Use the known period of 27 1/3 days for the motion of the moon about the earth a

Question

Use the known period of 27 1/3 days for the motion of the moon about the earth and the distance from the earth to the moon of 3.84 x 108 m to calculate the radius of the orbit of an earth satellite that stays above the same point on the equator. (Hint: Use Kepler’s third law) (b) Determine the force the sun exerts on a kilogram of water on the earth’s surface at a point nearest the sun and at a point farthest from the sun. (c) Do the same for the force exerted by the moon. (d) Explain why the tides are associated with the motion of the moon

Explanation / Answer

Given, time period of moon (Tm) = 27 1/3 days = 2.35872 x 107 seconds

Distance of moon from earth (Rm) =3.84 x 108 m

We know that time period of earth (Te) = 3.156 x 107 seconds

(Tm)2/(Te)2 = (Rm)3/(Re)3

(Re)3       =             (Rm)3(Te)2/(Tm)2

                =             (3.84x108)3(3.156x107)2/(2.35872x107)2

                =             4.6627 x 108 m

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