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When the Moon is directly overhead at sunset, the force by Earth on the Moon, $F_{\mathrm{EM}},$ is essentially at $90^{\circ}$ to the force by the Sun on the Moon, $F_{\mathrm{SM}},$ as shown below. Given that $F_{\mathrm{EM}}=1.98 \times 10^{20} \mathrm{N} \quad$ and $F_{\mathrm{SM}}=4.36 \times 10^{20} \mathrm{N}, \quad$ all other forces on the Moon are negligible, and the mass of the Moon is $7.35 \times 10^{22} \mathrm{kg}$ determine the magnitude of the Moon's acceleration.

$0.0065 \mathrm{m} / \mathrm{s}^{2}$

Physics 101 Mechanics

Chapter 5

Newton's Laws of Motion

Motion Along a Straight Line

Motion in 2d or 3d

Applying Newton's Laws

Cornell University

University of Michigan - Ann Arbor

University of Sheffield

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question number 15 This is the moon and force by our on the moon is like this and at 90 degrees and other force by the sun on the moon is acting so the resultant off these two forces is magnitude off the resultant off these two forces is equal to under route 1.98 in 2. 10 to the power 20. Well, the square Plus they didn't wreck it. Four point 36 into 10 to the power 20 Holy square Because we know that the resultant off two forces magnitude off the resultant off two perpendicular forces is a call to F as, um is Squire plus as e m is square. So solving this, we will get the magnitude off the resultant force four point 78 into 10 to the power 20 Newton. This is the magnitude off the result in fourth F. Now let us find the X elation. X elation is equal to the net. Force F upon the mass off the moon, that force is four point 78 into 10 to the power contained Newton's. They were aided by the mosque seven point 35 in tow tend to the power 22 kg. This will give us the X elation off the moon. 6.5 Indo 10 to the power minus three meter per second Squired. This is the required X elation off the moon.

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