1.   We know

    g = GM/R²

since m cancels on both sides in the equation mg = GMm/R²

Now you are given R_x = 1/2R_y or R_x/R_y = 1/2

The densities are equal

To find the ratio a_x to a_y you can first express 'a' in terms of G, R and rho, the density. a = GM/R²

= G [4/3 (pi)R³ (rho)] / R²

= 4/3 (pi)G(rho)R

Since 4/3, (pi), G and (rho) are constants, when you divide the constants cancel and you'll be only left with the radii as shown below:

The ratio of a_x/a_y = R_x/R_y = 1/2.

   Q.E.D. (from Latin, quod erat demonstrandum, which is a nice way to write, hence proved!)

2.   The earth's gravity is indeed centered at the core of the planet but you have to take into account the shape of the earth/planet. Our earth is not spherical but flattened at the poles. The acceleration due to gravity can be shown to be

g = GM/R²

Clearly it is inversely proportional to the radius of the earth. Since the earth's radius is lesser at the poles by a little over a 20 km, you'll expect the 'g' to be greater at the poles.

You're right about the invariance of 'g' at the poles. 'g' is affected by the rotation of the earth but only at the equator. The value of 'g' at the poles is independent of the rotation of the earth. The change in the value of 'g' with rotation can be shown to be

g' = g - Rw²cos²(Ø)

where 'Ø' is the latitude at the place under consideration.

I'm not sure who came up with this first but this is commonly studied in all undergrad programs.

3.   You first have to know the relation between gravitational force and distance. It is what a physicist would call an "inverse square proportion".

From Newton's law of universal gravitation

F = GMm/r²

where F = force
    G = universal gravitational CONSTANT
    M = mass
    m = mass
    r = distance between the masses

F is INVERSELY proportional to (distance)²

If the distance between the object is tripled, then the force F must be 1/(3)² = 1/9

The force DECREASES nine fold. For example, if the force was 27 N at first, after the distance is tripled, the force must become 3 N!

4.   Not exactly. It is widely believed that the man of the millenium Albert Einstein died a sad man on account of not being able to resolve this. His theory of general relativity attempts to understand this.

Unfortunately I do not have that strip of cartoon or remember the exact words but remember the essence of it. Have you come across Dibert created by Scott Adams? In one of his encounters with Dogbert, he is asked the above question and he answers saying because the objects have mass. The argument gets repeated when he is asks how do you detect mass and Dogbert points out that he seems to be going around in circles with his answer. Dilbert quips "Rather why don't you look at it as leaving no loose ends". That sums it up I think.

There have been several theories and I thought I could suggest some intersting answers to the question that are available on the URL links suggested below without delving into the merits of each. Happy browsing.

    http://www.sciam.com/askexpert/physics/physics14.html

    http://www.curtin.edu.au/curtin/dept/phys-sci/gravity/

    http://gravity.ontheinter.net/

    http://home.earthlink.net/~danielemilio/a_shifting_theory_of_gravity.html

5.   a. Draw a diagram as usual so that you can visualize this. The potential energy of the in its orbit U = -GMm/r

where M is the mass of the earth, m the satellite, and r is the orbital radius.

The kinetic energy is

K = GMm/2r

The total energy before collision is

T = U + K
    = -GMm/2r

for one satellite. For both satellites, the energy would be doubled and therefore

T = -GMm/r

b. Let V be the velocity of the wreckage after the collision. Using the law of conservation of momentum you'll have

mv + m(-v) = 2mV

giving you

V = 0

Obviously the wreckage has no kinetic energy since V = 0 but has only potential energy given by

T = U + 0 = U
    = -GM(2m)/r

c. Since there is no velocity with the wreckage, there would no longer be any centripetal force and therefore the wreckage would fall down towards the centre of the earth under the action of the earth's gravitational attraction (however small).

6.   
a. Since the gravitational force must provide for the centripetal force

GMm/r² = mv²/r

giving you

v = [GM/r]^0.5
    = [gR²/r]^0.5

since GM = gR²

= [(9.81)(6.4 x 10⁶)²/(7.04 x 10⁶)]^0.5
    = 7.6 kms^-1

b. The time period

T = 2(pi)r/v
    = 2(3.14)(7.04 x 10⁶)/(7.6 x 10³)
    = 5817 s

c. Let y be the height of the satellite after its 1500^th revolution, and at this altitude let the velocity be V.

The total energy at the original height will be

U + K = -GMm/2r
    = -mgR²/2r

Similarly T at the end of the 1500^th revolution will be

U + K = -mgR²/2y

The loss of energy = 1/2[mgR²][1/y - 1/r]

Given this loss of energy is = NE
    = (1500)(1.4 x 10⁵)
    = 2.1 x 10⁸ J

Substituting all the values and simplifying you should get y = 6.8 x 10⁶ if my calculations are right.

d. Using the formula we used in a., for finding the velocity you should have

V = [gR²/y]^0.5
    = 7.7 kms^-1

e. And the time period

T = 2(pi)y/V
    = 5.5 x 10³ seconds

f. Using the work energy theorem

work done = loss of energy = F.(distance)
    = 2.1 x 10⁸ = F.(2.4 x 10⁵)

giving the value for

F = 8.8 x 10² N

g. Yes it should be since there is no other external force and you've neglected air resistance.