A satellite is placed 39875 m above the Moon's SURFACE. (M=7.35e22 kg, R=1.74e6 m) Find the Period of the satellite in seconds. (assume a circular orbit)

Answer 1

r = .039875*10^6 +1.74*10^6 = 1.78*10^6
G = 6.67*10^-11
M = 7.35*10^22

F = m v^2/r

G M m /r^2 = m v^2/r

v^2 = G M /r

T^2 = (2pi)^2 * r^2/v^2

T^2 = (2pi)^2 * r^3 /GM

T^2 =
39.5*5.64*10^18/(6.67*10^-11*7.35*10^22)

T^2 = 4.54 * 10^7 = 45.4 * 10^6

T = 6.74 * 10^3 = 6740 seconds (about 1.8 hours)

Answer 2

thank you so much, now i understand