Falling Through The Earth

[DJBruce]

So our instructor posed this question to use at the beginning of the year. What would happen if you where to drill a hole straight through the center of the Earth and jump in (all technicalities aside including air resistance).

 

Well just thinking about the problem I think that you you would make it to the exact other side of the Earth because gravity would be decelerating you for the same amount of time as it would be accelerating you. But I want to be able to show this mathematically. So I have came up with this. I would appreciate if people would say if this is right and if it is wrong could you point me in the right direction.

 

V=[math]\sqrt{2GH}[/math]

You would be accelerating till you reach the center of the Earth so you could use the above equation to figure out how fast you are going when you reach the center of the Earth.

V=[math]\sqrt{(2)(9.8)(6356750)}[/math]=3529.74

No I figured once you pass the center of the Earth you could say that you are falling up (or begin shot upwards) and so you could use this equation to figure out how high you will get.

-1/2(G)(X[math]^{2}[/math])+(V[math]_{o}[/math])(x)+H[math]_{o}[/math]

-1/2(9.8)(X[math]^{2}[/math])+(3529.74)(X)+0

When you figure out the maximum you find that the maximum height you could reach is 635666.55 which I feel is with in the realm of error of say that you would make it to exactly the other side of the Earth.

[DrP]

There was a thread on this a while ago - I don't know if this helps?

 

http://www.scienceforums.net/forum/showthread.php?t=32327&highlight=hole+earth

[Mr Skeptic]

You would fall in an unusual orbit, and almost certainly hit the side of the hole.

[swansont]

Is the acceleration due to gravity always going to be 9.8 m/s^2?

[hermanntrude]

You would fall in an unusual orbit, and almost certainly hit the side of the hole.

 

I think the roation of the earth is one of the "technicalities" we're supposed to disregard

[Kyrisch]

You'd yo-yo back and forth through the tunnel, really.

[ecoli]

Is the acceleration due to gravity always going to be 9.8 m/s^2?

 

I don't think so... not if we treat the earth as a point mass. wait, or would it?

[Mr Skeptic]

The mass of the earth that is attracting you would decrease as the cube of (distance from center of the earth divided by radius of the earth). Gravity would become weaker rather than stronger the closer you get to the center.

[swansont]

I don't think so... not if we treat the earth as a point mass. wait, or would it?

 

You can treat the mass located at r<R as a point. The mass r>R is irrelevant. (You are at R). This is from Gauss's law. As Mr Skeptic deduces, this means it's not just the r^2 variation from the formula, because you have to account for M®.

 

But the OP uses a constant, which leads to an incorrect answer.

[hermanntrude]

would the outside of the earth be exerting a gravitational force on you when near the centre? essentially slowing your fall?

[D H]

would the outside of the earth be exerting a gravitational force on you when near the centre? essentially slowing your fall?

No. This is the Shell Theorem first used by Newton. As Swansont mentioned, this also results from Gauss' Law, which is equivalent to Newton's law of gravitation. Wiki article: http://en.wikipedia.org/wiki/Gauss%27_law_for_gravity

[DJBruce]

So our instructor posed this question to use at the beginning of the year. What would happen if you where to drill a hole straight through the center of the Earth and jump in (all technicalities aside including air resistance).

 

Well just thinking about the problem I think that you you would make it to the exact other side of the Earth because gravity would be decelerating you for the same amount of time as it would be accelerating you. But I want to be able to show this mathematically. So I have came up with this. I would appreciate if people would say if this is right and if it is wrong could you point me in the right direction.

 

V=[math]\sqrt{2GH}[/math]

You would be accelerating till you reach the center of the Earth so you could use the above equation to figure out how fast you are going when you reach the center of the Earth.

V=[math]\sqrt{(2)(9.8)(6356750)}[/math]=11162.0921

No I figured once you pass the center of the Earth you could say that you are falling up (or begin shot upwards) and so you could use this equation to figure out how high you will get.

-1/2(G)(X[math]^{2}[/math])+(V[math]_{o}[/math])(x)+H[math]_{o}[/math]

-1/2(9.8)(X[math]^{2}[/math])+(11162.0921)(X)+0

When you figure out the maximum you find that the maximum height you could reach is 6356669 which I feel is with in the realm of error of say that you would make it to exactly the other side of the Earth.

 

I do apologize but I was checking my math and noticed a few errors. I have correct my mistakes.

 

But Kyrisch you are saying that the 81 meter difference is not error but instead will be what causes you to yo yo back and forth until you reach the center of the Earth. My question is why would this happen. Even if gravity changes as you approach the centre of the Earth gravity would be pulling you down (accelerating you) just as long as it is pulling you back to the center (decelerating you). Wouldn't it.

[ecoli]

yeah... that part would be no different from throwing a ball in the air.

 

What would keep you oscillating past the center though... momentum, kinetic energy?

[swansont]

I do apologize but I was checking my math and noticed a few errors. I have correct my mistakes.

 

But Kyrisch you are saying that the 81 meter difference is not error but instead will be what causes you to yo yo back and forth until you reach the center of the Earth. My question is why would this happen. Even if gravity changes as you approach the centre of the Earth gravity would be pulling you down (accelerating you) just as long as it is pulling you back to the center (decelerating you). Wouldn't it.

 

 

You're still using 9.8, and this is incorrect. g isn't constant, it is a function of r.

[Country Boy]

It's easier use "conservation" of energy.All the way from the surface of the earth to the center, your speed, and so kinetic energy is increasing and potential energy decreasing. From the center of the earth on, you speed is decreasing. It will be 0 (and your kinetic energy will be 0) when your potential energy is exactly what it was to begin with- and so you will be the same distance from the center of the earth.

(That is ignoring "technicalities" like air resistance or friction against the side of the hole as well as the fact that the earth is not a perfect sphere. Not to mention what would happen if the other end of hole was under water!)

[DJBruce]

You're still using 9.8, and this is incorrect. g isn't constant, it is a function of r.

 

Yes I understand the 9.8 is incorrect but I do not think it effects the answer becuse when gravity changes on the way down it will change equally as much when you are attempting to escape gravity (the second half of your fall) so you would go just as far.