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The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force $\overline{\mathbf{F}}_{\text { SM }}$that the sun exerts on the moon is perpendicular to the force $\overrightarrow{\mathbf{F}}_{\mathrm{EM}}$ that the earth exerts on the moon. The masses are: mass of sun $=1.99 \times 10^{30} \mathrm{kg}$ , mass of earth $=5.98 \times 10^{24} \mathrm{kg}$ , mass of moon $=7.35 \times 10^{22} \mathrm{kg}$ . The distances shown in the drawing are

$r_{\mathrm{SM}}=1.50 \times 10^{11} \mathrm{m}$ and $r_{\mathrm{EM}}=3.85 \times 10^{8} \mathrm{m} .$ Determine the magnitude of the net gravitational force on the moon.

$5.67 \times 10^{-5} \mathrm{N},$ right

$3.49 \times 10^{-5} \mathrm{N},$ right

$9.16 \times 10^{-5} \mathrm{n},$ right

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Cornell University

Numerade Educator

University of Washington

McMaster University

to deter a mind The magnitude of the gravitational force on the moon we have to determine first war is the gravitational force that the sun exerts on the moon on. Then what is the reputational force that the earth exerts on the moon? We will begin the completing their force that the sun exerts on the moon F S m We can conflict that force by using the following equation as M is the gravitational force, then it is a cost to G times the mass off the sun times the mass off the moon divided by the distance between the centre off the sun on the center, off the moon squared. And that distance is R S M, which is close to 1.5 times 10 to the 11. 71.5 banks 10 to 11 squared now plaguing the other values that were given by the problem We get f S M as being equals to 6.6 to 7 Time stands for buying is 11 times 1.99 times. Stand to the authority times The mass off the moon, which is 7.35 time stand to 22 and this is divided by 1.5 times. Stand to the 11 squared and this gives us a gravitational force off a magnitude of approximately four 0.34 times. Stand to the 20 new tones. Now we proceed to companies the magnitude of the gravitational force exerted on the moon by the earth. We do so by calculating F e M. It's very similar to what we had just don't want. The only thing that changes is that now we will use the earth Mass and the distance between the centre off the earth and the center off the room So we have F e m is equals to the neutrons. Constant times the mass off the earth times the mass off the moon, divided by the distance between the center of the earth in the center of the book squared by plugging the values that were given by the problem, we got 6.67 times stand to minus 11 times the mask off the earth off 5.98 times. Stand to 24 kilograms and the mast off the moon off 7.35 times. Stand to 22 and this is divided by the distance between the centre off the earth and the center off the boom, which is tree 0.85 times. Stand to the eight squared and this gives us gravitational force off the magnitude of approximately 1.98 times. Stand to the 20 new terms. Now, let me organize the board to finish the question. Okay. Now, to finish the question, we must calculate what is the magnitude of the resulting gravitational force of the moon That resulting gravitational force results from the addition off F as m and F e m. Note that we are ardent factors. So we have to use the Gregorian theory, um, to get the magnitude off the resulting force so the resulting force I will call f it's equals to the square it off f S m squared plus f e m squared. Why is that? Because we are adding F S m with F e m and these results in the force that points in this direction and we have a rectangle triangle and then we just have to apply the Ivorian fuel him to get the value off f By plugging the values that we have. We got that f is it close to the square it off 4.34 Find stand to the 20 squared plus 1.98 times, then to the 20 squared. And this gives the resulting gravitational force with a magnitude of approximately four 0.77 time stand to the 20 new stones.

Brazilian Center for Research in Physics