# Net Force acting on Earth

by laxanimal
Tags: acting, earth, force
 P: 3 Hi all, I've been working on this problem for a long while and just can't get the answer. When the Earth, Moon, and Sun form a right triangle with the Earth located at the right angle, the Moon is approaching its third quarter. (The Earth is viewed here from above its north pole.) Find the magnitude and direction of the net force exerted on the Earth. I found the magnitude to be 3.52e22 N but can't figure out what the direction is (been using tan**-1(force exerted by moon/force exerted by sun). Thanks in advance.
 HW Helper P: 661 Independently find the gravitational forces acting on the Earth due to the Sun and Moon. Then add the vectors together, which is in effect a pythagoreas in magnitude because of the right triangle geometry.
 P: 3 Thanks, I found the magnitude, but can't find the direction (the angle clockwise from the ray from earth to sun) any suggestions?
P: 882

## Net Force acting on Earth

This is a gravity question I assume, where the forces acting on the Earth are due to the forces of gravitational attraction between the Earth and the Sun/moon.

To find the resultant force acting on the Earth, you found the force caused by the moon, then force caused by the sun. These two forces are perpendicular to eachother along the sides of right triangle, so you are able to calculate the resultant using the pythagorean theorem.
But there is another nifty thing you can employ since it is a right triangle, you know that the tangent of the angle is equal to the opposite "leg" of force / the adjacent "leg" of force.
or that,
Sin (angle) = opp / resultant, or