# Homework Help: Value of g on Earth with a smaller radius

1. Nov 1, 2009

### Jimmy25

What would be the value of g if the radius of the earth was decreased by half while maintaining the same mass.

This is what I did

g = MG / R^2

g = MG / (0.5R)^2

g = MG / 0.25R^2

0.25g = MG / R^2

I know this is not correct and the actual answer is 4g because as you get closer to the center of the Earth the gravitational force will increase but I cannot figure out why this attempt is wrong.

2. Nov 1, 2009

### tiny-tim

Hi Jimmy25!

(try using the X2 tag just above the Reply box )
Why did you put that 0.25 on the LHS?

3. Nov 1, 2009

### Pengwuino

What you have initially is the gravitational acceleration due to the Earth while you're at the surface, or at a distance R. You then manipulated the equation to ask not what the value of g would be at the surface if the radius was half as large, but what would the value of the acceleration be at that same distance R if the radius was half as large.

What I would do is setup a ratio, g' and g, where g' is at the surface of some planet with radius R' with the same mass, but then take R' = 0.5R and divide the two equations.

4. Nov 1, 2009

### willem2

you confuse the values of g before and after the shrinking of the earth

you have g_old = MG/R^2

now g_new is MG/(0.5*R)^2

and (0.25)*g_new = MG/R^2 = g_old so

g_new = 4*g_old, as expected

5. Nov 1, 2009

### Jimmy25

Ok I see. So gold was equal to one quarter gnew

Thanks, was racking my brain over this one. Sometimes the simplest question can be the most challenging.