Work to Bring Object up Mineshaft to Earth Surface

[sean80439]

Here is the first part of the question:
If there is an object of mass m at the center of the earth, how much work does it take bring that object up a shaft to the surface? Edit: Earth is uniform density.

Do I integrate g(r) = -GMr/R^3 with respect to 'r' to get the change in gravity from the center of the Earth to the surface and then solve from R-0. If so I get a solution of -GM/2R, and that doesn't look at all right to me.

The second half of the question relates back to the first part and asks if mass m is dropped from the surface down a mineshaft to the center of the earth, what will its speed be when it reaches the center. Again, since gravity is changing with respect to the radius, how do I get a solid answer?

 

[TMM]

What's the force at the center? Half-way to the surface?

Use the fact that the mass of the shell outside the radius from the object to the center does not exert a net force.

Can you get a general force equation from these (for objects within the radius of the Earth)? Now how can you get work knowing the force?

 

[sean80439]

The net force would be zero at the center. Halfway it would be F=-GMm/2R^2 (r = R/2). It would be the F Normal = mg(r) wouldn't it, with W = mg(r)R? Why does it matter what the force is halfway to the surface or am I missing something?

 

[TMM]

Well the point is to see that the force is linear with distance as it increases from the center. From that you can integrate and find the work done.

Notice that the entire mass is not acting on the object when it is at the halfway point. In fact, the mass that is acting on it is the mass of the Earth enclosed by a sphere of radius equal to the distance of the object from the center. Does that make sense? Your current expression for the second bit is incorrect for this reason.

 

[sean80439]

So the equation should look like: W = ∫Fr δr or
W = ∫GMmr/R^3 δr and evaluate from 0 to R correct?

Also as I understand it, the relationship is linear inside the sphere and changes to the inverse square when outside correct? When you say second expression are you referring to the original integral in the initial post?

Last edit: Yes I think I know what you mean, the only mass acting on the object is the mass of the Earth at that radius ignoring all mass from every radius of r > r_object.