To determine how much tidal force the Sun has on the Earth compared to the Moon, we need to understand the concept of tidal forces. Tidal force is proportional to the mass of the celestial object and inversely proportional to the cube of the distance from the Earth.
1. Mass of the celestial objects:
- The mass of the Sun is significantly larger than the mass of the Moon.
- Mass of the Sun [tex]\( M_s \)[/tex]: [tex]\( 1.989 \times 10^{30} \)[/tex] kg
- Mass of the Moon [tex]\( M_m \)[/tex]: [tex]\( 7.34767309 \times 10^{22} \)[/tex] kg
2. Distance from the Earth:
- The Sun is much farther from the Earth than the Moon.
- Distance to the Sun [tex]\( D_s \)[/tex]: [tex]\( 1.496 \times 10^{11} \)[/tex] meters
- Distance to the Moon [tex]\( D_m \)[/tex]: [tex]\( 3.844 \times 10^{8} \)[/tex] meters
3. Tidal Force Calculation:
- The tidal force ([tex]\( F \)[/tex]) exerted by a celestial object is proportional to its mass divided by the cube of its distance from the Earth.
4. Tidal Force of the Sun ([tex]\( F_s \)[/tex]):
[tex]\[ F_s \propto \frac{M_s}{D_s^3} \][/tex]
5. Tidal Force of the Moon ([tex]\( F_m \)[/tex]):
[tex]\[ F_m \propto \frac{M_m}{D_m^3} \][/tex]
6. Calculate the ratio of the tidal forces:
[tex]\[ \text{Tidal Force Ratio} = \frac{F_s}{F_m} = \frac{\frac{M_s}{D_s^3}}{\frac{M_m}{D_m^3}} \][/tex]
Given all the relevant values:
1. Mass of the Sun [tex]\( M_s = 1.989 \times 10^{30} \)[/tex] kg
2. Distance to the Sun [tex]\( D_s = 1.496 \times 10^{11} \)[/tex] meters
3. Mass of the Moon [tex]\( M_m = 7.34767309 \times 10^{22} \)[/tex] kg
4. Distance to the Moon [tex]\( D_m = 3.844 \times 10^{8} \)[/tex] meters
After performing the calculations, it is determined that the ratio of the Sun's tidal force to the Moon's tidal force is approximately 0.46.
This means the Sun's tidal force on Earth is less than that of the Moon. Specifically, the Sun's tidal force is about 0.46 times (or less than half) that of the Moon's tidal force.
So, the correct answer is:
- 1/2 as much
