open-Response Homework Problem 24.4 A space shuttle, after delivering supplies t
ID: 1401176 • Letter: O
Question
open-Response Homework Problem 24.4 A space shuttle, after delivering supplies to a lunar base, is preparing to return to Earth. The mass of the moon is M. 7.35 x 102 kg, and the radius of the moon is r 1.738 x 105m. (a)[7 pt(s) At what minimum speed does the space shuttle have to travel after launch to escape the gravitational pull of the moon (b)[5 pt(s) How many Joules of potential energy does a 200kg object-moon system have, if the object is 800km above the surface of the moon? (c)[6 pt(s) How fast would the space shuttle have to travel if it needed to place a satellite into a stable, circular orbit by just releasing it 800km above the surface of the Moon? (d) 6 pt(s) If the satellite has inertia 200kg, what is its angular momentum in this circular orbit? (e)[6 pt(s) If the speed were not quite correct because of an error in the mass of the moon and the orbit started to decay just due to the pull of gravity, what would the angular momentum be when the satellite was at 200km above the surface. Why? (f)[5 pt(s) alculate gmoon.Explanation / Answer
on the surface of the moon the potential energy = PE = -GMm/R
kinetic energy KE = 0.5*m*Ve^2
total energy TE = -GMm/R + 0.5*m*ve^2
after escaping the gravitational pull TE = 0
from energy conservation TE on the surface = total energy after reaching the escaping point
GMm/R = 0.5*m*Ve^2
Ve = sqrt(2GM/R)
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while the satellite is revolving in orbit at a height h from the surface with speed Vo
gravitation force = centripetal force
Fg = Fc
GMm/r^2 = m*vo^2/(R+h)
Vo = GM/(R+h)
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(a)
Ve = sqrt((2*6.67*10^-11*7.35*10^22)/(1.738*10^6))
Ve = 2375.2 m/s <<------answer
(b)
PE = -GMm/(R+h)
PE = -(6.67*10^-11*7.35*10^22)/((1.738*10^6)+800000)
PE = -19.31*10^5 J <<------answer
(c)
Vo = sqrt(GM/R+h)
Vo = sqrt((6.67*10^-11*7.35*10^22)/((1.738*10^6)+800000))
Vo = 1.4*10^3 m/s
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(d)
angular momentum L = m*vo*(R+h)
L = 200*1.4*10^3*(((1.738*10^6)+800000))
L = 710640000000 kg m ^2 /s
(e)
the total angular momentum of the moom satellite system remains same
from conservation of angular momentum IN the absence of eternal torque the total angular momentum is conserved
Here gravitational force is an internal force
No net external torque acts on the system
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(f)
g = GM/R^2
g = (6.67*10^-11*7.35*10^22)/(1.738*10^6)^2
g = 1.633 m/s^2 <<------answer
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