Value of g on Earth with a smaller radius

[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.

 

[tiny-tim]

Hi Jimmy25!

(try using the X² tag just above the Reply box )

What would be the value of g if the radius of the Earth was decreased by half while maintaining the same mass.
…
g = MG / 0.25R^2

0.25g = MG / R^2

Why did you put that 0.25 on the LHS?

 

[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.

 

[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

 

[Jimmy25]

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

g_new = 4*g_old, as expected

Ok I see. So g_old was equal to one quarter g_new

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