If the Earth were compressed in such a way that its mass remained the same, but the distance around the equator were just one-half what it is now, what would be the acceleration due to gravity at the surface of the Earth?

If the distance around the equator is reduced by half, then the radius is also reduced by half.

Since the acceleration due to gravity is proportional to 1 / (radius²),

the acceleration changes by a factor of 1 / (1/2) ² = 1 / (1/4) = 4.

The acceleration due to gravity ... and also the weight of everything on Earth ...

becomes 4 times what it is now.

G=GM/r^2

Here are 2 ways to look at this:

1. Since the r is both inverted and squared on the right while acceleration due to gravity (g) is on the top on the left, any factor applied to r will be inverted and squared as applied to g. In this case, r is multiplied by 1/2, so g will be multiplied by (2/1) ^2, which is 4. Assuming your teacher uses 9.8 m/s^2 as the acceleration due to gravity, it would be 39.2 m/s^2.

2. Let r (1) be the radius of the earth, r (2) the smaller radius, g (1) is the acceleration due to gravity on earth and g (2) is the new acceleration due to gravity. In this case, r (2) = (1/2R (1)) ^2

You can re-arrange the equation for things that will not change on one side and things that are changing on the other.

This would be gr^2=GM. If GM is not changing, we can say GM = g (1) r (1) ^2 = g (2) r (2) ^2. Plug in what you know, 9.8*r (1) ^2=g (2) * (1/2r (1) ^2)

square the 1/2: 9.8*r (1) ^2=g (2) * 1/4r (1) ^2

the r (1) ^2 cancels out: 9.8=g (2) * 1/4

then multiply by 4 on each side: 4*9.8=g (2) = 39.2 m/s^2