Calculate the magnitudes of the gravitational forces exerted on the Moon by the Sun and by the Earth when the two forces are in direct competition, that is, when the Sun, Moon and Earth are aligned with the Moon between the Sun and the Earth. This alignment corresponds to a solar eclipse.) Does the orbit of the Moon ever actually curve away from the Sun, toward the Earth? (Please give your answer to three significant figures.)
I used G m1 m2/r^2 to get the forces, for the force between the Earth and the moon I did:
((6.67e-11)(5.97e24)(7.35e22))/(1.737e6 - 6.37e6)^2
and I got 1.36e24 N
To get the r value I substracted the radii... do I have to add them?
For the force between the Sun and the Moon, I used the distance from earth to the sun and substracted it from the distance between moon to earth.
((1.737e6 - 6.37e6)- (6.96e8 -6.37e6))=6.95e8 m
and then for the moon to sun force I computed:
((6.67e-11)(7.35e22)(1.9891e30))/(1.737e6 - 6.37e6)^2
and I got 1.40e34 N...
These answers are not right but I'm not sure what I'm doing wrong...
To calculate the magnitudes of the gravitational forces exerted on the Moon by the Sun and the Earth, we need to use the formula for gravitational force:
F = G * (m1 * m2) / r^2
Where:
- F is the magnitude of the gravitational force
- G is the gravitational constant (6.67e-11 N m^2/kg^2)
- m1 and m2 are the masses of the two objects
- r is the distance between the centers of the two objects
Let's start by calculating the force between the Earth and the Moon. The distance between the Moon and the Earth is the sum of their radii:
r = (1.737e6 m + 6.37e6 m) = 8.107e6 m
The mass of the Earth is 5.97e24 kg and the mass of the Moon is 7.35e22 kg. Substituting these values into the formula:
Fearth-moon = (6.67e-11 N m^2/kg^2) * (5.97e24 kg * 7.35e22 kg) / (8.107e6 m)^2 = 1.981e20 N (rounded to 3 significant figures)
For the force between the Sun and the Moon, we need to calculate the distance between the Moon and the Sun. Since the Moon is aligned with the Earth and the Sun, and the Sun is much farther away from both of them, we can approximate this distance by subtracting the distance between the Earth and the Sun from the distance between the Moon and the Earth:
r = (1.737e6 m - 6.37e6 m) - (1.496e11 m - 6.37e6 m) = -1.498e11 m
The negative sign indicates that the Moon is on the opposite side of the Earth compared to the Sun. However, we are only interested in the magnitude of the force, so we can ignore the negative sign:
Fsun-moon = (6.67e-11 N m^2/kg^2) * (1.9891e30 kg * 7.35e22 kg) / (-1.498e11 m)^2 = 5.732e17 N (rounded to 3 significant figures)
Therefore, the magnitude of the gravitational force exerted by the Earth on the Moon is 1.981e20 N, and the magnitude of the gravitational force exerted by the Sun on the Moon is 5.732e17 N.
Now, regarding the question of whether the orbit of the Moon ever actually curves away from the Sun, toward the Earth, the answer is no. The Moon's orbit is always curved towards the center of mass of the system, which is located inside the Earth but not exactly at its center. This means that the Moon orbits around the Earth and the Earth-Moon system orbits around the Sun, but there is no scenario where the Moon's orbit curves away from the Sun towards the Earth.