\[ F' = G \frac{{m_1 \cdot m_2}}{{(R/2)^2}} \]\)

3. Simplifying the Expression:

\(\[ F' = G \frac{{m_1 \cdot m_2}}{{R^2 / 4}} \]\)
\(\[ F' = G \frac{{m_1 \cdot m_2}}{{(R^2 / 4)}} = G \frac{{m_1 \cdot m_2}}{{R^2}} \cdot 4 \]\)
\(\[ F' = 4 \left( G \frac{{m_1 \cdot m_2}}{{R^2}} \right) \]\)
\(\[ F' = 4F \]\)

So, when the distance between the two objects is cut in half, the new gravitational force \(\( F' \)\) becomes four times the original gravitational force \(\( F \)\).

Therefore, the correct answer is:

C. It increases to 4 times its original magnitude.

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