When the Earth, Moon, and Sun form a right triangle, with...

When the Earth, Moon, and Sun form a right triangle, with the Moon located at the right angle, as shown in FIGURE $12 \cdot 34,$ the Moon is in its third-quarter phase. (The Earth is viewed here from above its North Pole.) Find the magnitude and direction of the net force exerted on the Moon. Give the direction relative to the line connecting the Moon and the Sun.

Key Concepts

Right Triangle Geometry in Force Analysis

When celestial bodies form a right triangle, specific relationships between their distances and angles simplify the analysis of vector directions and magnitudes. Understanding this geometry helps in decomposing the gravitational forces into perpendicular components and then recombining them to determine the net magnitude and the direction relative to a specified line, such as the line connecting the Moon and the Sun.

Vector Addition of Forces

Gravitational forces are vector quantities, meaning they have both magnitude and direction. In problems where multiple gravitational forces act on a single object, these forces must be added vectorially. This involves using geometric or trigonometric methods, such as resolving forces into components or applying the parallelogram law, to find the resultant net force.

Newton’s Law of Universal Gravitation

This law states that every two masses in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. It provides the mathematical framework to calculate the gravitational pull exerted by celestial bodies on each other, which is fundamental when determining the forces acting on the Moon from both the Earth and the Sun.

Transcript

00:01 When the moon is in its third quarter phase, the earth, moon, and sun form a right triangle as shown.

00:07 The problem wants us to find the magnitude and direction of the net force exerted on the moon.

00:13 Here on the right side, i've listed the masses of the sun, the moon, and the earth, as well as the distances between the moon, the earth, and the sun, and the earth.

00:24 I've also drawn out the right triangle indicating which leg is the distance between the moon and the earth.

00:33 The earth and the distance between the sun and the earth form the hypotenuse.

00:39 So first, let's go ahead and write out what we're trying to find, which is the force on the moon.

00:47 Here this will be equal to force on the moon by the sun plus the force on the moon by the earth.

00:57 And for the force on the moon by the sun, we're going to need the distance rmf.

01:08 Which will simply be the pythagorean theorem using rse and rme.

01:15 Here rms will be the square root of the hypotenuse, rse squared, minus the leg that we have, rme, and that gives us about 1 .48 times 10 to the 11 meters.

01:38 And notice how this is essentially the same thing as rse because the moon and the earth are very close together in comparison to the distance between the sun and the moon and the sun and the earth.

01:52 So let's go ahead and find the components of our force vector.

01:56 So for fms, the force on the moon by the sun, the sun is pulling the moon in the negative x direction...

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