Questions: How many times greater is the force exerted by the sun on the moon then the force exerted by the earth on the moon?

Solution Steps

We need to compare the gravitational force exerted by the Sun on the Moon with the gravitational force exerted by the Earth on the Moon.

The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by: \[ F = \frac{G m_1 m_2}{r^2} \] where \( G \) is the gravitational constant.

Let:

• \( M_s \) be the mass of the Sun
• \( M_m \) be the mass of the Moon
• \( d_{sm} \) be the distance between the Sun and the Moon

The force exerted by the Sun on the Moon is: \[ F_{sm} = \frac{G M_s M_m}{d_{sm}^2} \]

Let:

• \( M_e \) be the mass of the Earth
• \( d_{em} \) be the distance between the Earth and the Moon

The force exerted by the Earth on the Moon is: \[ F_{em} = \frac{G M_e M_m}{d_{em}^2} \]

To find how many times greater the force exerted by the Sun on the Moon is compared to the force exerted by the Earth on the Moon, we take the ratio: \[ \text{Ratio} = \frac{F_{sm}}{F_{em}} = \frac{\frac{G M_s M_m}{d_{sm}^2}}{\frac{G M_e M_m}{d_{em}^2}} = \frac{M_s}{M_e} \cdot \frac{d_{em}^2}{d_{sm}^2} \]

Using approximate values:

• \( M_s \approx 1.989 \times 10^{30} \) kg
• \( M_e \approx 5.972 \times 10^{24} \) kg
• \( d_{sm} \approx 1.496 \times 10^{11} \) m (average distance from the Sun to the Earth)
• \( d_{em} \approx 3.844 \times 10^{8} \) m (average distance from the Earth to the Moon)

\[ \text{Ratio} = \frac{1.989 \times 10^{30}}{5.972 \times 10^{24}} \cdot \frac{(3.844 \times 10^{8})^2}{(1.496 \times 10^{11})^2} \]

\[ \text{Ratio} = \frac{1.989 \times 10^{30}}{5.972 \times 10^{24}} \cdot \frac{1.477 \times 10^{17}}{2.238 \times 10^{22}} \] \[ \text{Ratio} = \frac{1.989}{5.972} \cdot \frac{1.477}{2.238} \times 10^{(30-24+17-22)} \] \[ \text{Ratio} = 0.333 \cdot 0.66 \times 10^1 \] \[ \text{Ratio} \approx 0.22 \times 10^1 \] \[ \text{Ratio} \approx 2.2 \]

Final Answer