Question

In this problem, we will calculate the tangential speed of the moon vm (in units of m/s) two different ways, then compare the two. a) The moon's orbital time period is 27.32 days (for exactly one revolution). Given its orbital radius of 3.84 x 108 meters (essentially the earth-moon distance), find the speed of the moon as it orbits earth. b) Newton tells us that the necessary centripetal force acting on the moon is supplied by gravity: vm= square root of Gme/r

which can be rearranged/simplified to solve for the moon's speed: Solve for the moon's speed. The earth's mass is listed in Appendix C, Table C3 in the text. c) Compute the percent difference between the two values. In your opinion, did Newton get it right?

Answer 1

(a) v = distance / time

assuming the orbit is circular, distance = circumference of circle

v = 2*pi*r / time

v = 2*pi*3.84e8 / 27.32*24*60*60

Notice I have converted time from days to seconds

v = 1022.15 m/s

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Now by second method

mv²/ r = GMm / r²

v² = GM / r

Here M is mass of earth = 5.98e24 Kg

v = sqrt (GM/r)

here G is universal gravitational constant and has a value 6.67e-11 m³ kg^-1 s^-2

v = 1019.17 m/s

so, yes we can say that, Newton was right. To be precise, he was extremely close.

percent difference = |1019.17 - 1022.15| / (1019.17 + 1022.15 / 2) * 100

percent difference = 0.2919 %