If Earth was half its size what would happen to the value of gravity?

Answer 1

That depends a great deal on what you mean by 'half its size':

1) If you mean that Earth would now have two times smaller radius, while retaining its mass, its gravitational force would remain the same on distant bodies when they can both be considered point masses. However, the gravitational acceleration on its surface (g) would now be greater due to smaller radius and would now equal:

gnew = GMEarth / (rEarth/2)2 = 4 GMEarth / rEarth2 = 4 gold, ie. it would be 4 times greater than it is now.

2) If you mean that Earth's volume was to be half of what it is now, while still retaining the same mass, new Earth's radius would have to be:

rnew = rold / 21/3.

While the first thing said in point 1) would still be true, gravitational acceleration on Earth's surface (g) would now be:

gnew = GMEarth / (rEarth/21/3)2 = 22/3 GMEarth / rEarth2 ~= 1.587 gold, ie. it would be approximately 1.587 times greater than it is now.

Also, both in cases 1) and 2), the tidal forces would be greater.

3) You could also mean that Earth would have to be now half it's current radius(or volume) while retaining its density and therefore having lower mass. In that case, Earth would exert lower gravitational attraction on distant bodies, proportionally to the mass loss.

In case of half radius, same density, g would remain as it is now:

ghalfradius = GMEarth / 4(rEarth/2)2 = GMEarth / rEarth2 = gold,

while in case of half volume, same density, g would be:

ghalfvolume = GMEarth / 2(rEarth/2)2 = 2GMEarth / rEarth2 = 2gold, ie. twice its current measure.