![[FnG]magnolia](/styles/overclockers/avatars/9.png){kind=avatar}
Caporegime
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Yeah, it was. Sorry 
(How much) does it matter when you weigh something?
- Thread starter Angilion
- Start date
Sorry for posting physics magnolia, should I do religion as well?
Christ, people moan about a lack of intelligence, but when someone tries to start a half decent debate on physics they get jumped on.
Although you have since apologised, which is fair enough.
Christ, people moan about a lack of intelligence, but when someone tries to start a half decent debate on physics they get jumped on.
Although you have since apologised, which is fair enough.
Tbh magnolia is right, when did people become so convinced of the value of their opinions? Additionally, you're on the ******* internet, the largest most easily searchable resource of information in human history. But rather than go and perhaps learn something, or know when to STFU, we get posts of peoples thoughts or assumptions (I'm talking about topics like this one in general now rather than this one specifically).
I don't know the answer to the specific question off hand, but it wasn't exactly hard to find the tools to calculate it.
The gravitational acceleration at earth due to the moon (using average distance) is;
6.6742x10^-11 * (7.3477×10^22 / (384399000^2))
That's the gravitational constant multiplied by the mass of the moon over the distance between the earth and moon squared.
= 0.00003318739161 ms^-2
So a 100kg person would have a weight of 982.2N due to earth's mavity alone.
They would be only 982.196681260839N if the moon was in front of them.
Basically you are 0.00337888328% lighter or heavier if the moon is directly in front or behind you respectively, and this is only rough as the distance between the earth and moon changes, not to mention the force of mavity due to the earth changes depending where you are on it as it's not a uniform sphere.
Was that really so hard?
I don't know the answer to the specific question off hand, but it wasn't exactly hard to find the tools to calculate it.
The gravitational acceleration at earth due to the moon (using average distance) is;
6.6742x10^-11 * (7.3477×10^22 / (384399000^2))
That's the gravitational constant multiplied by the mass of the moon over the distance between the earth and moon squared.
= 0.00003318739161 ms^-2
So a 100kg person would have a weight of 982.2N due to earth's mavity alone.
They would be only 982.196681260839N if the moon was in front of them.
Basically you are 0.00337888328% lighter or heavier if the moon is directly in front or behind you respectively, and this is only rough as the distance between the earth and moon changes, not to mention the force of mavity due to the earth changes depending where you are on it as it's not a uniform sphere.
Was that really so hard?
What's even more mind-blowing is that, because we don't know the absolute conditions needed for life - and because everything in the universe projects an infinite gravitational pull, no matter how small - then life may exist on this planet only because of the absolute positioning of everything else in the entire universe! Freaky, eh?
Disagree, I'm not sure you are quite appreciating how vast the Universe is and how small some of those effects are.
2 1 kg weights 1 light year apart would attract each other at 8.33106526×10^-75 ms^-2
That means it would take 6.15159817x10^31 years for each to move 1 planck length (16.162×10-36 metres) closer to the other (ignoring expansion of the universe or any other effects). For contrast the universe is only currently 1.37×10^10 years old.
Even really really heavy things are likely too far away to have any effect.
They do have an effect, the inverse square law means the resulting effect will never be zero, it will tend to zero but will become infinitely small.
I am aware how large the universe is, but to say those effects are zero is wrong
I am aware how large the universe is, but to say those effects are zero is wrong
[FnG]magnolia;21020963 said:
The educational standard of these boards does occasionally look bleak. It doesn't follow from the "hey guys' what's a fraction?" threads that there's no one on here who can describe mavity. The Newtonian approach outlined by bam0 is perfectly adequate.
I doubt I'm the only one who knows the best model currently available is mass induced space time curvature, i.e. general relativity. The mathematical implementation involves applied tensors. That realistically has no place on these boards, and I'm entirely unwilling to attempt to apply it to the OP. Try xkcd forums for a more likely crowd.
If you, magnolia, would like to reduce your ignorance on the subject, then the Feynman lectures are probably the place to start. Best of luck with them.
This is why we can never find the absolute mass of something, because, no matter where in the universe it is, it will still be affected by mavity of some sort
This is why we quote the measurement with error values.
however the effect would be indirect, not direct.?..yes?
Each object having an effect on its neighbour and then that effect alters the effect that object has on it's next neighbour and so on.....so while the effect of a distant object may not directly affect the Earth, indirectly it would?
We're talking truly vast distances for some of these objects though, and on that scale the tidal forces on Earth (tiny diameter in comparison) are going to be a small number at the end of a lot of zeros. Not forgetting that these same objects are also exerting attractive forces on the whole solar system.
A useful example; astronauts aren't weightless because they've escaped earth's mavity, it's because they are in freefall (both them and the vessels they are in).
Whilst I'm unable to contribute any further to this thread, I find this kind of discussion fascinating. So even though we don't have Stephen Hawking on hand to tell us the exact answer of everything, keep threads like this coming as I'm certainly reading.
I was speaking in relative terms though.....of course the influence of a body a 100m lightyears away has no direct effect on the Earth, however as a Universal concept it does have an indirect effect, as does everything within the Universe have on everything else....
I think that was the point being made, not that everything has a direct effect on any given point.
A variation of The butterfly effect if you like.
I am not physicist or mathematician, but it seems reasonable to assume that while the black hole the otherside of the Galaxy has no direct effect on me, it does have a direct effect on its neighbours which in turn alters the effect it has on its further neighbours and so on....all the way to that influence, however small, however indirectly, however immeasurably having an effect on me.
A Universal Connection if you like, where eveything in the Universe has an effect that alters that Universe in someway.
I'm probably not explaining it very well tbh.
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I see what you're saying "that every action has a reaction" but by the time those reactions reach us, it's at space foam level.
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I can sort of see what you're saying, but I still disagree. The problem for me is a matter of scale, both in space and time. The forces are so small that the amount of time they need to have any effect is overwhelmingly swamped by local effects.
You mention this Galaxy so I looked at some numbers (strictly Newtonian for time and simplicities sake). We're somewhere between 25-28 thousand light years away from the centre of our galaxy so I'll take the average of those two.
The total mass of the Milky Way is estimated at 2-3E42 kg, so lets call it 2.5 for now.
Assuming the entire mass of the Milky Way acting as a single point it would take 12 years for mavity to pull you 1 metre closer. 1 metre out of 26500 light years.
The amount of time it takes for these tiny forces to compound into large effects is so far beyond the human lifespan I just cannot consider them as relevant.
It would also take them a whole year before they started accelerating.
Assuming they were instantly teleported there, sure. Although I think we can gloss over 1 year in 60 million trillion trillion.
Assuming they were instantly teleported there, sure. Although I think we can gloss over 1 year in 60 million trillion trillion.
Fair enough.
About the same time they signed up for a discussion forum?
For a lot of people, yes. We're all good at some things and bad at others. I found your answer to the OP's question genuinely interesting, and also something I probably couldn't have done myself even you pointed me in the right direction.
Your attitude, however...
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Not sure if it's been answered yet, but anyway:
The Earth's gravitational field-strength varies over its surface. There are relatively small-scale variations, due to mountain ranges and various other features of the planet's crust. By far the biggest difference in mavity comes from the ellipsoidal shape of the Earth (the planet is spinning, and so "bulges" around the equatorial region).
The magnitude of the tidal forces due to the sun and moon is minute in comparison. You could estimate the difference my examining the difference in sea-level at high and low tide, and comparing that against the Earth's radius.
Wikipedia, as usual, has more specific info: (link).
So, due to the shape of the Earth, an object which "weighs" 1Kg (1000g) at the North pole will weigh 994.7g at the equator. On top of this you can add local variations due to geology, but these are of much smaller magnitude.
So, an object weighing 1Kg at high tide would weigh about 999.9998g at low tide. This difference would be immeasurable with all but the most sophisticated instruments.
The Earth's gravitational field-strength varies over its surface. There are relatively small-scale variations, due to mountain ranges and various other features of the planet's crust. By far the biggest difference in mavity comes from the ellipsoidal shape of the Earth (the planet is spinning, and so "bulges" around the equatorial region).
The magnitude of the tidal forces due to the sun and moon is minute in comparison. You could estimate the difference my examining the difference in sea-level at high and low tide, and comparing that against the Earth's radius.
Wikipedia, as usual, has more specific info: (link).
wikipedia said:
So, due to the shape of the Earth, an object which "weighs" 1Kg (1000g) at the North pole will weigh 994.7g at the equator. On top of this you can add local variations due to geology, but these are of much smaller magnitude.
wikipedia said:
So, an object weighing 1Kg at high tide would weigh about 999.9998g at low tide. This difference would be immeasurable with all but the most sophisticated instruments.