Posted by Bon Oncle on .
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?
Physics (please check!!!) - drwls, Friday, April 29, 2011 at 1:42am
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.
I don't get what I have to do... Do I have to add the radii instead??