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View Full Version : Do you weigh more at sea level?.....

Iamme
7th December 2003, 06:52 PM
...than up in the mountains? The atmospheric pressure is greater at sea level. Also, I think it is the closer you get to the center of the earth, the more gravity there is? (I think I read this, and this went contrary to a belief I had on this subject). When *I* lived at sea level...it seems to me that I weighed about 10 more. And I was even smoking cigarettes at the rate of 3 packs a day then!! Then I move up north (higher elevation)...quit smoking...and lose 10 pounds I have NEVER been able to get back. (sigh)....I'm thin...just like a buzzard man that I know. :D

plindboe
7th December 2003, 08:12 PM
I wouldn't think the air pressure would do any difference. The pressure of the air affect your entire body, so while it pushes you down it equally pushes you up, and the effect is negated.

The gravitational pull would be smaller the further away from the center of the Earth you get, but the difference would be minute, and I don't think a normal scale would notice the difference. The radius of the Earth is about 6.378km, so if you was placed on a 4km high mountain for instance, the difference in range would only be around 0,01%, which if I calculated correctly would mean that the difference in weight would be about 0,03%.

A bigger difference would be if you moved to the Equator, since the centrifugal force would affect you most there. But still a small undramatic difference.

The Don
8th December 2003, 03:37 AM

The strength of the force of gravitational attraction between two bodies is proportional to the masses of the bodies and the square of the distances btween their centres of mass.

In the case of our mountain, being say 10000m above sea level (top of Everest) would reduce the force to (6368/6378)^2 or 99.7% of its value at sea level.

This however assumes that the Earth is completely uniform in density. In fact different types of have very different densities and therefore will have an impact on the gravitational force. Geologists use this to determine the density of parts of the Earth's crust using a gravitometer.

What you may find is that the local variations in density far outweigh the impact of altitude.

andre
9th December 2003, 07:20 AM
Another yes here, based on the inverse proprotionality of the radius between the masses but the story within the Earth is slightly different. An hypothetical object within the Earth is only pulled down by the mass below it. The mass on top of it is counteracting and it is pulling it up.

After a couple of pages of math it is provable that the gravity within the Earth decreases linear with the radius. In the centre of the Earth gravity is zero.

Michael Redman
9th December 2003, 07:47 AM
Originally posted by plindboe
I wouldn't think the air pressure would do any difference. The pressure of the air affect your entire body, so while it pushes you down it equally pushes you up, and the effect is negated.The effect of the pressure is negated, but the denser the air, the more boyancy you would gain by displacing it.

Andonyx
9th December 2003, 08:22 AM
I just happened to have my handy dandy book called, "The Cosmological Milkshake," right here at the office, the book answers this exact question with an essay on page 124 called, "How much does your Weight Vary on Earth."

This is a great book and can be found here. (http://www.amazon.com/exec/obidos/tg/detail/-/0813520460/qid=1070986987/sr=1-1/ref=sr_1_1/104-9635496-6496721?v=glance&s=books)

Atop 5.5 mile-high Mount Everest, you would be about 1.0016 Earth Radii from the Earth's center. According to the inverse Square law, the strength of Gravity atop Everest should be 1/1.0016 or 0.9968 times its surface value - a0.32 percent reduction. Of course we only get an approximate answer using the inverse square law since it basically assumes you are suspended 5.5 miles above a spherical Earth. A more realistic estimate would take into account the gravitational pull of the mountain itself, which might reduce your weight loss to perhaps 0.25 percent, depending on the shape of the mountain.

So it's not that you would weigh "more" at sea level. Since sea level is determined to be one Earth Radii from the center of the planet, but that you would weigh less everywhere else.

As a 180LB. Man, I could expect a rough weight reduction of 0.45 pounds, at the top of Everest.

Not a lot, but a sufficiently sensitive scale should easily pick that up.

The equator suggestion is a good one, as the book says:

Due to this circular motion a point [on the Earth's eqautor] falls off the straight line tangent by by 0.054 feet in one second, making its acceleration 0.35 percent that due to gravity. As a result of this acceleration a scale at the Equator reads 0.35 percent less than it would if the Earth didn't rotate.

Which again for me, translates to a weight loss if 0.63 Lbs.

Andonyx
9th December 2003, 08:27 AM
Originally posted by andre

After a couple of pages of math it is provable that the gravity within the Earth decreases linear with the radius. In the centre of the Earth gravity is zero.

You are correct as this amazing book mentioned above also has an essay on page 127 titled, "What would you weight at the center of the Earth?"

He uses the method of comparing the mass of a constant density sphere which decreases by 8 fold as you move halfway into the center, but the inverse square law increases the gravitational attraction by 4 fold....etc etc to demonstrate this.

Iamme
9th December 2003, 06:26 PM
(Andonyx)

You are correct as this amazing book mentioned above also has an essay on page 127 titled, "What would you weight at the center of the Earth?"

He uses the method of comparing the mass of a constant density sphere which decreases by 8 fold as you move halfway into the center, but the inverse square law increases the gravitational attraction by 4 fold....etc etc to demonstrate this.

------------------------------------------------------------------------------

That said; what then could a 180 lb. man expect to weigh at the center of the earth?

WildCat
9th December 2003, 08:53 PM
Originally posted by Iamme

That said; what then could a 180 lb. man expect to weigh at the center of the earth?
Assuming that the earths mass is uniform (which it isn't) you would be weightless, as earths gravity would pull on you from all directions.

Of course, I'm a complete ignoramus when it comes to physics so I may be completely wrong.

Andonyx
9th December 2003, 10:46 PM
Originally posted by WildCat

Assuming that the earths mass is uniform (which it isn't) you would be weightless, as earths gravity would pull on you from all directions.

Of course, I'm a complete ignoramus when it comes to physics so I may be completely wrong.

Sorry, Lamme, I didn't conclude my thought really, but Wild Cat's quote above is exactly correct.

He illustrates it another way. If Earth was a hollow sphere and you climbed through a hatch to the inside, you would be weightless anywhere inside the sphere because distance to each other part of the sphere would be inversely proportional to its pull on you. So you could float around at will with a push off the inside of the shell.

In Real Earth just think of yourself as at the innermost hollow sphere of a series of concentric hollow spheres. The conditions are the same and you are weightless.

(And so am I...)

((The 180 LB man))

rockoon
10th December 2003, 10:36 PM
Originally posted by Andonyx

He illustrates it another way. If Earth was a hollow sphere and you climbed through a hatch to the inside, you would be weightless anywhere inside the sphere because distance to each other part of the sphere would be inversely proportional to its pull on you. So you could float around at will with a push off the inside of the shell.

Umm... no. Correct statements involved but incorrect conclusion. You will not be equally distant to all points of the shell of the hollow sphere. You will be greatly attracted to the points nearest you and the points farthest from you will have negligable effect.

The Don
10th December 2003, 11:50 PM
So if you're right in the middle you're OK, but as soon as you get off centre, you start accellerating towards the edge of the sphere until it subjects you to rapid decelleration (in a splat sort or way)

That doesn't sound very much fun at all, in fact I'll make a point of not doing it.

xouper
11th December 2003, 01:00 AM
rockoon: Umm... no.Umm... yes. :) Andonyx is correct that within a hollow sphere, the net force of gravity is zero at any given point. This has been discussed previously, here (http://www.randi.org/vbulletin/showthread.php?s=&threadid=29262)

Terry
11th December 2003, 09:03 AM
Originally posted by rockoon

You will not be equally distant to all points of the shell of the hollow sphere. You will be greatly attracted to the points nearest you and the points farthest from you will have negligable effect.

Three words: inverse square law.

--Terry.

phildonnia
11th December 2003, 09:15 AM
A fascinating fact I learned not too long ago; The source of the Mississippi River is closer to the center of the Earth than the delta.

Andonyx
11th December 2003, 10:21 AM
Originally posted by rockoon

Umm... no. Correct statements involved but incorrect conclusion. You will not be equally distant to all points of the shell of the hollow sphere. You will be greatly attracted to the points nearest you and the points farthest from you will have negligable effect.

You should read what I said.

When you climb into the shell of the hollow Earth if you are near the edge of the shell. You will be greatly attracted to the points closest to you. However, the effects of the points farther away will NOT be negligible, they will be reduced inversely proportional by the square of their distance. Since there are SO MANY more points away from you they will, if you do the math actually balance out. From inside the sphere you will be weightless at any point within the sphere.

Andonyx
11th December 2003, 10:26 AM
Originally posted by xouper
Umm... yes. :) Andonyx is correct that within a hollow sphere, the net force of gravity is zero at any given point. This has been discussed previously, here (http://www.randi.org/vbulletin/showthread.php?s=&threadid=29262)

I didn't see that you had adressed this before my post above. (Teach me to post before finishing the thread.)

But really, It is not I who am correct but Robert Ehrlich, the guy who wrote the book, "Cosmological Milkshake", that I have been so vehemently hawking.

(Don't know why, I just think it's a great book that if everyone read it around here, we would all have a really good common frame of reference for some of our more fanciful questions and discussions.)