Equator, weight

[Ian C]

the earth bulges at the equator (like me) and so anyone there is further away from the Earth's centre of mass.

 

The gravitational force is inversely proportional to the square of the distance between the centres of mass of the two bodies (the earth and the person standing on it in this case).

 

Yep but centre of mass doesn't really count, it's the amount of mass underneath you, and as also already said, with more rock and stuff under your feet, there will be a stronger gravitational pull due to a greater concentration of mass. Otherwise with a big enough planet you'd be weightless

 

-Ian

[Dash Rendar]

Your mass, in grams, however would stay the same because 'grams' is a measure of the resistance of a body to being moved and has nothing to do with acceleration or gravity. Your mass in kilograms would remain the same. It is common for people to use 'pounds' and 'grams' interchangeably but they are not.

 

A good argument, but still with erroneous detail. It's a common misconception that the pound is a unit of weight, but it is in fact - like grams - a unit of mass. (It is often erroneously described in US literature as a unit of weight, which might explain the NASA reference, but even the US officially recognise the pound as a unit of mass now.) The explanation would have been less flawed if they had described the hypothetical person's weight in Newtons...

 

Good question!

[juanchan]

If a plane were to try to take off facing West on the equator, would the earths spin act like a conveyor belt?

 

NASA launch their shuttles in a westward direction for that reason don't they?

 

From NASA's site

 

http://image.gsfc.nasa.gov/poetry/ask/a11511.html

 

I retract my previous statements because they are WRONG!

 

A good argument, but still with erroneous detail. It's a common misconception that the pound is a unit of weight, but it is in fact - like grams - a unit of mass. (It is often erroneously described in US literature as a unit of weight, which might explain the NASA reference, but even the US officially recognise the pound as a unit of mass now.) The explanation would have been less flawed if they had described the hypothetical person's weight in Newtons...

 

Good question!

 

If they'd said pounds force and not just pounds it would make sense. Pounds are equivalent to grams, just a different scale. Pounds force are equivalent to Newtons.

[Hermit]

Gravity... is lumpy Check out the GRACE research.

 

So if you want to weigh less, go to India!

 

http://www.piggynap.com/wp-content/uploads/2009/03/howcoolisthis.jpg

[JustGav]

Gravity... is lumpy Check out the GRACE research.

 

So if you want to weigh less, go to India!

 

http://www.piggynap.com/wp-content/uploads/2009/03/howcoolisthis.jpg

 

That's it, completely confuse the whole thing and add a new variable... Where on the equator would the OP be standing

[Max Headroom]

I buy all my toiletries from the local shop

[L33]

 

If you fill a bucket with water... cut a whole in the bottom of it.. and then swing it in a circle really fast... you would expect the water to fall out the bottom.. but it doesnt.

l

 

i tried spinning a bottle of squash with the lid off at home once but i slowed down to see if it was stil in there in mid rotation and covered my mums kitchen i was 10

[Scott]

No

[Hermit]

That's it, completely confuse the whole thing and add a new variable... Where on the equator would the OP be standing

 

Well there was bound to be someone popping up to make it complicated... happy to oblige

 

Think I might have read the colours wrong - blue might be high gravity, meaning lower sea level.

 

Either way, the two extremes along the equator are roughly in India and the Phillipines...

 

http://www.unavco.org/edu_outreach/tutorial/graphics/geoidvsellipsoid.gif

 

(Geoid height is true sea level, which is determined by gravity).

[Tannhauser]

Yep but centre of mass doesn't really count, it's the amount of mass underneath you, and as also already said, with more rock and stuff under your feet, there will be a stronger gravitational pull due to a greater concentration of mass. Otherwise with a big enough planet you'd be weightless

 

-Ian

 

Your logic sounds right, and yet I'm pretty sure your conclusion is wrong, and that the bulging equator does result in loss in weight, quite apart from rotational effects. I know that at the top of mountains, there is a measurable loss in weight due to altitude, so it seems that in this case that the extra mass under the feet does not offset the increased distance from the centre of mass.

 

My Maths and Physics aren't good enough to figure out why that should be. I think it might be something to do gravitational attraction to mass that is not directly underneath you. The total gravitational pull of the Earth is the attraction to every bit of matter on the Earth, right? So maybe at the equator, you have slightly more mass directly under your feet attracting you, but your distance to other parts of the Earth that are also providing a net pull downwards is increased because of the bulge and this more than offsets the pull directly downwards.

 

I'm not sure I'm explaining what I mean well. I need Digsy.

[Tannhauser]

Ian:

 

Here's a more parsimonious hypothesis. I think someone has already mentioned this above:

 

If the upper layers of the Earth contribute more to the bulge than the lower layers, that could also account for the decrease in weight at the equator.

 

The inner layers like the core are the densest, so increasing perpendicular distance from this might have more of an effect than adding extra mantle or crust. This assumes, however, that the inner layers are more perfectly round than the outer layers.

 

This is a bit of a post hoc reasoning, as I'm working backwards from a conclusion that I believe has been demonstrated empirically.

[MrRalphMan]

NASA launch their shuttles in a westward direction for that reason don't they? .

 

Yep - http://www.qrg.northwestern.edu/projects/vss/docs/navigation/2-why-launch-from-equator.html

[Digsy]

I need Digsy.

 

I'm flattered

 

There's a rather nice Wiki page on just this subject. Its not something that I've ever really got into but the maths for gravitational attraction aren't hard:

 

F = G x (m1 x m2) / r^2

 

F is the force attracting two masses, m1 and m2. r is the distance between the two masses. The key to the question at hand is how to define the distance, r.

 

In a simple model this can be the distance between the centres of gravity (centres of mass) of m1 and m2, so when considering the weight of somebody at the equator, r would be the distance between the position of the centre of gravity of the earth (not the centre of the earth) and the surface. The position of the centre of gravity would be affected by the composition of the whole earth, not that just under your feet. Likewise the direction that gravity if acting along may not be straight downwards, which means that not all the force would be acting through your weighing scales.

 

Consider a planet which is not sperical but cylindrical. Standing on one of the poles (i.e. on the circular surface at either end) the planet's centre of mass and all of its volume will be directly underneath you, so it would be easy to determine how much you would weigh.

 

Now, if you could walk over the edge of the pole and along the surface of the cylinder(let's assume that the planet is sufficiently big that you don't know which part you are standing on), you would suddenly find that your scales were telling you that you weighed a lot less. This is because the direction vector between your c of g and that of the planet is now predominantly in front of you, so the majority of the attractive force is now no longer acting "downwards" through your feet. Of course living in such a place your concept of what id "down" might be radically different than if you lived at one of the poles

 

If you walked along the side of the cylinder, as you approached the middle you would find yourself feeling very heavy indeed. This is because (a) the distance between you and the planet's c of g is getting smaller and also the direction along which the force is acting is acting more and more through your feet. When right at the middle, you would still have the planet's c of g directly under your feet (the same as it was when you were at the pole) but now you would be much closer to it.

 

This demonstrates that your apparent weight varies not only with your distance from the c of g of the planet, but also with local variations in the planet's gravitational field. In the crazy cylindrical Digsy-world, these effects are very obvious, but on a predominantly spherical planet they will be much more subtle. I would imagine that even with the variations in planet density, if we could see where the earth's c of g was it would be pretty much under our feet wherever we were.

 

Considering the distance (the "height" above the c of g, because the r term in the equation is squared, it will tend to dominate. If you increase your distance by a factor of 2, the for force will decrease by a factor of 4. The question was posed that is that does standing on a mountain counteract the "altitude effect" by concentrating more mass under you?

 

There is an uncited reference on the Wiki page that states that you weigh 0.28% less at the top of Everest than you do at sea level. Without doing the maths yourself its difficult to prove, but it makes sense when you consider that if the earth was perfectly spherical apart from Everest, then as well as being heavier the centre of gravity of the earth would be offset towards the mountain by a very small fraction of its height. However in order to add enough mass to make an effect the mountain would have to be very tall indeed, and by the time you had made it that tall, you would be even further away from the planet's c of g. In short, it makes sense to me In fact I'd be tempted to say that you would weigh more if you stood on the other side of the planet, at sea level, directly opposite the mountain.

 

To take Ian's hypothesis that "on a big enough planet you would be weightless", disproving this assumes that the planet has a more or less uniform density. As the planet gets bigger, your distance to its centre of gravity increases but so does its mass. The volume of a sphere (and indirectly, its mass) is 4/3 pi.r^3 so although the radius term in the equation would reduce by r^2 the mass term would increase by r^3. Overall the gravitational force would increase by a factor of r because (r x r x r) / (r x r) = r

 

However, if the planet was a gas giant with an iron core, then it is likely that the bigger the gas part got, the lower the gravity at the surface would be.

 

By the way, I love the idea that cetrifugal force won't make water won't fall out of a hole in the bottom of a bucket. If this was true then half the pumps in the world (including those in our Supra engines, and turbos as well) wouldn't work.

[Tannhauser]

 

(Lucid reasoning)

 

 

Thanks for that. I had actually drafted a reply using GmM/d^2 and referring to the fact that the effect of d increases geometrically (if that's the term), but I couldn't articulate it properly.

 

I like the Cylinder World example, and I followed it until your assertion that weight would be massively increased standing on the cylinder's curved surface directly over the centre of mass. It occurs to me that if the cylinder is relatively thin, then there are significant amounts of mass pulling at an angle to perpendicular, rather than straight down. How does this affect weight?

 

I can only assume that the forces towards Cylinder World's mass to the left and right can be considered two 'resultants' that could be resolved into a horizontal component and a vertical component. The horizontal components balance each other out leaving only the vertical components, which is what is registered on the scales as weight. So these attractional forces would act like guy ropes on a tent.

 

Is this right?

 

I posit that if this is true, on a banana-shaped world, weight at the 'equator' on the convex surface would be somewhat greater than on the concave surface. AT the latter, one would experience attraction towards the poles curving above.

[Ian C]

I'll get my coat.

 

-Ian

 

PS just read the link in your sig, Cliff. Best crucifiction for about 2000 years

[tbourner]

I like the Cylinder World example, and I followed it until your assertion that weight would be massively increased standing on the cylinder's curved surface directly over the centre of mass. It occurs to me that if the cylinder is relatively thin, then there are significant amounts of mass pulling at an angle to perpendicular, rather than straight down. How does this affect weight?

 

I can only assume that the forces towards Cylinder World's mass to the left and right can be considered two 'resultants' that could be resolved into a horizontal component and a vertical component. The horizontal components balance each other out leaving only the vertical components, which is what is registered on the scales as weight. So these attractional forces would act like guy ropes on a tent.

 

I assume if you go from the flat round surface at one end, over the edge onto the rounded sides, it would then be like walking down a mountain, you'd be 'pulled' toward the c of g of the planet, so similarly to standing on a (smooth surface) mountain you'd have trouble setting up your scales and standing on them! as Digs suggested your view of what 'down' means would have to be significantly different on that world.

[Digsy]

I can only assume that the forces towards Cylinder World's mass to the left and right can be considered two 'resultants' that could be resolved into a horizontal component and a vertical component. The horizontal components balance each other out leaving only the vertical components, which is what is registered on the scales as weight. So these attractional forces would act like guy ropes on a tent.

 

Is this right?

 

Exactly right as far as I can tell. The guy ropes analogy is spot on. It would be very easy to do the maths if you broke Cylinderworld up into, say 10 equal segments and calculated the magnitude and direction of the pull from each of them, summed them up and compared them to the pull from a single-piece Cylinderworld. I reckon they would end up being the same.

I posit that if this is true, on a banana-shaped world, weight at the 'equator' on the convex surface would be somewhat greater than on the concave surface. AT the latter, one would experience attraction towards the poles curving above.

 

Bananaworld is an interesting one. If the c of g isn't actually within the planet, then what happens? What you are suggesting makes sense but I can't be 100% sure that is what would happen.

 

The reason for my uncertaintly is that on the same Wiki page there is a section which mentiones the adjustment to gravity for a person that is under the planet's surface. It mentions that the mass of the "shell" of the planet which is outside of the radius where the person is standing does not count towards the mass producing the gravitational attraction to the planet's core. I found this to be counter-intuitive, as the c of g of the shell would still be at its center. There is a link to the maths associated with the spherically symmetrical earth which explains this, but I havn't got into it yet.

 

I would guess that this would not apply on Bananaworld, as it is neither spherical nor totally symmetrical, but as I don't understand the maths yet I'll not stick my head on that particular chopping block

[tbourner]

If you expand the bananaworld idea to a dyson sphere type world, where the person is inside the hollow planet, if it were a perfect sphere with a perfectly hollowed out interior and no rotation at all you would imagine gravity would be non-existant rather than holding the person in the middle - it may attract towards the edges even suggesting the c of g runs throughout the entire crust! I can't get my head around the principle and I'm too dumb to read into the maths!

[Digsy]

If you expand the bananaworld idea to a dyson sphere type world, where the person is inside the hollow planet, if it were a perfect sphere with a perfectly hollowed out interior and no rotation at all you would imagine gravity would be non-existant rather than holding the person in the middle - it may attract towards the edges even suggesting the c of g runs throughout the entire crust! I can't get my head around the principle and I'm too dumb to read into the maths!

 

That's exactly what the shell theorem says - each point mass on the shell produces an outwards attraction whose net effect is to cancel itself out. The mathematically odd thing is that this happens as soon as you pass inside the shell, not just when you reach its centre.

[tbourner]

So you're attracted to every single piece of mass? Therefore it will only 'cancel itself out' in terms of the whole planet's gravity, if there is an inner planet to be attracted to (ie: the remaining inner planet beneath you is the gravity you feel), I get that bit, but if you remove the inner planet the net gravity of the shell will be an attraction to the shell? So in the middle you'd be weightless but on the inside wall you'd have gravity relative to the thickness of the shell?

[4packet]

I was quite intrigued to find out that... a friend of mine, who studied physics at uni and now has a degree in it.. he works as an engineer for some posh aircraft company... the first thing they were taught in uni was that centrifugal force doesnt exsist. Only the principal does...

 

If you fill a bucket with water... cut a whole in the bottom of it.. and then swing it in a circle really fast... you would expect the water to fall out the bottom.. but it doesnt.

 

All the way through uni, all of his lecturers spent their time convincing people that it doesnt exsist.... lol

 

Agreed. I have been taught the same. The force only acts a perpendicular plane to the instantaneous direction of motion. To put another way it acts along the tangent of the circle.

 

The only 'real' force is called the centripetal. This essentially acts in the oppposite direction to what we think of the centrifugal force. This force is essentailly your arm holding onto the bucket.

[Digsy]

Nope, as soon as you move onto the inside boundary of the shell, or any point further inside the hollow part, the net force on you is zero. In effect you would be weightless once inside the shell. Gravity would only exist while you were in a tunnel from the outside surface to the inside surface, or when you are outside (on the outer surface or above it).

[Digsy]

The only 'real' force is called the centripetal. This essentially acts in the oppposite direction to what we think of the centrifugal force. This force is essentailly your arm holding onto the bucket.

 

I misread the initial post on this. Presumably the experiment with the bucket with a hole in bottom was never actually demonstrated?

[RedM]

OT:

 

Digsy, Tannhauser - What do you two do for a living? I'm loving this thread. I don;t get most of it but it's so well written.

[tbourner]

Nope, as soon as you move onto the inside boundary of the shell, or any point further inside the hollow part, the net force on you is zero. In effect you would be weightless once inside the shell. Gravity would only exist while you were in a tunnel from the outside surface to the inside surface, or when you are outside (on the outer surface or above it).

 

So I was right the first time? Well sort of.

Seems very counter intuitive though, like you have some mass there with no gravity associated with it! I guess it's related to the idea that if you drilled a hole through the Earth at ANY angle, it would take 42 minutes for gravity to get you to the other side (assuming no friction/air resistance etc), whether it be from Londond to Sydney or London to Surrey, it'll take 42 minutes!