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?

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

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.