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
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...
Does these look right?
I don't understand how you are coming up with your r values. They should be center-to-center. You seem to be subtracting radii of the bodies themselves, to get a surface-to-surface distance.
For the sun-moon distance in a solar eclipse configuration, subtract the earth-moon distance from the earth-sun distance. You say it the other way around, but since it gets squyared, it doesn't matter.
The answer to your last question is "yes". The moon moves in and out compared to the orbit of the earth around the sun, sometimes moving away from the sun. Imagine a circle with 12 waves in it, in and out.
To calculate the magnitudes of the gravitational forces exerted on the Moon by the Sun and by the Earth, you used the formula for gravitational force: F = G * (m1 * m2) / r^2. The formula is correct, but there are some errors in your calculations.
Let's first calculate the force between the Earth and the Moon. The mass of the Earth is approximately 5.97 x 10^24 kg, and the mass of the Moon is approximately 7.35 x 10^22 kg. The average distance between the Earth and the Moon is approximately 3.84 x 10^8 meters.
Using the formula, the force between the Earth and the Moon is:
F_earth-moon = (6.67 x 10^-11 N*m^2/kg^2 * (5.97 x 10^24 kg) * (7.35 x 10^22 kg)) / (3.84 x 10^8 m)^2
Calculating this, we get:
F_earth-moon = 1.985 x 10^20 N (to three significant figures)
Now let's calculate the force between the Sun and the Moon when they are in direct alignment with the Earth. For simplicity, we will assume that the distance between the Moon and the Earth is equal to the average distance mentioned earlier.
The average distance between the Sun and the Earth is approximately 1.496 x 10^11 meters. So, the distance between the Moon and the Sun, when they are aligned with the Earth, is:
distance_moon-sun = (1.496 x 10^11 m) - (3.84 x 10^8 m)
Calculating this, we get:
distance_moon-sun = 1.492 x 10^11 meters
Now, let's calculate the force between the Sun and the Moon:
F_sun-moon = (6.67 x 10^-11 N*m^2/kg^2 * (7.35 x 10^22 kg) * (1.9891 x 10^30 kg)) / (1.492 x 10^11 m)^2
Calculating this, we get:
F_sun-moon = 3.503 x 10^20 N (to three significant figures)
Therefore, the magnitude of the gravitational force exerted on the Moon by the Sun is approximately 3.503 x 10^20 N, and the magnitude of the gravitational force exerted on the Moon by the Earth is approximately 1.985 x 10^20 N.
Now, let's address the second part of your question. The orbit of the Moon does not actually curve away from the Sun towards the Earth. The Moon orbits in an elliptical path around the Earth, and the distance between the Moon and the Earth can vary throughout its orbit. However, the gravitational force exerted by the Sun on the Moon is always pulling it towards the Sun, causing its orbital path to be influenced by both the Earth and the Sun.
Therefore, the statement "the orbit of the Moon never actually curves away from the Sun towards the Earth" is true.