Speed through the Center of the Earth

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1. Apr 4, 2015

Okazaki

1. The problem statement, all variables and given/known data
Suppose we had a straight tunnel, through Earth's center, to a point on the opposite side of the planet, and used it to deliver mail to the other side. With what speed would our packages pass through Earth's center.

2. Relevant equations
ag = (GM)/R2
Mins = (4/3)πR3

3. The attempt at a solution
So, I'm assuming (but not completely sure) that the question is asking for the man's speed at the center (but again, I could be interpreting the question completely wrong. Either way, I'm still stuck on how to solve it.)

I figured that I could take the integral of ag with respect to R (since the distance from the center of the earth is continuously changing as you pass through the center.)

So: ag = ∫[G*ρ*(4/3)π *R3]/R2
(from 0 to R)

My problem here is that ρ is just M/V which is 3M/(4πR3), which basically cancels out a whole bunch of R's and leaves you with:

ag = GM∫[1/R2
(from 0 to R...I'll worry about signs later)

AND:
ag = GM/R, which doesn't actually work out since you end up having m2/s2, which is not acceleration.

I know I probably messed up the integral, but I'm not sure how else you would solve this problem.

2. Apr 4, 2015

collinsmark

There are two radii to consider (this keeps much of the cancellation from happening):
• There's the average density* calculation which you have already done, $\rho = \frac{M}{V}$, where $V = \frac{4}{3} \pi R^3$. Here, this $R$ is constant, and doesn't change as the position of the mail package progresses. You have already calculated this in your original post. (And of course, the mass of the Earth is constant.)
• Then there's the mass of Earth "underneath" the mail package. And the volume of earth "underneath" the package is $\frac{4}{3} \pi r^3$. This $r$ changes with the mail package's position, however. So this mass underneath is this volume times the previously calculated $\rho$. We're making the assumption that this $\rho$ density is constant and does not change as a function of the mail package's position.*
What I haven't discussed is how the "shell" of earth above the mail package fits into the picture, if at all. Can you comment on what the gravitational forces felt by an object inside a massive hollow, spherical shell?

*In this problem we are assuming that the Earth has a uniform density. This assumption isn't true, but it's not too horrible an approximation. Since we're already ignoring the Earth's rotation, friction and such anyway, it should be okay.

Last edited: Apr 4, 2015
3. Apr 4, 2015

Okazaki

Ok, that makes sense. I'll try to work that out now.

I've only briefly discussed shells. But in the case of the earth, once you get into the innermost shell (of the Earth, in this case) the net force acting on the object is 0. So in a hollow, spherical shell, once you get past the outer layer, the net force is 0?

4. Apr 4, 2015

Okazaki

I tried it again, it still doesn't work. No matter what I try, too many R's end up cancelling out:

Even if I set it so that Mins = ρ * (4/3)π*r3
w/ ρ = 3M/(4πR3)

ag = (4/3) π Gρ ∫ r3dr/r2
(from 0 to R)

ag = (4/3) π Gρ ∫ rdr
ag = (4/3) π Gρ * R2/2
ag = (4/3) π * G * 3M/(4πR3) * R2/2

...Do you see what I mean?

5. Apr 4, 2015

Staff: Mentor

In order to determine velocity by integrating acceleration you must integrate w.r.t. time. Integrating acceleration as a function of position is not correct.

Rather than deal with the chain rule and mucking about with resulting equations, why not consider a conservation of energy approach? What's the work done in moving an object from the surface to the center? You still need to deal with the mass of the planet below the current position in order to determine the force as a function of position.

6. Apr 5, 2015

Okazaki

I'm just really confused about how specifically to deal with the mass of the planet below the current position.

7. Apr 5, 2015

Staff: Mentor

How so? The mass below the current position is the mass contained within the spherical volume with radius equal to that of the current position.

8. Apr 5, 2015

Okazaki

I mean, how would I integrate with time when I don't even know the time it will take to get down to the center of the earth?

9. Apr 5, 2015

Staff: Mentor

That is why I suggested using conservation of energy (work-energy theorem) instead. Then time is not involved.

10. Apr 5, 2015

Okazaki

...So you don't use an integral, then? But how do you deal with the changing mass below you?

11. Apr 5, 2015

Staff: Mentor

No, you must use an integral. You want to compute the work done by gravity while moving a test mass from the surface to the center.

12. Apr 5, 2015

Staff: Mentor

It is closely related to the approach via energy conservation - and I agree that it is the easiest approach here.

13. Apr 7, 2015

Okazaki

I still am completely lost. I don't understand how to make this integral, nor how to integrate it.

14. Apr 7, 2015

Staff: Mentor

What is it you need to integrate to find work?

15. Apr 7, 2015

Okazaki

The potential energy of an object on the earth's surface down until the center of the earth?

16. Apr 7, 2015

Staff: Mentor

Not quite.

Work = Force x Distance, right? The work done in moving a test mass from one position to another without allowing it to accelerate will be what you want. (Think of the test mass as being moved at a constant velocity, applying a force that exactly balances the gravitational force so no acceleration takes place)

So the work done by gravity is equal to the integral of $F_G(r) \cdot dr$. You need to find an expression for $F_G(r)$ to integrate. This is where the mass of the planet below the current position comes into play so you can use Newton's Law of Gravitation to find that force at any radial position.

17. Apr 7, 2015

Okazaki

So:
W = ∫(FG(r)*dr
(from 0 to R...I'll figure out the signs later)

W = ∫(GMm)dr/r2
W = ∫(GρVm)dr/r2
W = Gρm ∫(4πr3)dr/(3r2)
W = Gρm * (4π)/3 * ∫rdr
W = Gρm * (4π)/3 * R2/2
W = Gm * (ρ) * (2π)/3 * R2
W = Gm * (3M)/(4πR3) * (2π)/3 * R2
W = Gm * M/(2R)

...And that's the work?

18. Apr 7, 2015

Staff: Mentor

Looks okay to me. How does that amount of work translate to a velocity for an object that undergoes the same change in location via free-fall?

19. Apr 7, 2015

Okazaki

Can't you use the kinetic energy theorem (0.5mv2)?

20. Apr 7, 2015

Staff: Mentor

Try it and find out. What answer do you get?