Value of g at the center of the earth

  • #26
Simon Bridge
Science Advisor
Homework Helper
17,857
1,654
I'm afraid I don't see the answer [in wikipedia]. I'll try to answer it then.
You realize that your question has been answered by post #5 right?
Or did you have a different question I missed?

Gravity acceleration. I should have said "force" I guess.
"acceleraton" is fine. "acceleration of gravity" is better. "free-fall acceleration of gravity" is best.
You could also have talked about gravitatonal potential (not to be confused with potential energy).

The force of gravity, by comparison, varies with mass as well as separation, so it is not as useful for your purposes: you be constantly have to qualify every statement.

So since every molecule is static, that means acceleration vectors cancel at every point inside a planet, not just the center?
Only for a uniform hollow spherical shell. Only inside the hollow part.

The calculation will have some cancellations in it inside, pretty much, any mass distribution though.

To appreciate it you really need calculus - you divide the mass up into lots of small sub-masses and work out the gravity, on a test mass, at the point of interest, due to each sub-mass by itself. Then add them all up (remembering that the force of gravity is a vector) and divide by the test mass. This gets you the acceleration due to gravity at that point.

Without calculus you can see the effect by doing the same calculation for different arrangements of discrete masses.
 
  • #27
Simon Bridge
Science Advisor
Homework Helper
17,857
1,654
Aside:
Now that's funny, because once you have arrived at the other end, the process should work the other way around, taking you back to the initial end just by the force of gravity, and then back to the other end and so endlessly. Does not this sound like a perpetual motion machine moving you constantly to and fro without any energy being spent?
No. In order to be a machine it has to do useful work. A perpetual motion machine, therefore, must do useful work in perpetuity with no energy input.

This is what is meant by "perpetual motion of the second kind". There is nothing intrinsically non-physical about perpetual motion of the first kind - motion in perpetuity without doing useful work. I'm not keen on these terms because they make me sound like a UFO nut. Presumably perpetual motion of the 3rd kind is where the machine becomes sentient and gives you advise about stopping nuclear proliferation, and the fourth kind is where you vanish into the Alaskan winter never to be seen again...

Anyway:

The initial energy input is that required to raise the falling mass to the surface of the "Earth" - we don't normally see that energy input because it was provided when the Earth was first formed. But more important is the "work in perpetuity" part.

To extract work from an object falling along such a tunnel as described, you could put a wheel (say) in the center of the path. As the mass passes, it strikes the wheel, turning it... you've seen water-wheels, I don't have to bore you with the details. There are other approaches, we could drop a magnet and put a coil at the center, it does not matter, anything that extracts some of the energy from the motion results in the falling mass not quite reaching the other end of the tunnel. Back and forth it goes with a lower amplitude each time until it ends up stationary at the center all energy exhausted.

Notice that this is under the impossible idealized conditions that there is no friction or other loses in the system to bother us.

Discussions of specific pmm proposals is banned - I hope this discussion is general enough.
It is legitimate, however, to ask how much energy can be generated this way?

Doing the maths: it is about 4x107J/kg (U/m=GM/R=g(R)R)

Energy used globally, from all sources, passed 500x1018J per year in 2012.

So, back of envelope, each dropped kg is enough to supply the entire Earth for about four and a half months... assuming total conversion. Of course the energy would not be available that quickly anyway but it's fun to compare. i.e. the original description implies dropping ("you") a human - who shall we drop? If we dropped the average US American (~90kg apparently) you get energy equivalent to 30-odd years at the 2012 consumption rate. The sales would probably just about offset the damages in the ensuing lawsuit. "Gee we'd like to rescue him but he's running my dishwasher..." See? Fun!
 
  • #28
315
11
That was very interesting, thanks!

Something else though, the original quote said:
"if you drilled a hole between any two locations, you could jump in and gravity would accelerate you the first half, decelerate the second half, and you arrive at the other end. Perhaps more interesting is that the hole needn't pass through the center of the earth (sphere), and also that every trip, no matter what locations the hole connects, take the SAME TIME to complete."

I guess that this would only happen if the starting location is farther from the center than the end location. Intuitively I can not imagine that if you were dropped in a location close to the Earth's centre gravity would transport you to any location closer to the surface. Even if the connecting tunnel passed through the center, you would only reach up to the same distance from the center as you left from, never further.

And so, my consideration of you getting transported continuously to and fro would only happen when the two locations are at equal distance from the center, otherwise you will just stop at the location closest to the center, right?
 
  • #29
Simon Bridge
Science Advisor
Homework Helper
17,857
1,654
That was very interesting, thanks!

Something else though, the original quote said:
"if you drilled a hole between any two locations, you could jump in and gravity would accelerate you the first half, decelerate the second half, and you arrive at the other end. Perhaps more interesting is that the hole needn't pass through the center of the earth (sphere), and also that every trip, no matter what locations the hole connects, take the SAME TIME to complete."

I guess that this would only happen if the starting location is farther from the center than the end location.
The end location has to be at equal or lower radial distance than the start location.
The quote is talking about any two locations on the surface of a sphere.
If the start was at the bottom of the ocean, and the end was at the top of a mountain, you'd still have a climb ahead of you.

This is where you have to keep your wits about you when reading stuff like this - a lot depends on context.

And so, my consideration of you getting transported continuously to and fro would only happen when the two locations are at equal distance from the center, otherwise you will just stop at the location closest to the center, right?
Conservation of energy would have you pop up no farther from the center than you started - assuming uniform mass distributions of course. This is correct.
 

Related Threads on Value of g at the center of the earth

  • Last Post
2
Replies
27
Views
4K
Replies
34
Views
5K
  • Last Post
Replies
12
Views
2K
Replies
1
Views
3K
Replies
5
Views
682
Replies
187
Views
43K
Replies
9
Views
757
Replies
21
Views
3K
Replies
4
Views
2K
Top