How does the weight of a person change when he moves from deep mind to the surface of the earth?

• (Advanced) >" onclick="window.open(this.href,'win2','status=no,toolbar=no,scrollbars=yes,titlebar=no,menubar=no,resizable=yes,width=640,height=480,directories=no,location=no'); return false;" rel="nofollow">
  
• 

In response to the question "Does gravity vary across the surface of the Earth?", you stated that the force of gravity would be less at the top of Mt. Everest. You use the formula 1/R2 as the basis for your answer. It seems to me that that equation only works as you leave the surface of the earth. I say this because as you sink into the earth, the force of gravity decreases until you reach the center where it is 0. And, if you were to add mass to the overall surface of the earth, as it grew bigger, even though you moved further from it's center, gravitational force would increase. I assume that you are referring to a theoretical surface of the earth where gravity is at it's maximum. In that case, since Mt. Everest is above that surface, the gravitational pull would be less. My question is this (finally): Would I weigh less in Death Valley than at sea level (all other things being equal)?

The force of gravity you feel standing on the surface of Earth depends on two things. They are:

1. Your distance from the centre of the Earth, R
2. The mass enclosed within the radius R, M(R)

Point 2 is actually quite subtle. It is only mass inside the radius at which you are at which affects the force of gravity you feel. That is not immediately obvious, perhaps, but there is a neat theorem which undergraduate physics majors go through which proves it (at least in the case of a spherically symmetric object).

So, if you tunneled through the Earth to the centre, right at R=0 there would be no mass enclosed, so there is no net force of gravity. Once you go above the highest mountain peak all the mass of the Earth is interior to your position and so the mass no longer changes with radius and the 1/R2 law directly applies. In between those two extremes there is a play-off. As you move further from the centre of the Earth the gravity falls like 1/R2, but the mass enclosed within R also increases slightly, so the net change in gravity will be something different.

For a sphere with a uniform density, the mass increases with radius like R3, so as you move through a sphere like that the net change of gravity is to increase in proportion to R. Once you get outside the surface of the sphere the gravitational force then drops off like 1/R2.

The Earth is not a uniform sphere. It is quite centrally concentrated, so the change of mass with radius is much less than R3 once you get outside of the core. Above the surface of the Earth the change of the amount of mass with you elevation is very small compared to the total mass, so the 1/R2 law works quite well - but you are right in saying that it is a simplification.

In order to work out if you weigh less in Death Valley than at sea level or the top of Mount Everest, you then need to know how much of the Earth's mass is enclosed within those different radii. I would bet that the difference is so small as to be negligible, in which case you can just use the 1/R2 law and you therefore do weigh more in Death Valley than at sea level. If the mass change were significant you would need to factor that into the calculation, and so if you go significantly deeper than Death Valley eventually you will again weigh less than at sea level.

Thank you very much for your explanation. You answered my question quite well. Unfortunately, I now owe a co-worker a coke (-:

The question came from a discussion on the fact that there is no gravitational pull at the center of the earth. I surmised that, if one were to drill a hole completely through the earth and suck out all the air, a person could jump in the hole and emerge on the other side of the earth. This of course presupposes no friction. If, on the other hand, there was friction, from air or from brushing the sides of the tunnel, then you would end up with the ultimate bungee jump.

I am curious though, if you care to figure it out, what the velocity would be as you passed through the center and how long it would take to make the trip (again assuming no friction).

Actually this is an interesting question and one that I had on my Physics finals as an undergrad! :) There we assumed that the Earth was a uniform sphere which is mathematically easy (but not very realistic). If you drilled a hole through the Earth and jumped in, you would actually oscillate back and forth, much like a pendulum oscillates on a string! (This of course assumes no friction or air resistance.) With a mass model for the Earth (i.e. a formula that says how the mass changes with depth) you can fairly easily figure out how long that would take, and the velocity at the centre.

The physics and math of this hypothetical hole through the Earth are discussed in more detail at the following pages:

This page was last updated on January 30, 2016.

Karen was a graduate student at Cornell from 2000-2005. She went on to work as a researcher in galaxy redshift surveys at Harvard University, and is now on the Faculty at the University of Portsmouth back in her home country of the UK. Her research lately has focused on using the morphology of galaxies to give clues to their formation and evolution. She is the Project Scientist for the Galaxy Zoo project.

Twitter:  @KarenLMasters
Website:  http://icg.port.ac.uk/~mastersk/