Q: If the earth had twice its present mass, but the same radius what would be the value of g?

Answer 1

First of all, recall:
g = -32 ft/s² or -9.8 m/s²

Now, using the gravitation formula:
g = (m1*m2)/D²

In this case, one mass is the mass of the earth, so it doubles. The other mass is the mass of the object being weighed, so it stays the same. The distance does not change, since the radius does not change.

The new value of g would be:
g' = (2m1*m2)/D² = 2g

So, the force of gravity would double:
g' = -64 ft/s² or -19.6 m/s²

Answer 2

Gravitational Force = G m1 m2 / r^2
At the more local level, Gravitational Force = m * g

Setting those two equations equal to each other,

mg=Gm (2M)/r^2
When m2=M (before changing the mass of the earth), g =GM/r^2.
After changing the mass of earth, g = G(2M)/r^2= 2*9.8= 19.6m/s^2

Answer 3

I believe it would be 2 g, because the radius is the same.

Answer 4

g is related to the square of the mass

Answer 5

2x the current g. In other words, 19.6 m/s^2.

Answer 6

twice as much as it has now