**(1)** Hold up a pencil. Where is its center of mass? **The center of
mass of a pencil is very close to the middle of the pencil - the balance point.
**

**(2a)** A 180lb-astronaut weighed about 1/6 his usual weight when he was
walking on the Moon. What was his mass on the moon? Same / more / less? **
SAME mass - weight is the FORCE of gravity on that mass. Weight changes, mass
stays the same. **

**(b)** Which of the following are ways to loose weight? (i) diet so that
your mass decreases, (ii) go to a planet that has less mass than the Earth
(iii)
climb to the top of a very tall building **All are ways in which you would
have less weight - the FORCE of gravity would be less because in (i) Mass of
object is less, (ii) Mass of planet is less, (iii) Distance is MORE (so force
is less). **

**(3)** The Earth is not quite round--the radius at the pole is 6357 km
and 6378 km at the equator.

**(a)** Where on the surface of the Earth would you weigh the most?**
Where force of gravity is greatest will be where the distance between the
center
of the Earth and you is LEAST - at the poles.**

**(b)** Where on the surface of the Earth would you weigh the least?
**
Conversely, you weigh most at the equator. **

**(c)** Calculate the following ratio:

- Maximum Weight
**Force of gravity @ poles**

**------------------------- =**---------------------------

Minimum Weight**Force of gravity @ equator**

GM

= -------------------------

GM

[R_{equator}^{2}]

= ---------------

[R_{pole}^{2}]

= [R_{equator} / R_{pole} ]^{2}

**= [6378/6357] ^{2} = 1.0066
**

** this means the difference in weight at the poles is only 0.66% higher
than at the equator - but less than the change in weight produced by
drinking a cup of coffee.**

**(4)** **(a)** Take the RATIO of these two forces:

F_{Earth}

-------- =

F_{Moon}

GM_{Earth}M_{apple}/ R_{Earth}^{2}

= -------------------------

GM_{Moon}M_{apple}/ D_{Moon-Earth}^{2}

**(b) **Cancel and re-arrange terms to make the expression as simple as
possible.

**
**

M_{Earth}/ R_{Earth}^{2}

= -------------------------

M_{Moon}/ D_{Moon-Earth}^{2}

M_{Earth}D_{Moon-Earth}^{2}

= ------ X --------------

M_{Moon}R_{Earth}^{2}

= [M_{Earth}/M_{Moon}] x [D_{Moon-Earth}/R_{Earth}]^{2}

**(c) **Work out how many times greater the force on the apple due
to the Earth is compared to the force on the apple due to the Moon using
the fact that the Earth's mass is 81 times that of the Moon and that the
distance to the Moon is 60 times the Earth's radius.

F_{Earth}

-------- =81 x 60^{2}~ 300,000

F_{Moon}

**(6) **Now consider an astronaut on the Moon dropping a hammer-- the Earth
is shining above, pulling the hammer up, the Moon is pulling the hammer down--which
wins? Follow the same procedure as the previous question--calculate the ratio
of forces. (The Earth's radius is 3.7 times the Moon's radius.)

F_{Moon}

-------- =

F_{Earth}

**
**

M_{Moon}/ R_{Moon}^{2}

= -------------------------

M_{Earth}/ D_{Moon-Earth}^{2}

M_{Moon}D_{Moon-Earth}^{2}

= ------ X --------------

M_{Earth}R_{Moon}^{2}

= [M_{Moon}/M_{Earth}] x [D_{Moon-Earth}/R_{Earth}]^{2}

= [1/81] x [60 x 3.7]^{2}= 608

**The force of the Moon on the hammer is 608 times greater than the force of the
Earth on the hammer. So, the hammer falls to the Moon.
**

**(7)** Using G= 6.67 x 10^{-11}, M_{E} =
6 x 10^{24}
kg and R_{E}= 6.4 x 10^{6} meters, you can work
out the value
of the acceleration due to the Earth's gravity on the Earth's
surface--g. The
units should be in m/s^{2} (meters per second per second)
since an acceleration
is a change in speed over time and speed is measured in m/s
(ok--sometimes we
use km/hr for high speeds). You should get an answer of 10 m/s^{2}.

**
**

g = G M_{Earth}/R_{Earth}^{2}

= 6.67 x 10^{-11}x 6 x 10^{24}/ (6.4 x 10^{6})^{2}

= 10 m/s^{2}

**(b)** What will be its *average speed* over that one
second?

**(c)** How far will it have fallen in that one second?

**(d)** What if you had dropped a lead brick instead of a book?
**EXACTLY the same acceleration will have occured**

**(8)** Now plug in numbers and get the value of the acceleration due to
gravity on the Moon compared with the acceleration on the surface of the Earth
(using that the Moon has 1/81 times the mass of the Earth and that the Earth
is 3.7 times bigger than the Moon). Hint: you only need to square a number and
multiply 2 numbers to get an answer that looks like g_{Moon} = NUMBER
x g_{Earth}
**
**

g_{Earth}= G M_{Earth}/R_{Earth}^{2}

g_{Moon}= G M_{Moon}/R_{Moon}^{2}

g_{Moon}/g_{Earth}

= [M_{Moon}/M_{Earth}] x [R_{Earth}/ R_{Moon}]^{2}

= [1/81] x [3.7]^{2}= 1/6

**So, g _{Moon} = 1/6 g_{Earth}
**

**(9)** When you are standing on the Earth you experience the gravitational
forces of all the objects in the universe! Luckily, the Earth's gravity wins
and you do not float off into space. But the effect of the gravity of these
other objects is real (it causes tides, for instance). In question 3 you compared
your weight when standing at the pole compared with the equator. Now consider
the effect of the Sun's gravity pulling on you at the same time as the Earth.
At what time of day will you weigh the most? And the least?

**Here we want to consider the effect of the Sun's gravity on the pull of
the Earth (weighing you down). For the gravitational forces of the Sun and Earth
to be in opposite directions - making you feel lighter - the Sun must be pulling
you "up" (away from the Earth) - which happens at noon. At mid-night
the forces of the Sun and Earth add together to make you weigh more - but the
amount is pretty negligible!**

**(10)** In questions 4 and 5 you compared the gravitational forces due
to the Earth and the Moon. Close to the Earth the gravitational force of the
Earth was stronger--but on the Moon the force of the Moon was stronger than
that of the Earth.. This means that there must be a place between the Earth
and the Moon where the gravitational forces of the Moon and Earth cancel. Where
is this location? Half way between them? Closer to Earth? Closer to the Moon?

**If you have seen Space Cowboys, this may be familiar to you. But the movie
got it wrong - the gravity of the Moon does not take over half way to the Moon
from the Earth - you need to go 9/10ths of the way to the Moon - because the
Moon is much less massive than the Earth.**