Setting gravitational potential energy to zero at infinity, with MMars = 6.42×10
ID: 1495666 • Letter: S
Question
Setting gravitational potential energy to zero at infinity, with MMars = 6.42×1023 kg and RMars = 3.4×106 m, (a)[5 pts] How many Joules of potential energy does a 150 kg object-Mars system have, if the object is one Mars radius above the surface of Mars? (b)[6 pts] What speed would a 150 kg object have to have at this height to have a total energy of zero? What is this speed called? (c)[4 pts] Assuming the object’s velocity is perpendicular to the radius, what would be the angular momentum of the object at this speed and location? (d)[5 pts] Given the gravitational force of Mars being the only force on the object, at what speed would the object be able to maintain a stable circular orbit at this height above the surface of Mars? (e)[4 pts] Suppose you give the object the right speed for a circular orbit, but you got the direction of the velocity wrong, so the object (let’s call it a satellite) ends up in an elliptical orbit instead, falling towards Mars. If the initial angular momentum was Li, how will the angular momentum of the satellite compare to Li when it is at a height of 300 km above Mars’ surface? Carefully explain how you found your answ
Explanation / Answer
potential energy U = -GMm/r
(a)
r = R+h
h = R
r = 2R
U = -(6.67*10^-11*6.42*10^23*150)/(2*3.4*10^6)
U = -944589705.8 J
++++++++++++++++++++++++++
b)
KE = 0.5*m*v^2
total energy = 0
0.5*m*v^2 -944589705.8 = 0
0.5*150*v^2 - 944589705.8 = 0
v = 3548.88 m/s
c)
L = m*v*r = 150*3548.88*(2*3.4*10^6) = 3.6198576*10^12 kg m^2/s
(d)
Fc = Fg
m*v^2/r = GMm/r^2
v = sqrt(GM/r)
v = sqrt((6.67*10^-11*6.42*10^23)/(2*3.4*10^6))
v = 2509.43 m/s
final speed v = sqrt((6.67*10^-11*6.42*10^23)/((3.4*10^6)+(300*10^3)))
v = 3402 m/s
Lf = m*v*r = 150*3402*(3.4*10^6)+(300*10^3)) = 1.88811*10^12 kg m^2/s
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