Question

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(10 Points) The moon revolves about the earth in a circle (very nearly) of radius R =239,000 mi or 12.6 X 108 ft, and requires 27.3 days of 23.4 X 105 sec to make a complete revolution. (a) What is the acceleration of the moon towards the earth? (Ans: a = 8.96 X 10-3 ft/sec) (b) If the gravitational force exerted on a body by the earth is inversely proportional to the square of the distance from the earth's center, the acceleration produced by this force should vary in the same way. Therefore, if the acceleration of the moon is caused by the gravitational attraction of the earth, the ratio of the moon's acceleration to that of a falling body at the earth's surface should equal the ratio of the square of the earth's radius (3950 mi or 2.09 X 106 ft) to the square of the radius of the moon's orbit. Is this true?

Answer 1

Sol As we know centripetal acceleration Q = w²R Here Ra 239000 miles : 1261920000 feet W: 27 -1 2 a rad day- 27.3 27 rad/sec 235 8720 ac -(235 6100) x 12 61920000 2358720 0.0089544 ac - qc = 8.95 x10-37 m/s2 the data which you have provided would not give Exact Answer. 5 g Earth = 32.2 f/sec? Om = 8.95 x1003 f) seca

ge Im 32.2 8.95 X1003 X103 ge 3.597 X10 9m ū 3.6 x103 Rm2 (1261920000)? (20903520)² 2 RE Rm² 3.6 XI0 RE² equal so they Exact values are too 3.59 7 X10 and 3.6uu X103 which I have rounded oft.