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 Motorcycle Safety
 Physics and the theoretical
 Hypothetical gravity question
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scottrnelson
Advanced Member
6882 Posts
[Mentor]


Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 12/02/2013 :  7:09 AM                       Like
I woke up this morning wondering about this - I don't know why.

If the earth wasn't hot in the middle, and was solid, not liquid everywhere inside, and we had some tunnels going down to the middle, and we had a 4000 mile elevator that could take you down all the way to the center....

I'm thinking that you would have zero gravity right at the center, but I can't give you the math explaining why. So if this were the case, where would the gravity be 0.5G? 2000 miles down?

What would the air pressure be at the center if it's 14.7 psi here?

Bonus question - how much more power would a 100 hp motorcycle have down at the center with that added air pressure?

And if there were a 100 mile diameter smooth "air sphere" down at the center, how fast would you have to ride that motorcycle to maintain 1G of outward force to hold you against that reverse surface down there? (What would the "orbit speed" have to be?)


That is all...

greywolf
Male Moderator
1484 Posts
[Mentor]


Evanston, IL
USA

Suzuki

DL650AL2

Posted - 12/02/2013 :  8:46 AM
In the center of the earth, the gravitational pull is equal in all directions. An object in the center in your theoretical column would float there as the sums of all earthly gravitational forces would be zero. Every location in the mass of the earth causing a gravitational pull will have a location of equal mass, distance, and opposite direction pulling the other way. Air pressure on the surface of the earth is due to the weight of the column of air above the surface due to gravity. That should mean the air pressure in the center would also be zero. The deeper into the hole you go, the more the mass above you is pulling up until equilibrium is reached at the center.

The point where you reach .5g of gravity and .5 atmospheres of pressure is best determined using calculus and I gave that up in college almost 40 years ago.
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James R. Davis
Male Administrator
17279 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 12/02/2013 :  9:38 AM Follow poster on Twitter  Join poster on Facebook as Friend  
I suspect that for all practical purposes, the deeper you go, the less you will weigh. On the other hand, the deeper you go, the more dense the environment material will be, so at least for some depths, you will find gravity increasing as you get closer to that more dense environment.

No idea how to calculate the depth at which you would find 0.5g's of gravity other than to suggest that nothing humans can now do could possibly get you that deep below the earth's surface.
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Alan_Hepburn
Male Standard Member
191 Posts


San Jose, Ca
USA

Honda

1994 GL1500SE

Posted - 12/02/2013 :  1:13 PM
I read something a while back that was interesting - if you could drill an elevator shaft through the Earth from point A to point B then the elevator would not need a cable to move it. It would accelerate for the first 1/2 of the journey and then decelerate for the second half of the journey, arriving at the other end with a smooth stop, all powered by gravity. And the shaft does not need to go through the center of the Earth - any shaft drilled in a straight line, say from LA to NY, would work. I don't pretend to understand the science behind it all!
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scottrnelson
Advanced Member
6882 Posts
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Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 12/02/2013 :  1:24 PM
quote:
Originally posted by Alan_Hepburn

It would accelerate for the first 1/2 of the journey and then decelerate for the second half of the journey, arriving at the other end with a smooth stop, all powered by gravity.
If we didn't have that pesky thing called friction, this would be a great way to travel fairly cheaply.

I think people have already proposed using tubes with vacuums inside and some sort of magnetic suspension of train cars to significantly reduce friction. The biggest issue holding that up from being used is the cost of building the thing.
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Andrew Dressel
Male Standard Member
244 Posts


Milwaukee, WI
USA

Moto Guzzi

California Special

Posted - 12/03/2013 :  7:41 PM
quote:
Originally posted by scottrnelson

I'm thinking that you would have zero gravity right at the center, but I can't give you the math explaining why. So if this were the case, where would the gravity be 0.5G? 2000 miles down?



According to this article, your weight decreases linearly from the surface to zero at the center. Thus, since the radius of the earth is about 4000 miles, you would experience 0.5 g at about 2000 miles down.

Calculating air pressure is trickier, as this paper shows because of the dependence on temperature.

The bonus question depends on density, which is a function of pressure and temperature, I believe, so also hard to calculate.

Finally, the "orbit speed" is easy: a = v^2 / r so v = sqrt(g*r).
Using g = 32.2 ft/s^2 and r = 528000 ft yields v = 4123.3 ft/s or 2811.341 mph.
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Cash Anthony
Female Administrator
1466 Posts
[Mentor]


Houston, Texas
USA

Honda

Magna 750

Posted - 12/04/2013 :  8:08 AM Follow poster on Twitter  Join poster on Facebook as Friend  
Scott wrote
quote:
I think people have already proposed using tubes with vacuums inside and some sort of magnetic suspension of train cars to significantly reduce friction. The biggest issue holding that up from being used is the cost of building the thing.


Here's some information about the Hyperloop train, straight from Elon Musk, the originator of the idea in its full-blown version. If you want the details, he's put out a 57-page plan that you can read here, or a detailed summary of it here.

He describes it this way: "It's a cross between a Concorde, a rail gun, and an air hockey table." I can hardly wait to ride it!
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scottrnelson
Advanced Member
6882 Posts
[Mentor]


Pleasanton, CA
USA

KTM

990 Adv, XR650L

Posted - 12/04/2013 :  8:49 AM
quote:
Originally posted by Andrew Dressel

Finally, the "orbit speed" is easy: a = v^2 / r so v = sqrt(g*r).
Using g = 32.2 ft/s^2 and r = 528000 ft yields v = 4123.3 ft/s or 2811.341 mph.

Except that g = 32.2 would be way too strong at 50 miles from the center.
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James R. Davis
Male Administrator
17279 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 12/04/2013 :  9:21 AM Follow poster on Twitter  Join poster on Facebook as Friend  
LOL

Doing a quick mathematical treatment on a website invariably leads to criticism.

You did ask for the speed at which 1g would be obtained, did you not?

His quick math error was that he used the diameter of the space instead of the radius.

So sue him.
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JanK
Male Junior Member
76 Posts


Ljubljana, Ljubljana
Slovenia

BMW

F650CS

Posted - 12/04/2013 :  5:00 PM
Gravity would decrease linearly ONLY if the density of the Earth were uniform.

Gravitational acceleration on the surface of a spherical body is g = M/R^2, where M is the mass of the body and R is the distance from the centre of mass of the body (i.e., from the centre of the earth). Knowing the volume of a sphere, 4/3*Pi*R^3, you can express mass as density (rho) times volume: M = rho*4/3*Pi*R^3 and substitute it into the gravitational acceleration expression, obtaining g = rho*4/3*Pi*R. This expression is valid for any homogeneous spherically symmetrical object, as long as you are on the surface.

Another fact (obtained with a bit of calculus) is that the gravity field inside a spherical shell is 0. Not only in the centre of the shell, but everywhere in its interior.

Once you get below the surface, you can divide Earth into two regions, separated by a sphere with the radius (let's call it r) equal to your distance from the centre of the earth. The region above you is a spherical shell and does not exert any acceleration. The region below you is again a sphere with a smaller radius (r, as opposed to R). The acceleration is g' = rho*4/3*Pi*r. So the gravitational acceleration would fall off linearly and 0.5g would occur at half the radius.

But, if the Earth's density increased, the more you traveled towards the centre, then the acceleration will fall off less quickly than linearly and, as James has pointed out, could even increase at certain radii, depending on the details of density as the function of the radius.

A fun calculation: assume that the Earth is made up of only two materials
- an inner dense core, whose radius is half the Earth's radius, r=R/2, with density rho and
- an outer, less dense shell, whose density is such, that the average density is half the core's density, rho'=rho/2.
Then the gravity at the interface is g = rho*4/3*Pi*(R/2) and gravity at the surface is g = (rho/2)*4/3*Pi*R = rho*4/3*Pi*(R/2) = rho*4/6*Pi*R. In other words, they are the same. Which also implies that the gravity is the highest at some point between the surface and the core.

In fact, if you check out https://en.wikipedia.org/wiki/File:...vityPREM.svg, you will see that, if you use an actual model of Earth's density, the gravitational acceleration does increase (for the most part) until you descend to around 3500km from the centre. And at some point (around 3000km from the centre) it is the same as on the surface.

Edited by - JanK on 12/04/2013 6:13 PM
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James R. Davis
Male Administrator
17279 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 12/05/2013 :  6:58 AM Follow poster on Twitter  Join poster on Facebook as Friend  
Thank you! You are quicker and more rigorous in your explanation than I was - I intended to put forth only a 'shorthand concept' explanation.

Here is a diagram that attempts to clarify those concepts.



The image at the right side suggests that the Earth is NOT a hollow shell, but that for some distance below the surface it is less dense than it is towards its center.

In other words, gravity would TEND to increase as you move from the surface of the 'shell' until you reached the more dense domain.
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greywolf
Male Moderator
1484 Posts
[Mentor]


Evanston, IL
USA

Suzuki

DL650AL2

Posted - 12/05/2013 :  9:00 AM
Here's some idea of composition and measures.

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