Just to be a really annoying pendant, if we pretend that Cecil had been talking conditions of uniform density, and ignore non terrestrial gravitational sources, wouldn’t there be a differential in gravity across an individual’s height? Obviously this is minuscule for Earth, but presumably non-zero, hence your blood would imperceptibly slosh towards the extremity closest to the Earth’s center of mass. I’ll reserve the Moon as a tidal back-up to help speed things along, not that the energy dissipation would be speedy in any event.
Yes, this is exactly my point. There would be a differential in gravity, which is the tidal effect I’ve been talking about. Under the standard assumption, the gravitational force is exactly proportional to the distance along the tube. Thus, the difference in force between your head and your toes is a constant, a few milligrams, it doesn’t matter whether you are at the center of the Earth, the surface, or somewhere in between. The dissipation due to the tides is caused by the time variation of the tidal force. If you have a constant tide, it just stretches things out. It is the act of stretching that requires an input of energy.
Incidentally, the magnitude of this constant differential force is exactly the same as the tidal force you “feel” at the surface of the Earth. Inside the tube it is constant (first derivative of a linear function), outside the tube it goes down by the cube of the distance from the center of the Earth (first derivative of the inverse square law). The Moon’s effect on you is vastly smaller, since the the Earth is much more massive than the Moon, and more importantly, the cube of the distance from the center of the Moon is many orders of magnitude larger than the cube of the radius of the Earth.
But presuming you don’t rotate at the centre of the Earth, wouldn’t the blood slosh back and forth relatively speaking?
Why would it do this?
I’m not sure I understand the question. When you jump in the hole, your body is already stretched by the Earth’s tidal force. When you reach the center of the Earth it is still stretched by the same amount. When you reach the other side of the Earth, same story. The stretch doesn’t change, so no sloshing.
Going upright into the hole, your feet experience a greater amount of gravity than your head, so blood pools there. However past the Earth’s centre of mass the effect is reversed, so it would begin to pool in your head causing tidal heating. The differential force is constant in size, but would still lead to the dissipation of energy, unless I’ve misunderstood.
Captain,
First, let me apologize for making an error. I’ll explain it in the third paragraph, but it does not change the conclusion.
The tidal effect is a differential force, represented by a compression or tension, not by a direction toward or away from the Earth’s center. After you jump into the hole feet first, your feet feel a weaker pull than your head. The effect is to compress your body, as if your body was in a vise exerting a few milligrams of force* from head to foot. At the center of the earth, there is no net force, but there is still a tidal effect squeezing your body the same amount. When you reach the other side of the Earth, there is still a few milligrams of squeezing force. Since there is no change in the amount of squeezing, there is no dissipation of energy.
In my earlier posts, my mistake was to say that the effect was to stretch the body, rather than squeeze it, and to say that the tidal force was identical to that at the surface of the Earth when completely out of the hole. In fact, the two effects are opposite in sign (since gravity gets weaker with distance in the inverse square law and stronger with distance within the hole). There is also a factor of two difference in magnitude. Here are the formulae for the tidal effects:
Outside the hole: -2GM/x^3 (where x is the distance from the center of the Earth)
Inside the hole: GM/R^3 (where R is the radius of the Earth, notice no x dependence)
To get the squeezing or stretching force, you need to multiply these terms by the mass of your head and feet and by the distance between your head and feet . Depending on the mass distribution in your body, the squeezing or stretching force at the surface is smaller than your weight by a factor of roughly your height divided by the radius of the Earth, which accounts for my earlier estimate of a few milligrams*.
- Among friends, please forgive my use of mass and weight interchangeably. Physicists do this all the time. It is only physics teachers who scold their students for doing so.
Just to add one clarification:
It might be helpful to think about riding inside an elevator falling through the tube (a good idea if you hope to breath). Since you and the elevator are in freefall, you will be weightless for the entire journey and feel nothing different going past the center of the Earth or turning around near the surface. In principle, very careful measurements would show the constant tidal effect, but no pooling of blood in either your head or feet.
Thanks for your explainable, I think I’ve finally grasped it. Presumably going from non-linear compression jumping from the outside, to linear compression inside, would have a non-zero, but immeasurable tidal decay effect.
BobG_80918 said:
Except that it is inaccurate. There were at least 4 “people” at the tea besides Alice.
Mad Hatter, March Hare, Doormouse, White Rabbit.
Carry on with the physics.
Hatters of the day were mad because they suffered from mercury poisoning, which was used somehow in the process of making hats.
I’m fairly certain that you would just die from the heat of the earth’s core or the rapid change in atmosphere- but it might be a fun ride if mother nature didn’t shank you at the end…
This topic/cecil response was originally op’d in 2003. Since that thread has seniority, I have placed a note there.Falling trough the center of the earth? - #104 by Leo_Bloom