A tunnel through the earth

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Levi Tate
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Levi Tate
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Homework Statement



A tunnel is drilled straight through the earth from Detroit to Zxxiuw. At the middle of the tunnel it is at 1/2 earth radius from the center. Assuming that a mass m placed in the tunnel will slide without friction, show that it experiences a force that is directly proportional to its distance from the center of the tunnel. Give an expression for this force in terms of the distance and the weight mg at the surface. Assume the earth is not rotating, a bad approximation, but hey, you're not going to drill this tunnel anyhow.


I am thinking I need to use energy and then take the derivative to get the force, but I am having trouble finding out the right assumptions to make such that I can do that.


That there is the problem, I am on an iPad so in the attachment is a picture of the problem as well as what I've been able to work out.

Thank you in advance
 

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  • #3
xirow
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id be very impressed if anyone could build a tunnel through the earth, but if it was perfectly through the center of the earth id imagine one would get stuck in the middle
 
  • #4
davenn
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id be very impressed if anyone could build a tunnel through the earth, but if it was perfectly through the center of the earth id imagine one would get stuck in the middle

why do you think that ?

Dave
 
  • #5
xirow
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when you reached the very center of the tunnel the majority of gravitational force applied on you would be equal on all sides (theoretically speaking, if it was perfectly in the center) then as you moved to towards one end the majority of the gravity from the earth applied to you would be coming from the opposite side of the tunnel you were on, so you would begin to fall back the direction you came. (This is why the movie total recall confused me so much.)
 
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  • #6
SteamKing
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I wouldn't worry about that. You'd burn up or be fried by radiation long before you got to the center of the earth.
 
  • #7
ehild
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Homework Statement



A tunnel is drilled straight through the earth from Detroit to Zxxiuw. At the middle of the tunnel it is at 1/2 earth radius from the center. Assuming that a mass m placed in the tunnel will slide without friction, show that it experiences a force that is directly proportional to its distance from the center of the tunnel. Give an expression for this force in terms of the distance and the weight mg at the surface. Assume the earth is not rotating, a bad approximation, but hey, you're not going to drill this tunnel anyhow.

The problem is to prove that the force is directly proportional to the distance between the mass and the centre of the tunnel. What forces act on the mass when it is at a certain position in the tunnel? Recall that the force of gravity at distance r from the Earth centre is [STRIKE]equal[/STRIKE] proportional to the mass of the sphere with radius r, assuming homogeneous mass distribution of Earth.

ehild
 
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  • #8
Dick
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Homework Statement



A tunnel is drilled straight through the earth from Detroit to Zxxiuw. At the middle of the tunnel it is at 1/2 earth radius from the center. Assuming that a mass m placed in the tunnel will slide without friction, show that it experiences a force that is directly proportional to its distance from the center of the tunnel. Give an expression for this force in terms of the distance and the weight mg at the surface. Assume the earth is not rotating, a bad approximation, but hey, you're not going to drill this tunnel anyhow.


I am thinking I need to use energy and then take the derivative to get the force, but I am having trouble finding out the right assumptions to make such that I can do that.


That there is the problem, I am on an iPad so in the attachment is a picture of the problem as well as what I've been able to work out.

Thank you in advance

What you really need to know is some facts about Newton's law of Gravitation, and you need to assume the mass density of the earth is constant everywhere.
 
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  • #9
Levi Tate
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I posted this thread over in the homework help thing if a anybody wants to help me figure out how to solve the idealized problem!
 
  • #10
xirow
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Whoops I've been posting in the wrong page. This question reminds me of total recall. as you approached the end of the tunnel the force gravity has would start to fight against the object. because there is no friction it should just continue to accelerate then decelerate as it passes the center of the tunnel. the distance it is accelerating is equal to the distance it accelerates (similar to throwing a ball in the air and having it come back down but the reverse). so it would fall (or roll) all the way to the center while while its acceleration steadily decreases. then start decelerating at the center of the tunnel. it would continue to decelerate more and more frequently until it stops when its decelerating at its maximum, then start to do the same thing in reverse.
 
  • #11
xirow
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its acceleration towards the opposite direction its in is equal to its distance from the center of the earth divided by earths radius times 9.8m/s/s
 
  • #12
Dick
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its acceleration towards the opposite direction its in is equal to its distance from the center of the earth divided by earths radius times 9.8m/s/s

Nothing you've said proves that, now does it?
 
  • #13
Levi Tate
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Xirow, that was my intuition about the subject.

I know that the gravitation stems from the center of the sphere, and I know he initial PE because I know the radius, then I am just visualizing the radius of the sphere constantly decreases continuously, so that to approaches 1/2 radius of sphere.

I played around with the mass density, which I am not sure of, is it just d=m/v? Or the infinitesimal in case I end up integrating something? I'm trying to look at this equality here, because right at the center I reason the mass has zero speed instanteously, so..

MgRe= 1/2mv^2+mgr= 1/2mgRe

And somehow using that double equality, that is as far as I can think, but I don't see how that would fit in with the mass density, that seems to imply using an integral and force methods, but I could very well be wrong.
 
  • #14
Levi Tate
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I am referring to the mass, the potential energy of the mass decreases as it moves towards the center of the sphere, sorry for writing that so ambiguously.
 
  • #15
xirow
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Nothing you've said proves that, now does it?
sorry should have explained this more, before i was explaining the end result. The reason for the acceleration is because the force of gravity is based on the amount of mass on the other side of the object. From the surface of the earth there is an acceleration of 9.8 m/s/s as it moves closer to the center of the earth it decreases. then becomes zero at the center of the tunnel. then as it is at the other end of the tunnel it has an acceleration of 9.8 m/s/s in the other direction
 
  • #16
Levi Tate
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Right xirow that's the general idea of what's going on but the hard part is solving the problem.
 
  • #17
Levi Tate
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Okay so I just realized there is not zero kinetic energy at the center of the sphere, so that changes my energy equation, mgRe=1/2mv^2+mgr=1/2mgRe plus another kinetic energy term but those are both unknown velocities, the only thing I can think to do is cancel the kinetic energy and integrate, even though that doesn't give the force. I'm lost here.
 
  • #18
Dick
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Right xirow that's the general idea of what's going on but the hard part is solving the problem.

Then pay attention to ehild's more detailed clue. The acceleration of a mass at radius r is only coming from the mass inside of r. There is a calcuation to do.
 
  • #19
ehild
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The gravitational acceleration is not g inside the Earth, the force of gravity is not mg, and the gravitational potential energy is not mgr.

Derive the expression for the force of gravity at distance r from the Earth centre.

ehild
 
  • #20
Levi Tate
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I have that expression and tried to integrate using the second law, but I got a pretty ridiculous solution to the differential equation
 
  • #21
Levi Tate
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Cos I have to write v=dv/dt in one dimension don't i?
 
  • #22
ehild
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I have that expression and tried to integrate using the second law, but I got a pretty ridiculous solution to the differential equation

What expression do you have?

Cos I have to write v=dv/dt in one dimension don't i?

v can not be equal to dv/dt, they are not of the same dimension.

Is "t " time? What do you call "v" at all?

ehild
 
  • #23
Levi Tate
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I just used Newton's second law, GMem/R^2 = mdv/dt, then separated time to the LHS and integrated, but that this point I'm just kind of playing around with it, I actually got so mad about it I went onto another problem.
 
  • #24
Levi Tate
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V is a velocity in 1d, t is time.
 
  • #25
Levi Tate
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V=dr/dt I meant there
 
  • #26
Levi Tate
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Or excuse me a=dv/dt. I left the problem for a moment, but I feel like I should be using energy. I don't see a way to express the force unless I use spherical coordinates but that makes me a very hard problem and I am not if I should be in 3d.
 
  • #27
Dick
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Newton says that if you are inside of a spherically symmetric mass distribution at radius r, like an ideal earth, then you can calulate the force due to gravity by treating all of the mass below r as concentrated at the single point r=0 and you can ignore the mass above r. So the force is F=GM(r)m/r^2, where M(r) is the mass of the material below radius r. Assume the earth has a constant density ρ. What's M(r)? There are no differential equations involving v and a to solve here.
 
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  • #28
ehild
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If you mean R the radius of Earth than GMem/R^2 is the force of gravity on the surface of Earth, and dv/dt is the acceleration on the surface of Earth, that is g≈9.8 m/s2. The problem does not ask velocity, why do you want to integrate it?

You need the force of gravity inside the Earth. Proceed according to Dick's hint.

ehild
 
  • #29
Levi Tate
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Thanks a lot fellows, I am going to try to play around with this here after I eat breakfast and see what I come up with.
 

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