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[Scood]

how much would 30 pounds (at sea level ) of dirt(or any material at all) weigh at 100 kilometer in the air (from sea level on the equator).


I'm talking weight, not mass. Mass would be the same, weight would not be the same.

I'm asking for the sake of asking.

[mainiac]

If you are going to ask about such a small difference, then it matters where on earth you are talking about.  There are non-trivial differences in gravity in different parts of the globe (rule of thumb is things are heavier closer to the equator).  I'm assuming that from 100 km in the air you mean 100km above sea level.

[Heron TSG]

Is it above a flattish-surface, such as the ocean? Or is it 100km above mount Everest? What is the latitude? What phase is the moon in?

[Scood]

i meant above sea level....:/ ill edit previous post

[Osmosis Jones]

Mass of the Earth M = 5.9737 * 10²⁴ kg (lot of disagreement on this one, going with wiki value)
Earth equatorial radius R = 6.3781 * 10⁶ m
Gravitational Constant G = 6.6730 * 10^-11 N m² kg^-2

F_g = G M m / R²

:. F (R) = 9.7990(m) N
:. F (R + 100000) = 9.4988(m) N

E.g. the weight force is about 3% weaker. The (m) is the mass of the object orbitting the Earth, if you leave it out, the answers are the actual gravitational acceleration in units of m s^-2.

NOTE, this doesn't take into account the lessening of the observed weight force due to circular acceleration.

For that;
Remember that for circular motion, the sum of ALL forces (grav + reaction force) = m v²/R, where v is the tangential speed of the object, and the reaction force is what we perceive as weight, which is equal and opposite to gravity in a non-rotating frame.

Earth rotates once every 24 hours = 86400 s

If we assume a geostationary orbit,
:. Distance travelled at equator = 2*pi*R = 4.0075*10⁷ m
:. V_R = 463.83 m/s
:. F_{C R} = 0.0337(m) N

Likewise for 100kms up,
:. Distance = 4.0703*10⁷ m
:. V_{R + H} = 471.10 m/s
:. F_{C R+H} = 0.0343(m) N

Obviously, these are pretty damn negligible, but for the sake of completeness

:.Weight Force  F_R = 9.7653(m) N
:.Weight Force  F_{R + H} = 9.4645(m) N

So again, just over a 3% difference between the two.

Now, in future, please don't use these forums to do your physics homework

If you are going to ask about such a small difference, then it matters where on earth you are talking about.  There are non-trivial differences in gravity in different parts of the globe (rule of thumb is things are heavier closer to the equator).

Other way round. Equator is further from the centre of the Earth, so things are lighter.

EDIT: One last thing... in reality, anything at 100 kms up won't weigh a damn thing, because there is nothing to push against and ergo, no reaction (weight) force. But hey, theoretical case and all that

[Fayrik]

how much would 30 pounds (at sea level ) of dirt(or any material at all) weigh at 100 kilometer in the air (from sea level on the equator).

Hehehe! Second time today I've had high atmospheric information given to me in both metric an imperial.

Though, of course, it seems Osmosis Jones factored in that discrepancy beforehand.

What phase is the moon in?

What time of the year is it, and which hemisphere are you in? The sun plays a factor too! 

[GamerKnight]

IT BURNS US!!! THE PHYSICS BURNS THE CHEMISTRY STUDENT!!! AHHHHHHHHHH!!!!

EDIT: Interesting, now that I have the burn ointment on.

[ein]

You chemists and your silly letters and numbers.
Us biologists can actually see what we're studying without the use of your fancy-schmancy scanning electromawhatsits.

[GamerKnight]

HA! I am also studying Biology next semester. SO I AM BIOCHEMIST!!! BOW DOWN BEFORE ME!!!

[G-Flex]

rule of thumb is things are heavier closer to the equator)

Shouldn't it be the other way around? At the poles, you're closer to the center of the Earth. Also, centripetal force has a greater effect near the equator (higher linear velocity of rotation), further decreasing your apparent weight there relative to the poles.

Also, here's some stuff: http://en.wikipedia.org/wiki/Gravity_of_Earth#Altitude

[Heron TSG]

What time of the year is it, and which hemisphere are you in? The sun plays a factor too! 

The tilt shouldn't change it too much, because the sun is just so far away. The moon, however, is pretty close by. So close that it pulls all the water and air on and around the Earth closer to it as it passes.

[Virex]

Mass of the Earth M = 5.9737 * 10²⁴ kg (lot of disagreement on this one, going with wiki value)
Earth equatorial radius R = 6.3781 * 10⁶ m
Gravitational Constant G = 6.6730 * 10^-11 N m² kg^-2

F_g = G M m / R²

Can you apply that formula here? I'm not a rocket surgeon, but it seems to me like that formula would only apply for point masses, or in other words, very long ranges. If you get closer to a planet you have to take into account that it's not a point and that the spherical shape means part of the gravitational force it exerts is going to be perpendicular, which then gets compensated by an equal force from the other side.

[Normandy]

So long as you're not inside the object itself, gravity acts as if the mass was a point mass concentrated at the center of mass of the object.

EDIT: Right. It's for a spherically symmetric object.

[G-Flex]

So long as you're not inside the object itself, gravity acts as if the mass was concentrated at the center of mass of the object.

The nature of vector math disagrees with you.

Wait. Apparently I'm wrong, although I have no idea why.

Oh wait, duh. The vectors not parallel to the line from you to the center of the sphere partially cancel out, but certain parts of it are also treated as being closer to you, I guess. Or farther away.

[Osmosis Jones]

As long as you are outside the surface of the Earth, yes, you can. Newton, back when he outlined the laws of gravitation, realised that you can integrate the position vectors of a planet across its entirety and approximate it to the centre of mass. There are cases where this does not apply, e.g. for very irregularly shaped objects (picture a giant cup shape, etc), but for spherical or near spherical objects, yes, it is a valid approximation.

EDIT: Also, seriously guys.. no posts for four hours, and then AS I AM WRITING MY RESPONSE, 2 ninjas? Bloody hell