Question

A spacecraft of 150 kg mass is in a circular orbit about the
Earth at a height *h* = 5*R*_{E}.

(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: *R*_{E} 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

Answer #1

**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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