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Allright...you Geology and Physics Guys
- Thread starter Lonesomedave
- Start date
The total tidal force vertical component at the surface of the Earth for a person weighing 150 pounds is in the region of one-seventh of an ounce. Not really noticeable.
Of course, there wouldn't be a totally abrupt change to half of this value the instant you entered an actual hole, because the hole itself would cause very slight distortions in the immediate gravitiational field. My general relativity mathematics is not anywhere even close to calculating this (in fact, it's non-existent). The figures in the table represent the Newtonian calculations for being on the surface of a uniformly solid perfect sphere, or actually inside a uniformly solid perfect sphere.
I would imagine that in practice, if you could measure the actual tidal force at the surface and then measure it again at some point down the hole, you would get values fairly close to these approximations.
john turner
You don't want to do that. Trust me.
actually, you guys have been correct with your end results but not precisely with your reasons. check out the shell theorem
the hollowed center of the earth is equivalent to a shell, so our thought experiment is covered by these rules. even for a deformable body made up of some collection of point masses (such as our explorer's body) every point mass within the "shell" would have no net force exerted on it.
i'm still not completely convinced that no net force is the same as no force (if i exert a (0,1g,0) force (conteracting gravity (0,-1g,0) on a piece of jello in my hand, it has no net force exerted upon it, but it will still deform, even at point of force application)
the hollowed center of the earth is equivalent to a shell, so our thought experiment is covered by these rules. even for a deformable body made up of some collection of point masses (such as our explorer's body) every point mass within the "shell" would have no net force exerted on it.
i'm still not completely convinced that no net force is the same as no force (if i exert a (0,1g,0) force (conteracting gravity (0,-1g,0) on a piece of jello in my hand, it has no net force exerted upon it, but it will still deform, even at point of force application)
Newtonian calculations....weights quoted in pounds and ounces....ouch
Stones and grains would have been historically correct?
I converted for the sake of our American pals. But I did the calculations using SI (see table above). I wouldn't know how to even attempt them otherwise!Newtonian calculations....weights quoted in pounds and ounces....ouch
But... what units do you think old Isaac would have used?
I converted for the sake of our American pals. But I did the calculations using SI (see table above). I wouldn't know how to even attempt them otherwise!
But... what units do you think old Isaac would have used?
Probably 'Aristotles'
But John, our explorer is not just "inside a shell". At a point, say, halfway from the surface to the centre, he is inside some of the Earth's "shells" but outside others, and will definitely experience a net force (towards the centre). For a solid sphere rather than a hollow shell, the only place where the net force is zero is at the exact centre. See post #45.
The size of those net forces on different parts of our guy's body are what we've been considering in recent posts (on and off since about post #63).
(And for the jello, it has a net force acting on some points of its mass during the deformation process. When it has settled into its new shape, there is, as you say, no longer a net force anywhere, as at each point, its weight is counteracted by contact with your hand or elasticity in the material. If it wasn't - for example if the stuff was too soft - it would continue to accelerate downwards under gravity and slop over the edges of your hand.)
Bill, I can just about forgive you using pounds and feet for the benefit of our American friends, but come on....jello? It's jelly and you know it
Sorry.Bill, I can just about forgive you using pounds and feet for the benefit of our American friends, but come on....jello? It's jelly and you know it
Wow. Flashback to the rescue of the Chilean miners
... only that was 2,000 feet. I'd've gone down it though - it was in hard rock.
To be honest, it wasn't so much the depth that prompted the memory, it was the comment about being in a very narrow bore-hole. I still vividly recall that vessel (the Phoenix?) that they sent down to pull them out one at a time. An engineering marvel.
So Bill, as they were brought towards the surface - at steady velocity - there will have been an increasing mass of Earth below them (concentric spheres again)...so their weight would have been increasing until they returned to the surface, yes?
Yes, true - but not so as you'd notice, of course. Definitely measurable, though.
2000 feet down the hole (610m) is just about 1/10,000 of the distance to the centre, so your weight would still be 99.99% of what it is at the surface. For someone weighing 68kg/150lb, that's a difference of about 7 grams or a quarter of an ounce.
I would submit that us Americans that are interested in the subject at hand are likely science-minded and not confused by SI units. I very much prefer seeing things worked out and reported in SI.
You're right. I don't think technical readers will be bothered at all by SI units, but for casual browsers of the thread, 1/4 ounce might means more than 7x10^-3kg, whether they're British or American.
I always work in SI but sometimes quote in other units just for context.
Anyway, Munji is way too old to get used to SI units now.
You're right. I don't think technical readers will be bothered at all by SI units, but for casual browsers of the thread, 1/4 ounce might means more than 7x10^-3kg, whether they're British or American.
I always work in SI but sometimes quote in other units just for context.
Anyway, Munji is way too old to get used to SI units now.
BigBangTheory - Sheldon makes a math mistake:
In about 1995, in the mark scheme published by the government department responsible here for the national Standardised Assessment Tasks in English, Maths and Science, they included a calculation that converted N/cm^2 to N/m^2 (Pascals) by dividing by 10,000 instead of (correctly) multiplying by 10,000.
So, in the answers given to schools nationally to enable the checking of pupils' basic skills in these subjects, the government got this bit wrong by a factor of 10^8, or ten billion percent error.
You're right. I don't think technical readers will be bothered at all by SI units, but for casual browsers of the thread, 1/4 ounce might means more than 7x10^-3kg, whether they're British or American.
I always work in SI but sometimes quote in other units just for context.
Anyway, Munji is way too old to get used to SI units now.
Cut me some slack, K? All of my physics classes used SI, and I'm quite comfortable with them. But now I don't have to use them.
Just kidding, as you know very well.
I know, but I have to maintain my grumpiness.
