The drawing (not to scale) shows one alignment of the sun, earth, and moon. The

Question

The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force SM that the sun exerts on the moon is perpendicular to the force EM that the earth exerts on the moon. The masses are: mass of sun = 1.99  1030 kg, mass of earth = 5.98  1024 kg, mass of moon = 7.35  1022 kg. The distances shown in the drawing are rSM = 1.50  1011 m and rEM = 3.85  108 m. Determine the magnitude of the net gravitational force on the moon.
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Explanation / Answer

Fsm = GMsMm/(rsm)^2

= (6.67x10^-11)(1.99x10^30)(7.35x10^22)/(1...

first just add the exponents when mutiplying and subtract when dividing. The exponent will be;

-11 + 30 + 22 -11 -11 = 19

The argument is (6.67)(1.99)(7.35)/(1.5)(1.5) = 43.36

Fsm = 43.36 x 10^19

Similarly;

Fem = GMeMm/(rem)^2

=(6.67x10-11)(5.98x10^24)(7.35x10^22)/(3...

= 19.78 x 10^19

F = SqRt[(Fsm)^2 + (Fem)^2]

= 47.66 x 10^19 N

I see it cut off my answer.
The denominator in the first part is;
(rsm)^2 = (1.5 x10^11)^2 = 2.25x10^22

and in the second part;
(rem)^2 = (3.85x10^8)^2 = 14.82 x 10^16

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