# Homework Help: Energy of an Object Falling Through the Earth (yes, all the way through)

1. Apr 8, 2010

### Brilliant

This is from an old AP test, and it's a multiple choice question, so I don't know why it's giving me so much hell, but here it goes:

1. The problem statement, all variables and given/known data
Suppose that a hole is drilled through the center of the Earth to the other side along its axis. A small object of mass m is dropped from rest into the hole at the surface of Earth, as shown [strike]above[/strike]below. If Earth is assumed to be a solid sphere of mass M and radius R and friction is assumed to be negligible, correct expressions for the kinetic energy of the mass as it passes Earth's center include which of the following?
[PLAIN]http://dl.dropbox.com/u/4439149/physucks_mc_hole_through_earth.jpg [Broken]
The two correct expressions are
$$\frac {1}{2}mgR$$
and
$$\frac {GmM}{2R}$$

2. Relevant equations
Ug=GmM/R

PS sorry the image is sideways.

3. The attempt at a solution
My first through was just the typical: initial potential equals final kinetic. That means that
GmM/R would be a proper expression of the kinetic energy, but that is not a correct answer.

I got to thinking, I bet the object still has some of its potential energy at the center. So I basically just considered all of the mass to be a distance of 1/2R away, the direction didn't matter since energy is scalar. That yields 2GmM/R, which is more than the original potential.

Then I started thinking, why does the specific Ug at the surface increase as h increases, in mgh, but Ug decreases as R increases when looking at the universal law. Then I realized I had severely confused myself over a multiple choice problem that is probably very simple.

Last edited by a moderator: May 4, 2017
2. Apr 8, 2010

### ehild

It looks strange, but the potential energy is not zero in the centre of Earth. It is chosen zero at infinity, and it is U(R)=-GmM/R at the surface of Earth.
To get the potential energy in the centre U(0), you have to go back to the definition of the potential: -grad U = F.

Assuming homogeneous mass distribution inside the Earth, the force is proportional to the distance from the centre:

$$F=-\frac{GMm}{R^3}r$$

To get the potential difference between the centre and the surface of Earth, you have to integrate -F.

$$U(R)-U(0)=-\int_0^R{Fdr}=\int_0^R{\frac{GMm}{R^3}rdr}=\frac{GMm}{R^3}R^2/2$$

$$\rightarrow U(R)-U(0)=\frac{GMm}{2R} \rightarrow U(0)=-\frac{3}{2} \frac{GMm}{R}$$

ehild

Last edited: Apr 8, 2010
3. Apr 9, 2010

### Brilliant

OK starting to understand, but there is something I'm missing. I don't understand what grad U = F means, and also where the R cubed came from. I think the two are probably related.

Maybe something with Fxr = -U? But I still don't see where R cubed will come from. Probably some calculus?

I understand everything you did after that, just now how you decided that
$$F=-\frac{GMm}{R^3}r$$

Thanks, I really appreciate the help!

4. Apr 9, 2010

### ehild

You probably learnt how the potential is related to work. If a force is conservative, the work of this force when a body moves from point A to B, is independent on the path which connect these points. For a conservative force, the work is equal to the negative potential difference: W(A,B)=U(A)-(U(B). The place where the potential is zero can be chosen arbitrarily, but in the case of gravity we chose the zero point of the potential energy at infinity.

The definition of the potential energy at a point A is the work done by the force on a body when it moves from A to the place of zero potential (to infinity, in case of gravity).

The force can vary with position. If this is the case, the work is calculated by integrating the force with respect the displacement,

$$W(A,B)=\int_A^B{\vec Fd\vec r}$$

and the potential is defined at point A with the integral:

$$U(A)=\int_A ^{\infty} {\vec F d\vec r}$$

The gravitational force of a homogeneous sphere of mass M and radius R, at a distance r from is centre is -GMm/r^, as if the whole mass M were concentrated at the centre of the sphere. (The - sign shows that the force points inward.) This is valid when r is greater or equal to the radius of the sphere. If a body is inside this big sphere, only that amount of mass counts which is enclosed inside the sphere of radius r. This can be proved but requires some amount of integration.

If the sphere is homogeneous, the volume of the inner sphere of radius r is V(r) =4/3 r^3 pi, while the volume of the whole sphere is V(R)= 4/3 R^3 pi. The mass of the spheres is proportional to the volume: so the mass of the inner sphere is

$$\frac {m_{sph}}{M}= \frac{r^3}{R^3}$$

It is still valid that the force of gravity is the same as if all the mass were concentrated at the centre, but this mass is msph now:

$$m_{sph}= \frac{r^3}{R^3}M$$

$$F= -G\frac{m_{sph}m}{r^2}=-GM\frac{r^3}{R^3}{r^2}=-\frac{GM}{R^3}r$$

ehild

5. Apr 9, 2010

### Brilliant

Wow

I see the light, though I feel like I should already know this. And you're ignoring the shell of mass that has a radius grater than the falling objects radius as it falls, and only using the sphere on which it sits, basically. I guess all of the mass in the outer shell exerts a net force of 0. That makes sense, I can see it conceptually.

Thanks so much for explaining it all, however, I still don't see how that can be solved in 80 seconds. I guess maybe if you know all this you could eliminate answers.

Anyway, Thanks!

6. Apr 9, 2010

### ehild

These are basic things, usually explained during classes.

ehild

7. Apr 9, 2010

### Brilliant

Well, we have learned all of the concepts, it's just I was having trouble putting all that together. And it seems a bit more in-depth than most other multiple choice problems.

I appreciate the help though.

8. Apr 9, 2010

### ehild

Well, these are not easy to understand, although the formulae are easy to remember. I would not include such question among multiple choice problems.

ehild