Preparation problems
9-1. Space explorers land on a planet with the same
mass as Earth, but they find they weigh twice as much as they would on
Earth. What is the radius of the planet?
Solution:
We start with the statement that the force of gravity acting
on the explorers on the unknown planet is twice that on Earth.
Mathematically, this looks like
Since both forces are gravitational, we can use the universal law of
gravitation to solve the rest of the problem. So substituting
we get
9-7. What is the approximate value of the gravitational
force between a 67-kg astronaut and a 73,000-kg space shuttle when they
are 84 m apart?
Solution:
We start with the universal law of gravitation, using
m1 = 73,000 kg, m2 = 67 kg, and r = 84 m. This gives us
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(6.67 ×10-11 N ·m2/kg2)(73,000 kg)(67 kg)
(84 m)2
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9-8. Compare the gravitational attraction of the Earth
for an astronaut on the surface of the moon with the gravitational
attraction of the moon for the astronaut.
Solution:
If we assume that the astronaut is on the side of the moon
closest to Earth, we can determine the gravitational force the Earth
exerts on the astronaut using the law of universal gravitation:
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FEa = |
G MEarth mastronaut
rEa2
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where MEarth is the mass of the Earth, mastronaut is the mass
of the astronaut, and rEa is the distance from the center of Earth
to the center of the astronaut. We can look up the distance betwee the
center of the Earth and the center of the moon in Appendix E of our book
rEm = 3.85 ×108 m. We can find the distance we want by
noting
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(3.85 ×108 m) - (1.74 ×106 m) |
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where rmoon is taken from Appendix E.
Now we can write down the force of the moon on the astronaut:
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Fma = |
G Mmoon mastronaut
rmoon2
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Finally, we can compare the two using a fraction:
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G Mmoon mastronaut
rmoon2
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Solutions translated from TEX by TTH, version 1.57.
