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Finally... => Creative Projects => Topic started by: Scood on March 28, 2011, 11:26:28 pm

Title: for you physicists or rocket scientists out there
Post by: Scood on March 28, 2011, 11:26:28 pm

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.

Title: Re: for you physicists or rocket scientists out there
Post by: mainiac on March 28, 2011, 11:31:01 pm

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.

Title: Re: for you physicists or rocket scientists out there
Post by: Heron TSG on March 28, 2011, 11:34:09 pm

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?

Title: Re: for you physicists or rocket scientists out there
Post by: Scood on March 28, 2011, 11:40:42 pm

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

Title: Re: for you physicists or rocket scientists out there
Post by: Osmosis Jones on March 29, 2011, 12:34:17 am

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 :P

Quote from: mainiac on March 28, 2011, 11:31:01 pm

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 :P

Title: Re: for you physicists or rocket scientists out there
Post by: Fayrik on March 29, 2011, 01:34:21 am

Quote from: Scood on March 28, 2011, 11:26:28 pm

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.

Quote from: Barbarossa the Seal God on March 28, 2011, 11:34:09 pm

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!  :P

Title: Re: for you physicists or rocket scientists out there
Post by: GamerKnight on March 29, 2011, 01:36:13 am

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

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

Title: Re: for you physicists or rocket scientists out there
Post by: ein on March 29, 2011, 01:38:37 am

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.

Title: Re: for you physicists or rocket scientists out there
Post by: GamerKnight on March 29, 2011, 01:42:11 am

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

Title: Re: for you physicists or rocket scientists out there
Post by: G-Flex on March 29, 2011, 01:43:16 am

Quote from: mainiac on March 28, 2011, 11:31:01 pm

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

Title: Re: for you physicists or rocket scientists out there
Post by: Heron TSG on March 29, 2011, 07:57:53 am

Quote from: Fayrik on March 29, 2011, 01:34:21 am

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

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.

Title: Re: for you physicists or rocket scientists out there
Post by: Virex on March 29, 2011, 04:18:22 pm

Quote from: Osmosis Jones on March 29, 2011, 12:34:17 am

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.

Title: Re: for you physicists or rocket scientists out there
Post by: Normandy on March 29, 2011, 08:11:46 pm

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.

Title: Re: for you physicists or rocket scientists out there
Post by: G-Flex on March 29, 2011, 08:12:36 pm

Quote from: Normandy on March 29, 2011, 08:11:46 pm

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.

Title: Re: for you physicists or rocket scientists out there
Post by: Osmosis Jones on March 29, 2011, 08:19:36 pm

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 :P

Title: Re: for you physicists or rocket scientists out there
Post by: Normandy on March 29, 2011, 08:23:14 pm

More specifically, he proved that since you can treat a single shell of mass as a point mass, you can treat a spherically symmetric object as a point mass.

http://en.wikipedia.org/wiki/Shell_theorem#Outside_the_shell (http://en.wikipedia.org/wiki/Shell_theorem#Outside_the_shell)

Title: Re: for you physicists or rocket scientists out there
Post by: Heron TSG on March 29, 2011, 09:40:59 pm

Quote from: Osmosis Jones on March 29, 2011, 08:19:36 pm

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.

As long as you are not within an object, its gravitational field is generally calculated from its center of mass.

Title: Re: for you physicists or rocket scientists out there
Post by: G-Flex on March 29, 2011, 09:43:33 pm

Quote from: Osmosis Jones on March 29, 2011, 08:19:36 pm

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.

Keep in mind that there is a world of difference between equivalency and a "valid approximation".

Title: Re: for you physicists or rocket scientists out there
Post by: Osmosis Jones on March 29, 2011, 09:56:54 pm

Quote from: Barbarossa the Seal God on March 29, 2011, 09:40:59 pm

As long as you are not within an object, its gravitational field is generally calculated from its center of mass.

Yes, unless you are close to a large object of an interesting shape (say, something not close to spherical?), in which case you can't. Hence the excessive verbiage of my earlier post. Back when I was a Physics 101 tutor, we gave students a range of examples where you could and could not do so.

Quote from: G-Flex on March 29, 2011, 09:43:33 pm

Keep in mind that there is a world of difference between equivalency and a "valid approximation".

The gravitational effects of a spherical object are equivalent to those from a point mass at the object's centre.  It's a valid approximation because Earth is very near spherical.

Title: Re: for you physicists or rocket scientists out there
Post by: G-Flex on March 29, 2011, 10:05:14 pm

Right, I see what you're saying. Of course, if the subject of discussion involves nitpicking gravitational variation with increasing elevation, the Earth's spin, oblongness, etc. are worth mentioning.

Title: Re: for you physicists or rocket scientists out there
Post by: Osmosis Jones on March 30, 2011, 01:14:39 am

That's good, because I'm not sure I follow you. It's oblong-ness (nice word, btw) was taken into account by the use of the equatorial radius as opposed to the mean radius, the Earth's spin was accounted for in the second part of my first post, and the differing heights was the whole point of the problem.

Everything that is

Spoiler: relevant to mention (click to show/hide)

Before anyone mentions the moon or sun, that was not included in the initial data set, and is as such assumed irrelevant (also, at the distances involved, whatever effects they have will be pretty much identical on either height scale; it may change their absolute magnitudes, but it won't really affect their relative ones). Of course we are also ignoring mountains and varying densities of the planetary composition, but those effects are really quite negligible (like quite a bit beyond beyond the 5 significant figures presented earlier), in addition to having insufficient data.

was accounted for.

So what are you asking for?

FAKEEDIT: Actually, I didn't twig to this before, and I should really test it mathematically first, but I'm pretty damn sure it's valid; the maths for a spheroid (e.g. the Earth) allows it's treatment as a point mass anyway, because we are looking at it directly above the equator.

Title: Re: for you physicists or rocket scientists out there
Post by: Virex on March 30, 2011, 03:35:25 am

Quote from: Normandy on March 29, 2011, 08:23:14 pm

More specifically, he proved that since you can treat a single shell of mass as a point mass, you can treat a spherically symmetric object as a point mass.

http://en.wikipedia.org/wiki/Shell_theorem#Outside_the_shell (http://en.wikipedia.org/wiki/Shell_theorem#Outside_the_shell)

Oh well, that settles it then. Guess I should've payed more attention during physics...

Title: Re: for you physicists or rocket scientists out there
Post by: Nadaka on March 30, 2011, 11:59:50 am

Quote from: mainiac on March 28, 2011, 11:31:01 pm

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.

I'm going to have to have to ask for a source on that. I know of two factors that will make a difference from pole to equator off the top of my head.
Distance from center of mass and the centrifugal effect of the earths rotation. Both of those will reduce the effective weight of an object at the equator due to higher velocity and the roughly oblate shape of the earth.

Title: Re: for you physicists or rocket scientists out there
Post by: Dr. D on March 30, 2011, 05:22:03 pm

I really wish I knew General Relativity for this.

Title: Re: for you physicists or rocket scientists out there
Post by: Osmosis Jones on March 30, 2011, 10:26:09 pm

Quote from: Dr. D on March 30, 2011, 05:22:03 pm

I really wish I knew General Relativity for this.

NO need :) This works fine under the regime of Newtonian physics. General relativity is more for dealing with cases with large objects moving fast or at large distances (so there is a significant timeframe for the gravity waves to get from one object to the next).