Weightlessness in the Center of The Earth?

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SUMMARY

Digging a tunnel to the center of the Earth results in experiencing weightlessness due to the equal gravitational pull from surrounding mass. As one descends, the gravitational force decreases because the mass above cancels out, while the force is proportional to the radius from the center. The discussion clarifies that the Earth’s non-uniform density and shape (oblate spheroid) affect gravitational experiences, but the Newtonian principle of concentric spherical shells holds true. Therefore, while one may feel slightly heavier initially due to proximity to denser material, true weightlessness occurs at the center.

PREREQUISITES
  • Understanding of Newton's law of gravitation
  • Familiarity with gravitational forces and mass distribution
  • Knowledge of Earth's structure, including density gradients
  • Concept of oblate spheroid geometry
NEXT STEPS
  • Research the implications of Newton's shell theorem in gravitational physics
  • Explore the effects of Earth's density variations on gravitational force
  • Study the characteristics of oblate spheroids in planetary science
  • Investigate gravitational effects within hypothetical structures like Dyson spheres
USEFUL FOR

Students of physics, geophysicists, and anyone interested in gravitational theory and Earth's internal structure will benefit from this discussion.

hexrd
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Is it true that if you dig a tunnel to the center of the Earth you will experience weightlessness? I heard that you will get heavier then become weightless.
 
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hexrd said:
Is it true that if you dig a tunnel to the center of the Earth you will experience weightlessness? I heard that you will get heavier then become weightless.

Yes, at the center of the earth, the gravitational pull due to the mass around you is equal in every direction and so you would be "weightless". (That's assuming you fill in the hole behind you. Otherwise there would be a miniscule pull in the direction opposite the hole.

Your second statement "you will get heavier" is NOT true. As you dig into the earth, the gravitational pull will be due to the mass of Earth still BELOW you. The mass above you will "cancel". Since the mass below you is proportional to r3 and the gravitational force is proportional to 1/r2, the net force is proportional to r3/r2= r. The force on you will decrease as r becomes smaller, becoming 0 when r= 0.
 
HallsofIvy said:
Yes, at the center of the earth, the gravitational pull due to the mass around you is equal in every direction and so you would be "weightless". (That's assuming you fill in the hole behind you. Otherwise there would be a miniscule pull in the direction opposite the hole.)

This statement is only valid if the Earth is a uniform shape, such as a perfect sphere. Earth is actually an oblate spheroid with the equatorial radius approximately 6378.5 km and the polar radius approximately 6356.9 km and its shape is not uniform. Without an even distrubtion of mass surrounding the individual, a nonzero net force would be present pointing in the direction of the greatest distribution of mass.
 
The other thing to consider is that the density of the Earth isn't uniform as you go down. You're getting closer to the 'massier' stuff, while the lighter crust gets cancelled.

So you'll get heavier, but not significantly so.
 
The Newtonian logic of concentric spherical shells still works with a density gradient.
 
Just a thought. If the core of the planet where nickel surrounded by a layer of molten magma (stereotype?), wouldn't the magnetic field interfere with a gravitational effect? and with that pretense a person or object would be sort of like a ping-pong ball at the end of a blower? the result being batted around resulting in no gravitationally neutral spot?
 
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selfAdjoint said:
The Newtonian logic of concentric spherical shells still works with a density gradient.

That doesn't make sense to me.

If that is true, then if you were inside a hypothetical Dyson sphere, the gravitational force of the sun would be the same whether you were right near the sun or out near the sphere itself. Right?
 
enigma said:
If that is true, then if you were inside a hypothetical Dyson sphere, the gravitational force of the sun would be the same whether you were right near the sun or out near the sphere itself. Right?
No. The "Newtonian logic of concentric spherical shells" that SelfAdjoint refers to is the fact (proved by Newton) that a uniform spherical shell of mass has a gravitational field that is zero everywhere inside the shell.

Of course that's only true if the shell is empty. :smile: Inside a Dyson sphere you have the sun, which exerts its usual gravitational force.
 
Yes, but SelfAdjoint said that if you've got a density gradient inside the shell the thing works.

He said that in response to my post that you will get a little heavier first.

If there is a light, thin crust, and a heavier mantle, then when you go down under the crust before you start to dig in the mantle you will first get a (very little) bit heavier as you are now closer to the heavier stuff while having the lighter stuff get zero'd out.
 
  • #10
I suspect SelfAdjoint was just making it clear that the argument for there being zero gravity at the center of the Earth (the original question), which is based on Newton's shells, still works even if the Earth's density depends on the distance from the center. I don't think he was disputing your statement.

Did my statement about the Dyson sphere make sense?
 
  • #11
Yes. You were making the same point which I was trying make by analogy. If you've got a Dyson's sphere, your distance from the sun does affect the gravitational force.

Similarly, if you dig through a lighter portion of the crust, getting closer to a heavier section after a discontinuity (which is what happens when you get to the mantle), your weight would increase a small amount before it starts to decrease.
 

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