Question

2. Recall that in Homework 1, we found that the Sun exerts a force on the Earth almost 200x that of the Moon on the Earth. However, the Moon is clearly primarily responsible for ocean tides! Sir Isaac Newton correctly realized that it wasn't the strength of this gravitational force that produced tides, but the fact that the Sun and the Moon pulled harder gravitationally on one side of the Earth than the other. Thus, the change in gravitational force over a region of space is a useful and important quantity for the determination of tidal forces. a) Derive an equation for the rate of change of gravitational force with respect to distance r. In other words, find dFc/dr. b) Using your result and the distances to the Moon and Sun, show that the variation of the Moon's gravitational force at the Earth's position is greater than that of the Sun's.

Answer 1

This is one of the early practical applications of theory of gravitation in understanding phenomenons of the planet Let us see how to approach the problem.

Let mass of Sun, Earth, Moon be Ms Me Mm .. Now, Granitational Force between two objects M&mat seperation or is faz amma Rate of change of fa with respecte to or da 2-2 am dr we have minus sign, because a decreases with increase . Rate GMm of decrease of d f g - 2

(rse) 3 & For system of Sun-Earth, Rate of decrease of fa (R se) - 2 G Me Ms Me = 6x po 2 kg Ms = 2X1030 kg - 2 x 6.678,0" x 2X103 % 6x1024 (1.5X101) 3 Mn = 7.3481022 Ice Rse = 4.748101 ose = 1.5x10" m sme = 3.84 x 100 ml. Similarly Rme = 2 a G me mm 286.67810" x 7.34X102 % 6x1024 (3.84 X10013 Rme = 1.03 x1018
Hence, we can observe
Rme > Re