Certainly!
Newton's Law of Universal Gravitation states that the gravitational force (\(\(F\)\)) between two masses (\(\(m_1\)\) and \(\(m_2\)\)) separated by a distance (\(\(d\)\)) is given by the equation:
\(\[ F = \frac{G m_1 m_2}{d^2} \]\)
where \(\(G\)\) is the gravitational constant.
Let's analyze what happens if we double one of the masses, say \(\(m_1\)\).
1. Initial Gravitational Force:
We start with the initial masses \(\(m_1\)\) and \(\(m_2\)\) separated by a distance \(\(d\)\). The initial gravitational force is:
\(\[ F_{\text{initial}} = \frac{G m_1 m_2}{d^2} \]\)
2. Doubling One of the Masses:
Now, let's double the mass \(\(m_1\)\), so the new mass becomes \(\(2m_1\)\).
3. New Gravitational Force:
With the doubled mass, the new gravitational force becomes:
\(\[ F_{\text{new}} = \frac{G (2m_1) m_2}{d^2} \]\)
Simplifying this, we get:
\(\[ F_{\text{new}} = 2 \cdot \frac{G m_1 m_2}{d^2} \]\)
\(\[ F_{\text{new}} = 2 \cdot F_{\text{initial}} \]\)
The new gravitational force (\(\(F_{\text{new}}\)\)) is exactly twice the initial gravitational force (\(\(F_{\text{initial}}\)\)).
By using the given values and the gravitational constant (\(\(G = 6.67430 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)\)), we see:
- The initial gravitational force is \(\(6.6743 \times 10^{-11}\)\) Newtons.
- When we double the mass \(\(m_1\)\), the new gravitational force becomes \(\(2 \times 6.6743 \times 10^{-11} = 1.33486 \times 10^{-10}\)\) Newtons.
Therefore, doubling one of the masses results in the gravitational force also being doubled.
