A spacecraft of 150 kg mass is in a circular orbit about the Earth at a height h

Question

A spacecraft of 150 kg mass is in a circular orbit about the Earth at a height h = 5RE.

(a) What is the period of the spacecraft's orbit about the Earth?
T = answer in hours

(b) What is the spacecraft's kinetic energy?
K = Units in J

(c) Express the angular momentum L of the spacecraft about the center of the Earth in terms of its kinetic energy K. (Use the following as necessary: RE for the radius of the Earth, K for the kinetic energy of the satellite, and m for the mass of the satellite.)
L =

(d) Find the numerical value of the angular momentum.
L = Units in J · s

Explanation / Answer

part A

T² = K R^3 where K = 4?² /GM
K = 4?² / (6.673e-11*5.977e24) = 9.89818e-14
Given R = 5Re + Re = 6Re = 6*6371e3 = 3.82e+7 m
T² = 9.89818e14*(3.82e+7) ^3
T = ? (9.89818e14*(3.82e+7) ^3
T = 7.425e+18 s

T = 2.06x10^15 hrs.
-------------------------------
Part B

K.E =0.5mv² = 0.5*150v² = 0.5*150*(GM/R)
K.E = 0.5*150*6.673e-11*5.977e24/1.9113e+7
K.E = 1.562e9 J
---------------------------
Part C

L =mvr
L² = (mvr) ²
L²/r² = (mv) ²

K = 0.5 mv²
2m K = (mv) ²

L²/r² = 2m K

L = r*? (2m K)

D
L = 3.82e+7*? (2*150* 1.562e9)
L = 3.82e+7*? (2*150* 1.56e9)
L = 2.61328e13 Js

Hope it will help you.

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