2. a A satellite of mass 80.0 kg is in a circular orbit around planet Q. Planet
ID: 1596173 • Letter: 2
Question
2. a A satellite of mass 80.0 kg is in a circular orbit around planet Q. Planet Qis spherically symmetric and has radius 3.00x106 m. The speed of the satellite is 5000 m/s and the radius of the satellite's orbit is 8.00x10 m What is the mass of the planet Q? Ans. 3.00 Xlo 14 kg b You travel to planet Qin a spaceship and land on the surface of the planet. While exploring the surface, you release a small rock from rest at a distance of 20.0 m above the surface of the planet. What is the speed of the rock just before it reaches the surface?
Explanation / Answer
In an orbit the gravitational force provide the necessary centripetal force
for the satellite
Fg = Fc
G*m*Q/r^2 = m*v^2/r
GQ/r = v^2
Q = v^2*r/G
Q = 5000^2*8*10^6/(6.67*10^-11)
Q = 3*10^24 Kg
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(b)
initial energy Ei = -G*Q*m/(R+h)
before touching the surface
final energy = Ef = -GQm/R + (1/2)*m*v^2
from energy conservaiton
Ef = Ei
-GQm/R + (1/2)*m*v^2 = -G*Q*m/(R+h)
(1/2)*v^2 = GQ*(1/R - 1/(R+h))
(1/2)*v^2 = 6.67*10^-11*3*10^24*(1/(3*10^6)-1/((3*10^6)+20))
v = 29.8 m/s
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(2)
initial kinetic energy Ki = 0
iniital potential energy = Ui
work done by fricition Wf = -uk*m*g*x
final potential energy Uf = 0
final kinetic energy Kf = 0
from work energy relation
W = dU + dK
-uk*m*g*x = Uf - Ui + 0
Ui = 0.3*5*9.8*4 = 58.8 J
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