The gravitational potential energy of a system consisting of a planet of mass M
ID: 1560193 • Letter: T
Question
The gravitational potential energy of a system consisting of a planet of mass M and and object of mass m is U=GMm/r , where G is the universal gravitational constant and r is the distance between the center of the planet and the center of the object, and assuming that the potential energy is zero when the objects are infinitely far apart.
(a) Show that the potential energy can be written in the form
U=mgsR(R/r)
, where
gs
is the acceleration of gravity at the surface of the planet and
R
is the radius of the planet.
(b) Determine the amount of work that must be done on a 120 kg payload to elevate it to a height of 1900 km above the Earth's surface. Note that the radius of the Earth is 6370 km. (Suggestion: enter your answer in MJ.)
(c) The orbital velocity of a small object around a planet is
vorb=
. Show that the orbital velocity can be written
vorb=
.
(d) Determine the amount of additional work that is required to put the payload into circular orbit at this elevation. (Suggestion: enter your answer in MJ.)
(e) One kilogram of liquid hydrogen releases 140 MJ of chemical energy. What is the mass of liquid hydrogen needed to put the payload into orbit?
Explanation / Answer
a) U = -G*M*m/r
= -(-G*M/R^2)*m*R^2/r
= -gs*m*R^2/r (since g = G*M/R^2)
= -gs*m*R*(R/r)
b) Workdone = Uf - Ui
= -G*Me*m/(Re + h) - (-G*Me*m/Re)
= -G*Me*m/(Re + h) + G*Me*m/Re)
= -6.67*10^-11*5.98*10^24*120/(6370*10^3 + 1900*10^3) + 6.67*10^-11*5.98*10^24*120/(6370*10^3)
= 1.726*10^9 J
= 1726 MJ
c) V_orb = sqrt(G*M/r)
= sqrt((G*M/R^2)*(R^2/r))
= sqrt(gs*R^2/r)
= sqrt(gs*R*(R/r))
d) amount of additional workdone = (1/2)*m*vorb^2
= (1/2)*m*(gs*R^2/r)
= (1/2)*120*9.8*(6370*10^3)^2/(6370*10^3 + 1900*10^3)
= 2.88*10^9 J
= 2880 MJ
e) mass of liquid hydrogen needed, m = 2880/140
= 20.6 kg
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