Question

FEMI (3%) Problem 4: When the Moon is directly overhead at sunset, the force by the Earth on the Moon, FEM, is essentially 90° to the force by the sun on the moon, FsM, as depicted in the image. Let FEM=1.99 x 1020 N, FsM= 4.61 x 1020 N, all other forces on the Moon be negligible, and the mass of the Moon be m= 7.35 x 1022 K 2 kg Moon ESM 50% Part (a) What is the magnitude of the net force exerted by the Earth and Sun on the Moon in Newtons? Fnet = Grade Summary Deductions 0%

Answer 1

$F_{EM}=1.99*10^{20}N$

$F_{SM}=4.61*10^{20}N$

As are in perpendicular direction, we can denote these two forces along x- and y-directions respectively.

So, the net force on the moon is given by,

$F_{net}=\sqrt{(F_{EM})^2 + (F_{SM})^2}$

So, $F_{net}=\sqrt{(1.99*10^{20})^2 + (4.61*10^{20})^2}=5.021*10^{20}N$ (Answer)