(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

GME Mme/ Rpole2
= -------------------------
GME Mme/ Requator2

[Requator2]
= ---------------
[Rpole2]

= [Requator / Rpole ]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:

FEarth
--------  =
FMoon

GMEarth Mapple/ REarth2
= -------------------------
GMMoonMapple/ DMoon-Earth2

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

MEarth/ REarth2
= -------------------------
MMoon/ DMoon-Earth2

MEarth     DMoon-Earth2
= ------ X --------------
MMoon      REarth2

= [MEarth/MMoon] x [DMoon-Earth/REarth]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.

FEarth
-------- = 81 x 602 ~ 300,000
FMoon

(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.)

FMoon
-------- =
FEarth

MMoon/ RMoon2
= -------------------------
MEarth/ DMoon-Earth2

MMoon     DMoon-Earth2
= ------ X --------------
MEarth      RMoon2

= [MMoon/MEarth] x [DMoon-Earth/REarth]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, ME = 6 x 1024 kg and RE= 6.4 x 106 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/s2 (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/s2.

g = G MEarth/REarth2
= 6.67 x 10-11 x 6 x 1024 / (6.4 x 106)2
= 10 m/s2
(a) So, if you drop the text book from the top of a tall building, what will be it's speed after it has been falling for 1 second?
After falling for 1 second the speed will be 10 meters/second

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

It started at rest and ended up at 10 meters/second so that the average speed over that second is 5 meters/second.

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

Moving at an average speed of 5 meters/second for one second, the book will have fallen 5 meters

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 gMoon = NUMBER x gEarth

gEarth = G MEarth/REarth2
gMoon = G MMoon/RMoon2
gMoon/gEarth
= [MMoon/MEarth] x [REarth/ RMoon]2
= [1/81] x [3.7]2 = 1/6

So, gMoon = 1/6 gEarth

(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.