Spring 2005

MWF 11:30

**Instructor**: Dr. Bruce Law, CW
327, Tel: 532-1618.

# REVIEW QUESTIONS OF Home Work 9

# R6. Assuming that this 1kg body is sitting
on the surface of the Earth, the gravitational force that acts on this body is
just it`s weight i.e. W = mg = 1x10 = 10N.

# R9. F = GMm/r^{2} therefore if r
double r^{2} must increase by a factor of 2x2 = 4. Hence the force F between
two bodies must be __smaller__ by a factor of 4 (because in the formula for
F you are dividing by r^{2}).

# R13. The force is zero when the object is infinitely
large from the Earth (i.e. r = infinity in the formula F = GMm/r^{2}
and when you divide by (infinity)^{2} then F =
0).

# R21. The Sun and moon exert a greater
gravitational force on one side of the Earth than on the other because they are
close to one side than the other.

#

# EXCERISES OF Home Work 9

# E7. This can never happen. Although the mass
of the Earth is much larger than the mass of the moon the rocks are at a much,
much closer distance (r) to the moon than they are to the Earth. Hence as F =
GMm/r^{2} the force between the rocks and the moon is much, much larger
than the force between the rocks and the Earth.

## E10. The force of gravity is give by F = GMm/r^{2}. The mass of the
Earth is much larger than the mass of the Moon,
therefore, the space pod must be much near to the Moon than the Earth in order
that the force of gravity from the Moon be equal to the force of gravity from
the Earth.

## E11. The weight of the Earth in the gravitational field of the apple is 1N
also, according to Newton`s third law.

**E12. The Earth
and Moon attract each other with the same force according to Newton's third
law. **

**E14. The force
of gravity is give by F=GMm/r**^{2}. Therefore if the radius of the Earth
(r) is increased, the force of gravity (and hence your weight) will decrease.
Similarly, if the radius of the Earth is shrunk (r smaller) then your weight
would increase.

**E17. F=GMm/r**^{2},
the mass of Juptier (M) is 300 time as large as that
of Earth, however, the radius (r) of Juptier is much,
much larger than that of Earth – this is the reason why an object on the
surface of Juptier scarcely weighs three times as
much as on Earth.

** **

# PROBLEMS OF Home Work 9

**P3. F=GMm/r**^{2}
where now r increases by a factor of 4, therefore, r^{2} increases by a
factor of 4x4 = 16 and therefore the force F must __decrease__ by a factor
of 16 because you are dividing by r^{2}. As a consequence the value of
g at this distance would be 9.8/16 = 0.6m/s^{2}.