How much work would be needed to raise the payload from the surface of the moon (i. e., x = r) to the "end of the universe"?

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

Explanation:

"Assume the weight move up at constant speed. With no net acceleration, the force applied is -(weight). Since the weight at height R is -P(R/x)² (minusbecause it's directed downward) the applied lifting force is P(R/x)², and the work done moving from x to x+dx is dW = P(R/x)² dx. Intetgrate this:"

W(x) = ?PR²/x² dx = -PR²/x + C

 The work done moving from x=R to x=2R is:

W(2R) - W(R) = -PR²(1/(2R) - 1/R) = PR/2

(b) The work done moving from 2R to 3R is:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation:

frequency or wavelenth of em radiation in a vacuum