| ToC, Chapter 12 | Course Calendar |
D12.1 Calculate the mass of Jupiter, given that its moon
Callisto has a mean orbital radius of 1.88 x 10^{6}
km and an orbital period of 16 days, 16.54 hours.
The force of gravity provides the centripetal force to keep Callisto
in its orbit.
We must find the linear speed of Callisto.
T = 16 d, 16.54 h = [(16)(24) + 16.54 ]
h = 400.54 h
T = 400.54 h [3600 s/h]
= 1.44 x 10^{6} s
F_{gravity} = F_{centripetal}
F_{gravity} = G M_{J} m_{C}/r^{2} = m_{C} v^{2}/r = F_{centripetal}
G M_{J} m_{C}/r^{2} = m_{C} v^{2}/r
M_{J} = r v^{2}/G
M_{J}
= 1.9 x 10^{27} kg
D12.2 Starting with the moon's period of 27.3 days,
calculate the radius of its orbit.
The gravitational force between Earth and our moon provides the centripetal
force,
Don’t try to solve for the radius
immediately for we know the velocity only in terms of the radius,
= m = m
T = 27.3 da (24 h/da) (3600 s/h) = 2.36 x
10^{6}
s
r^{3}
=
r^{3}
=
r^{3}
= 5.627 x
10^{25}
m^{3}
r = 3.83 x
10^{8}
m
r = 3.83 x
10^{5}
km
D12.3 The acceleration of a falling body near Earth’s surface, at a distance R from Earth’s center, is 9.80 m/s^{2}.
(a) Use a suitable proportion to calculate the acceleration toward Earth of a falling body that is 60 R from Earth’s center.
(b) Our moon is in an orbit of radius 60 R, with a period of revolution of 27.26 days. Show, as did Sir Isaac Newton, that the centripetal acceleration of the moon toward Earth agrees with your answer from part (a).
Earth’s radius is R = 6.38 x 10^{3} km.
D12.4 What orbital radius should a weather satellite have if it is to have a period of 6.0 hours?
r = 1.68 x 10 ^{7} m = 1.68 x 10 ^{4} km = 16,800 km
D12.5 From the data below, calculate the acceleration of free fall on the surface of
a) Jupiter,
b) Saturn, and
c) our Moon.
mass radius Jupiter 1 900 x 10^24 kg 71 400 kmSaturn
561 x 10^24 kg 60 000 km moon 0.0736 x 10^24 kg 1 740 kmF = m a = G
a)b)
c)
| ToC, Chapter 12 | Course Calendar |
(C) 2005, Doug Davis; all rights reserved