Math Help: Hole to the center of the Earth

Grandmaster Jogurt · Post by **Grandmaster Jogurt** » 2006-02-21 03:31pm

I'm having trouble wrapping my head around this problem. In it, there is a hole drilled to the center of the Earth, and an object is dropped into it. Assuming the Earth has constant density and there is no air resistance, what would be the object's velocity as it reached the center? I'm supposed to find it using conservation of energy as well as using a first-order differential equation involving dv/dr.

I understand how Gauss' Law affects the acceleration, in that the effective mass of the planet drops linearly as the object's distance from the center decreases, but I'm having trouble expressing this mathematically. Can anyone help?

drachefly · Post by **drachefly** » 2006-02-21 03:38pm

dv/dr? That's not how I'd go. Well, I trust you see how to do it with potential energy, right?

The differential energy gain from pulling out will be the force of gravity at that location times dr. You know how to get the force of gravity at any location using Gauss' law (you said as much). So integrate this differential energy gain. That's the energy method, basically.

dv/dr... I'm not seeing it. It has units of inverse time... a frequency. Well, that connects to another solution I was thinking of...
Once you do the previous part you'll note that the potential inside is that of a simple harmonic oscillator. So you can use the energy of an SHO given this 'spring constant' and amplitude.

Grandmaster Jogurt · Post by **Grandmaster Jogurt** » 2006-02-21 04:08pm

drachefly wrote:dv/dr? That's not how I'd go. Well, I trust you see how to do it with potential energy, right?

The differential energy gain from pulling out will be the force of gravity at that location times dr. You know how to get the force of gravity at any location using Gauss' law (you said as much). So integrate this differential energy gain. That's the energy method, basically.

dv/dr... I'm not seeing it. It has units of inverse time... a frequency. Well, that connects to another solution I was thinking of...
Once you do the previous part you'll note that the potential inside is that of a simple harmonic oscillator. So you can use the energy of an SHO given this 'spring constant' and amplitude.

dv/dr seems weird to me, too, but it's in the question.

Suppose that a hole has been drilled through the center of the Earth, and that an object is dropped into this hole. Write a first-order differential equation for the object's velocity, v as a function of the distance r from the Earth's center (i.e., an equation involvind dv/dr), and solve it to determine the speed the object achieves as it reaches the center of the Earth.

I feel silly for having to ask this, but how would one do CoE? The only thing I could think of is an integral of -G*m*4/3*pi*rho*r^2, since M as a function of r is (4/3)*pi*r^3. Is this correct?

drachefly · Post by **drachefly** » 2006-02-21 04:12pm

Not quite. r^3 / r^2 != r^2

Well, you can get a differential equation for the KE, right? So, substitute in V and there you go, you have a diff EQ for V in r.

SpacedTeddyBear · Post by **SpacedTeddyBear** » 2006-02-21 10:06pm

You actually don't need to do any calculus for this problem. Set the density of the earth and the density of some spherical element inside the earth equal to one another. Solve for the mass of the spherical element. Once you do that, plug that into the force between two masses. You should get something that is inversly proportional the radius of the Earth^3* the distance from the center. Then set that equal to m*a.

SpacedTeddyBear · Post by **SpacedTeddyBear** » 2006-02-21 10:21pm

Whoops, didn't read that about having to use energy conservation and Diff Eqs... gonna get back to ya on that.

Xenophobe3691 · Post by **Xenophobe3691** » 2006-02-21 10:34pm

Is the Diff Eq. you have to use a = v*(dv/dr)?

Eframepilot · Post by **Eframepilot** » 2006-02-21 11:31pm

Using conservation of energy, just use regular mgh as potential energy at the surface and equate to kinetic energy at the center, where the gravitational potential will be zero.

Grandmaster Jogurt · Post by **Grandmaster Jogurt** » 2006-02-22 01:40am

Eframepilot wrote:Using conservation of energy, just use regular mgh as potential energy at the surface and equate to kinetic energy at the center, where the gravitational potential will be zero.

Wouldn't that only be true if Earth was an actual point-source? The amount of force the object feels actually decreases as it gets closer to the center, not increases as it would normally, so I wouldn't think you can use the same equation.

Xenophobe3691 wrote:Is the Diff Eq. you have to use a = v*(dv/dr)?

I'm pretty sure I don't have to use a specific equation unless it's the only one that works. Is that the best way to get an equation involving dv/dr?

Grandmaster Jogurt · Post by **Grandmaster Jogurt** » 2006-02-22 03:12am

drachefly wrote:Not quite. r^3 / r^2 != r^2

Wait a second, isn't Gravitational PE -GMm/r? Where's the r^2 coming from?

Eframepilot · Post by **Eframepilot** » 2006-02-22 06:48am

Grandmaster Jogurt wrote:
Eframepilot wrote:Using conservation of energy, just use regular mgh as potential energy at the surface and equate to kinetic energy at the center, where the gravitational potential will be zero.
Wouldn't that only be true if Earth was an actual point-source? The amount of force the object feels actually decreases as it gets closer to the center, not increases as it would normally, so I wouldn't think you can use the same equation.

The best part about using energy is that it's always conserved under a conservative force like gravity, so the amount of potential energy of the object at the surface will be equal to the amount of kinetic energy of the object at the center. The equation P.E. = mgh is only valid at the surface, where g = GM/r^2, but that's the only place you need to use it.

Kuroneko · Post by **Kuroneko** » 2006-02-22 12:33pm

The acceleration is a = GM(r)/r², but since M(r) = 4πr³ρ, where ρ is the density of the Earth (assumed uniform), a = 4πGρr. Since dv/dr = (dv/dt)(dt/dr) = a/v = 4πρr/v, we have trivial separable ODE: v dv = 4πρr dr. Integrating this gives the specific kinetic energy in terms of r.

drachefly · Post by **drachefly** » 2006-02-22 02:42pm

Grandmaster Jogurt wrote:
drachefly wrote:Not quite. r^3 / r^2 != r^2
Wait a second, isn't Gravitational PE -GMm/r? Where's the r^2 coming from? :?

You are correct that grav PE is -GMm/r, but you were about to integrate over the Force, and that has an extra r in the denominator.

And of course Kuroneko is right. This time, his solution was genuinely obvious and simple. Not that I'd normally choose to do it that way anyway...