Two objects in deep space have a gravitational force between them. If one of the objects suddenly

doubles in mass and the distance between the two objects drops to half of its original, by how much is
the magnitude of the gravitational force changed?

Increased by 4 times

Decreased by 2 times

Decreased by 4 times

Increased by 8 times

1 answer

The gravitational force between two objects is described by Newton's law of gravitation, which can be expressed as:

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

where:

  • \( F \) is the gravitational force,
  • \( G \) is the gravitational constant,
  • \( m_1 \) and \( m_2 \) are the masses of the two objects,
  • \( r \) is the distance between the centers of the two objects.

Let's denote the initial masses as \( m_1 \) and \( m_2 \), and the initial distance as \( r \).

Initially, the gravitational force is:

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

Now, the first object doubles its mass, so its new mass is \( 2m_1 \). The distance between the objects is also halved, so the new distance is \( \frac{r}{2} \).

The new gravitational force can be calculated as follows:

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

Now, simplifying the expression for \( F_{new} \):

\[ F_{new} = \frac{G \cdot (2m_1) \cdot m_2}{\frac{r^2}{4}} = \frac{G \cdot (2m_1) \cdot m_2 \cdot 4}{r^2} = \frac{8G \cdot m_1 \cdot m_2}{r^2} \]

Thus, the new force \( F_{new} \) is:

\[ F_{new} = 8 \cdot \frac{G \cdot m_1 \cdot m_2}{r^2} = 8 \cdot F_{initial} \]

This means the gravitational force increases by a factor of 8.

Therefore, the answer is:

Increased by 8 times.