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Nigeyboy

A Question of Gravity

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OK – got a bit of  strange question! I’m pretty clued up on physics etc, and although I won’t be challenging Prof Hawkins any time soon I can’t work out the answer to this hypothetical question!!

Imagine you drilled a large hole right through the Earth, so it passed through the exact centre point of the planet. Also imagine that some amazing shielding protects you from the heat etc. Now imagine jumping down that hole . . . . .

 . . . . you would be falling down for quite some time. However, when you shot past the centre point you will be falling ‘up’ in relation to Earths gravity. This would then slow you down and pull you back. The question is, what would happen when you stopped shooting back and forth and were held at the exact centre point of earths gravity?

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You would sit there at the centre.

It would be a long climb out.

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I'd say in this scenario you wouldn't have any friction so you would be going back an forth indefinitely...., but if hypothetically you had some friction and eventually stopped in the center, then you wound be in a gravitational equilibrium getting pulled equally in every direction so you would have the same effect as if in zero G.. just float there.

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I guess the total force on your body would be zero. I don't know how to describe in english xD

ΣF=0 on every axis and direction. The center of the earth is the center of the gravity so if you are in the center the no force on your body. But I think that the center woulb be like a point of like 1cm.

Anyway, I think don't really know, I am not good at physics

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So theoretically your body would be pulled in each direction from the centre point at a force of 1g?

Would  make a fun ride!! I worked out (possibly incorrectly!!)  that if air resistance stayed the same all the way down it would take about 32 hours to reach the middle:

Radius of Earth (3950miles) / terminal velocity of a human in freefall (124mph) = 31.8hours

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Not so fast... (literally)

It would take a lot longer that 31.8 hours..

As you travel towards the centre, the pull of gravity to the centre would get less because lots of mass would now be above you.

Also, the deeper you travel, air pressure would increase, so increasing the density, leading to an increase in resistance, slowing you down even more.

If you remember the guy jumping out of a baloon from high up went supersonic, due to reduced air resistance?

Been a long day at work, too tired to attempt the maths, but happy for someone else to take a stab at it.

Would be interesting.

Gordon.

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Nigeyboy, that's not a daft question :smile: . I was contemplating a similar thing a year or so ago, and turning to the web I came up with the intriguing fact that the gravitational force inside a massive hollow sphere is everywhere zero, strange though it may seem! Outside the sphere, of course, the force of gravity is proportional to the mass of the sphere and falls inversely as the square of the distance from its centre (i.e. inverse-square law).

Inside a solid sphere, you can consider the problem in two halves. First, the force of gravity from all the matter above you will be zero (from the above). Second, for the material in the sphere below you, the force of gravity will be proportional to its mass. So if the sphere is of uniform density, the force will be less than the force from the whole sphere. The closer you get to the centre, the smaller the sphere beneath your feet, the smaller its mass, and so the gravitational force will be smaller also. In other words, as you progress towards the centre of the earth, so the gravitational force will reduce until, as others have said, you reach the centre when it'll be zero. One would imagine then that as you fall towards the centre, you would go slower and slower. So would you ever reach the centre? Or would your inertia cause you to overshoot and then slowly oscillate about the centre?

Ian

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Some really interesting explanations here, and things I hadn’t considered – like gravity decreasing as you descend!

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Actually, thinking about it some more, I'm not sure that you would go slower and slower. Hmm, I'd need to get my creaking brain around this

Ian

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I wonder what the effect of differing air pressure on each side of the globe would be, would there be a howling gale inside the hole or a partial vaccum or something realy weird?

Alan

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Interesting thought, but I don't think I'd want to speculate on that ☺. I can imagine all sorts of physics coming into play.

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As you descend the force upon your body steadily decreases. But the acceleration (and the velocity of course) is there. So I guess that you pass the center and then the gravity is steadily increasing towards again the center. 

At some point due to the friction with the air molecules you would stop, but not at 31.2 hours :x

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Interesting thought experiment, i foresee an issue of the center of mass been shifted due to the hole and thing you may collide with the side wall and not fall in a straight line. unlike a magnetic field which can hold you in a confined path i dont thing gravity is strong enough to do that.

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If I've read the article properly, I'd like to take issue with one of his conclusions.

As the mass in the sphere beneath your feet varies as , and the force between two masses varies as 1/, then as r (the distance you are from the centre of the earth) becomes smaller, the overall effect will be that the force of gravity will vary as r, which is the same as stated in the article. In other words, the acceleration due to gravity starts off at the earth's surface as 9.8m/s, and falls linearly to zero at the earth's centre. However, I do take issue with the determination of period, because the value of "g" used in the expression T=2π√R/g varies with r, so I don't think it is correct to use 9.8m/s for this value. The effect of the reducing force on you as you journey to the centre will be to increase the duration of the period. A good many years ago I'd be able to work it out, but my little grey cells are beyond repair now :sad2:.

Ian

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This is a standard physics question. If you could make a tunnel right through the Earth, and we assume there is no air in the tunnel (hence no air resistance), then once you jumped down the tunnel you would travel right to the other end, then fall back through again. You would continue in this oscillatory way. The period of oscillation would be the same as if you orbited the Earth at the height from which you made your initial jump (i.e. ground level - though it would be pretty hard to orbit the Earth at ground level). If there's air in the tunnel then you will gradually slow down due to air resistance, and will eventually come to rest at the centre. You'll then need to climb back out, having a choice which way to climb. If you were to do the calculation carefully, you would need to take into account the variation of air density as a function of distance from the centre. You would also need to decide what to do about temperature - the centre of the Earth is pretty hot, so you'd need to decide how "realistic" to make it. All of that would be pretty complicated and almost as tedious as climbing out, and I don't have a big enough envelope handy, so I'll pass.

Edit: Apparently it's called a "gravity train".

https://en.wikipedia.org/wiki/Gravity_train

According to the article:

  • The time of a trip depends only on the density of the planet and the gravitational constant, but not on the diameter of the planet.
  • The maximum speed is reached at the middle point of the trajectory.

On the planet Earth specifically, a gravity train has the following parameters:

  • The travel time equals 2530.30 seconds (nearly 42.2 minutes), assuming Earth were a perfect sphere of uniform density.
  • For a train that goes directly through the center of the Earth, the maximum speed is about 7,900 metres per second (28440 km/h).
Edited by acey
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I would imagine that unless your tunnel is pole to pole, you're going to get thrown against the side by coriolis force. Ray

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even if in a vacume, gravity is not an equal force (especially over the diameter of the earth), it will vary, which will casue movement in the tragerctory, is may settle into a pattern as long as no colssions occure with the side walls.  but it wont be a straight up and down.

Edited by Earl

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I would imagine that unless your tunnel is pole to pole, you're going to get thrown against the side by coriolis force. Ray

Coriolis effect is due to difference in angular momentum at different parts, so is only significant over large distances (e.g. weather systems). No effect on something the size of a person - or water going down a plughole.

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Coriolis effect is due to difference in angular momentum at different parts, so is only significant over large distances (e.g. weather systems). No effect on something the size of a person - or water going down a plughole.

Thanks for the feedback Acey, I appreciate the point you make, and I really should try and express myself a bit better.

However, this is a fascinating topic and I couldn’t help wondering what would happen if the tunnel were not pole to pole (the OP doesn’t stipulate this).

So what if there were such a tunnel, 10m diameter with it’s entrance and exit close to the equator?

If ‘Nigeyboy’ were stood at the western rim, happily co-rotating at about 7.27 x 10-5 radians per second, and we convinced him to make that one small step for man, would his angular momentum not cause him to drift towards the eastern side wall, making contact after falling a mere 6 km?

I think (although may be mistaken) that were he not confined by the tunnel wall, his trajectory in this case would be elliptical (Keplerian).

Would welcome a better explanation since my only experience with anything like this comes from watching gyroscopic inclinometers used to profile 5km deep boreholes.

What I think I understand came from:

https://www.bibnum.education.fr/sites/default/files/137-analyse-laplace-en.pdf

Ray

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Taking the question one step further: if after jumping into our tunnel and all the oscillating, we are now at rest at the centre of the earth. If I steady myself and have a small platform to stand on. How high could I jump in this "zero gravity" environment?

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Taking the question one step further: if after jumping into our tunnel and all the oscillating, we are now at rest at the centre of the earth. If I steady myself and have a small platform to stand on. How high could I jump in this "zero gravity" environment?

Interesting thought, 2 options I would guess:

You jump and at t=0 you have G=0 so off you go apparently forever.

As you get higher there is steadily more mass "below" you then above so you begin to slow down, in effect as gravity gets greater you get dragged back.

A slight difference to leaping off the earths surface as the higher you go the less the pull of gravity is.

The mathematics of your jump at the centre of the earth is likely to be dependant on r2 not r-2.

The other option is at any distance from the centre then you are no longer in a zero gravity environment, and lets leave it at that. :grin: :grin: :grin:

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I'm with Ray above.

The Coriolis effect will have our extreme base jumper rolling along the walls of the tunnel as he is forced against it by his ever changing angular momentum.

Not so much about the size of the object but the distance and rate of it's travel perpendicular to the direction of rotation. The magnitude of the Coriolis effect would be in proportion to the angle the tunnel makes with Earth's rotational axis. 

I don't expect it to be a violent exchange for our traveler (can't do the maths!) but, in an equatorial tunnel, he'll be slowed from a thousand miles per hour ( the rough velocity of an object sat at the equator by virtue of Earth's rotation) to zero on his inward fall. As he overshoots the center he'll be whipped back up to speed again!

I'd rather do the equatorial journey. I've had enough of lousy cold weather...

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