(How much) does it matter when you weigh something?

[Deadbeat]

WARNING! MATHS!

Awfully disappointing lack of maths in this thread for a forum of geeks, so I've had a bit of time to play with the equations and I've come up with a representation.

Apologies first:

1) Sorry it took so long. I had a bit of a problem sorting out the Windows calculator - couldn't remember how to raise indices to negative powers, couldn't figure out how to resolve brackets properly etc etc. Got it sorted in the end, though.

2) Sorry if it's slightly wrong. The first time I calculated the force due to earth I came out with 9.817N, but in later revisions I could only get it to 9.822N. From what I remember, it's supposed to be 9.780N - if anyone can see what I'm missing I'll be very grateful, but it's not too important for the purposes of demonstration. I've not really gone too deep into accuracy here (despite appearances).

3) Sorry for the formatting. I don't know how to properly reproduce formulas or type in subscript, so you'll have to bear with me.

Basic Formulas and constants

(1)F = m x a

(2)F = G((m1+m2)/r2) - alternatively F = G(m/r2) since the object in question [1kg mass] is negligible in comparison to the object whose gravitational force we're calculating [the moon].

(3)W = m x a

Any physicist will be able to tell you that these are essentially all the exact same formula. Weight is just a representation of the force due to mavity of a given object, so in higher mavity environments the same mass object would weigh more, and in low mavity the same object would weigh less (contrary to what Tekgun said above).

Constants include the universal gravitational constant G, the min, max and mean distance of the moon from the Earth and radius of the Earth visualised as a sphere of equal mass to its true elliptical shape.

G = 6.674 x 10^-11

Earth:
rEarth = 6.370 x 10^6 (visualised as a sphere)
mEarth = 5.974 x 10^24

Moon:
rmin = 3.564 x 10^8 - the square of this is 1.270 x 10^17
rmax = 4.067 x 10^8 - the square of this is 2.178 x 10^17
rmean = 3.844 x 10^8 - the square of this is 1.478 x 10^17
mmoon = 7.348 x 10^8

All distances in metres, of course.

Calculating force due to mavity

Firstly, we'll calculate the force of mavity of the Earth for future reference:

FEarth = G(5.974 x 10^24)/4.059 x 10^13 = 9.823 N [formerly calculated as 9.817N - dunno why it changed later on, but it did ]

Now, the force of the moon at mean distance:

Fmean = G(7.348 x 10^22/1.478 x 10^17) = 3.318 x 10^-5 N

Fmin is applied when rmin is in effect (Fmin in this case refers to the distance rather than the force, since the force is actually greatest in this case).

Fmin = G(7.348 x 10^22/1.270 x 10^17) = 3.861 x 10^-5 N

Fmax = G(7.348 x 10^22/2.178 x 10^17) = 2.252 x 10^-5 N

So, now we have the relative forces, we can calculate the applicable weights under each circumstance.

Calculating weight

We'll be using an object of mass 1kg to simplify everything. For the record, weight is not the same as mass, as seen in this section.

W = mg, and g = gEarth - gmoon. Without a moon:

W = 1 x 9.822733678 = 9.822733678kg

So an object of mass 1kg on the surface of our spherical Earth has a weight of about 9,822g - or 9.822kg. I'm using the expanded force resultants that I've got noted down but not typed up here so I can compare weight differences in reasonable units later. From here on in I'll be dropping the (1 x) function as it's redundant. Just remember that all the forces are multiplied by the mass of our object, which is 1kg.

Moon at distance:

Wmean = 9.822733678 - 3.318034641 x 10^-5 = 9.822700497kg

Wmax = 9.822733678 - 2.251632323 x 10^-5 = 9.822711162kg

Wmin = 9.822733678 - 3.861460787 x 10^-5 = 9.822695063g

Summary

Compared to the mean distance of the moon from the Earth, which we'll take as our reference frame, having the moon at its closest point to the object decreases its weight by 5.454mg. Having the moon at its farthest point increases the weight by 10.665mg.

Having the moon orbiting around us increases the weight of our object by 33.181mg compared to not having a moon.

All of the above calculations are considering the Earth as a solitary body in space or accompanied by nothing more than the moon - the Sun, other planets, stars and any other physical object in space is discounted. For the sake of completion, the gravitational attraction of the moon at mean distance is 3.318 x 10^-5 N and throws the weight of an object off by 33.181mg (about 0.0033%). The gravitational attraction of the sun, after some rough calculations, is about 2.518 x 10^-60 N, and would throw the weight off by about 0.0000003%. We're talking about weights in the order of a millionth of a mollionth of a mollionth of a mollionth of a mollionth of a mollionth of a mollionth of a mollionth of a mollionth of a thousandth of a gram.

TL;DR

The moon has a very, very, very small effect on the weight of an object. The sun has a negligible effect.

 

[Vonhelmet]

F=Gm1m2/r^2, not G(m1+m2)/r^2, as you've said in one of your lines above. At first blush, I'd say your errors in calculating g stem from using rounded figures in your calculations, but I don't know if you are actually using rounded figures or not without seeing what you're putting in your calculator.

EDIT: Disregard what I've just deleted, if you happened to see it. Apparently I'm awesome at physics, but terrible at reading.

 

[Deadbeat]

F=Gm1m2/r^2, not G(m1+m2)/r^2, as you've said in one of your lines above. At first blush, I'd say your errors in calculating g stem from using rounded figures in your calculations, but I don't know if you are actually using rounded figures or not without seeing what you're putting in your calculator.

I was calculating the relative acceleration between two bodies, not the (insert word I forget her... barysomething? It's been a while) acceleration. I resolved it in the final part by getting the net force acting on the object.

Edit: And of course it's negligible either way in this scenario - an addition of 1kg to a mass of 734 million kg won't really change things . Though I admit you may be right... it's been such a very, very long time since I actually studied this stuff that I can't really remember which formula is more appropriate.

 

[Vonhelmet]

I was calculating the relative acceleration between two bodies, not the (insert word I forget her... barysomething? It's been a while) acceleration. I resolved it in the final part by getting the net force acting on the object.

It's in point 2 in your basic formulas section, but yes - you're right, your later maths does assume a 1kg mass and it's clear you know what you're doing because you've not added it on to the mass of the earth, or said that you're not doing so due to it being negligible.

 

[Angilion]

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?

Not really. It's also not the right answer, since you have calculated the average strength of the moon's gravitational field at the centre of the Earth.

Obviously, being on "the ******* internet, the largest most easily searchable resource of information in human history." doesn't necessarily yield the right answer to a question.

I could go all GD about it and post something like this:

Dear GD. I have twisted my panties. How long should I microwave them for in order to untwist them? Also, spec me a fork to do it with.

Don't we have enough threads like that?

 

[bam0]

Not really. It's also not the right answer, since you have calculated the average strength of the moon's gravitational field at the centre of the Earth.

Obviously, being on "the ******* internet, the largest most easily searchable resource of information in human history." doesn't necessarily yield the right answer to a question.

You're trying to be awkward and that is what you go for? The difference between apogee and perigee is about 7 times greater than the radius of the earth. So at some point in the month my approximation would be right. Although obviously still not completely rights I used Newtonian equations to calculate it, the point being that for this level of discussion the answers we get are right enough.

 

[Vonhelmet]

Using point masses is a perfectly acceptable assumption when illustrating the point that it makes just a tiny, tiny difference.

 

[Angilion]

Using Wikipedia gave this answer:

[..]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.

Deadbeat's answer, using his own calculations (thank you for such a thorough answer to my middle of the night, spur of the moment question), got this answer:

Summary

Compared to the mean distance of the moon from the Earth, which we'll take as our reference frame, having the moon at its closest point to the object decreases its weight by 5.454mg. Having the moon at its farthest point increases the weight by 10.665mg.

They're quite different answers. A difference of about 0.2mg from Wikipedia and a difference of 16.119mg from Deadbeat. Maybe the criticism of "wiki-quoting" is justified?

 

[Angilion]

You're trying to be awkward and that is what you go for?

Since that is the fact that you didn't answer the actual question, I think it's the most relevant criticism.

If I was trying to be awkward, I would have led with the comment about you getting your panties in a twist. Why don't you just stick with "spec me an X" threads, or whatever it is you like, and save yourself the stress this thread has obviously caused you?

 

[[FnG]magnolia]

welp.

 

[Vonhelmet]

Using Wikipedia gave this answer:



Deadbeat's answer, using his own calculations (thank you for such a thorough answer to my middle of the night, spur of the moment question), got this answer:





They're quite different answers. A difference of about 0.2mg from Wikipedia and a difference of 16.119mg from Deadbeat. Maybe the criticism of "wiki-quoting" is justified?

Or maybe criticism of Deadbeat's answers is justified? Without seeing both answers it's hard to be sure who's right or wrong. Does the Wikipedia page give a reference for their statement? If so, maybe they have their maths and we can compare and contrast.

 

[Threepwood]

Dear GD. I have twisted my panties. How long should I microwave them for in order to untwist them? Also, spec me a fork to do it with.

Brilliant.

 

[Duff-Man]

Or maybe criticism of Deadbeat's answers is justified? Without seeing both answers it's hard to be sure who's right or wrong. Does the Wikipedia page give a reference for their statement? If so, maybe they have their maths and we can compare and contrast.

Deadbeat is assuming that the Moon exerts a linear gravitational pull upon the Earth. In fact, the Earth and moon are rotating around each other, around a common centre (which is closer to the Earth than the Moon, due to the difference in mass). The tidal forces are due to the difference in the Moon's gravitational pull over the Earth's surface, not due to the direct pull of the Earth, which is largely counteracted by the centripetal force exerted by the orbit. The gravitational change due to the tidal forces is significantly lower than you would find from a simple linear force-distance calculation.

As an analogy - consider a space-station orbiting the Earth. The "weight with Earth there vs weight without" argument deadbeat gives clearly shouldn't apply, because an astronaut sat in the satellite would feel weightless whether or not the Earth was there. When objects are rotating, you must always consider centripetal force.


This website explains it better, so I'll quote them:

Add the Moon, with 1/80 the mass of the Earth, orbiting in a perfect circular orbit once per 29.5 days at a distance of 300,000 km. Forget about the water for the time being.

In order for the system to balance, the Earth and moon must both orbit around a common center point called the barycenter. The barycenter is located 1/80 of the way from the Earth's center to the Moon's center, which turns out to be inside the Earth and about 2650 kilometers below the point on the surface where the Moon is directly overhead.

Because the Earth and Moon are orbiting the barycenter, they both have a centripetal motion, a circular motion that takes 29.5 days to complete one circle. There is also a corresponding centripetal acceleration. These are the orbital centripetal motion and orbital centripetal acceleration.

We are still assuming the Earth is rigid and solid, so every part of the Earth experiences the exact same amount of orbital centripetal acceleration.

However, the points on the Earth that are closer to the moon actually experience a bit greater gravitational pull from the Moon, and the points on the Earth farthest from the Moon experience less. An object sitting on the Earth's surface will experience this lunar mavity, which changes slightly throughout the day.

If the force were always exactly enough to equal the centripetal acceleration, there would be no variation throughout the day. But instead, the lunar mavity never exactly balances the centripetal motion, so there is a little bit of force left over. This little bit left over is the tidal force.

The tidal force is a small fraction of the total gravitational pull of the Moon, which in turn is a small fraction of the force from the Earth's mavity. It only exists when viewed from a frame of reference that is sitting on the surface of the rotating Earth.

The strength of the Lunar tidal force is 1/45 of the Moon's mavity, which in turn is 1/180000 of the Earth's mavity. In other words, the Lunar tidal force is 1/8,000,000 the strength of the Earth's mavity.

So Wikipedia suggested that a typical tidal variation was roughly 1/5,000,000 of Earth's mavity, and the above estimate shows 1/8,000,000 as the maximum magnitude.

 

[Deadbeat]

Aye, as stated it was pretty rough working with a lot of simplified assumptions - spherical earth, min/max distances, linear mavity etc etc. Wikipedia said it themselves in the linked article:

A perfect sphere of spherically uniform density (density varies solely with distance from centre) would produce a gravitational field of uniform magnitude at all points on its surface, always pointing directly towards the sphere's centre. However, the Earth deviates slightly from this ideal, and there are consequently slight deviations in both the magnitude and direction of mavity across its surface. Furthermore, the net force exerted on an object due to the Earth, called "effective mavity" or "apparent mavity", varies due to the presence of other forces, such as the centrifugal force of Earth's rotation. A scale or plumb bob measures only this effective mavity.

Parameters affecting the apparent or actual strength of Earth's mavity include latitude, altitude, and the local topography and geology.

Saying that, I can't find the quoted value of 999.9998g anywhere on the page. Anyway, they're diffferent calculations - Wikipedia gives the weight of an object at high and low tides, i.e. when the moon is (presumably) directly overhead and when it's perpendicular from the object using the centre of the Earth as zero coordinates, whereas I was taking the moon at its closest and farthest points directly overhead.

Anyway, it's only to show the sort of effect it has on weight, which is virtually negligible. You'd have to have a much more massive body for the effect to be anywhere near worth considering, and as shown, the Sun makes next to no difference at all.

I remember doing a similar calculation a few years back while poking fun at astrology. Turns out that, from a gravitational perspective, what your midwife ate for lunch has a much greater effect on you at the moment of your birth than the relative position of Jupiter.

 

[bam0]

Since that is the fact that you didn't answer the actual question, I think it's the most relevant criticism.

There are loads of reasons why my answer is not "right", but your criticism isn't really relevant, the Moon naturally cycles between apogee and perigee so the precise answer is dependent on where in that cycle it it (not to mention a whole host of other factors) the point was that even with the rough approximations made the answer was "very very little", you'll get bigger fluctuations just on earth.

 

[Duff-Man]

Aye, as stated it was pretty rough working with a lot of simplified assumptions - spherical earth, min/max distances, linear mavity etc etc.

The problem isn't with your approximations about Earth's shape, the orbital distance, or your model for mavity. It's that you didn't take into account the rotation. Remember the most basic principle of orbit:


The gravitational attraction between bodies is equal and opposite to the repulsive force due to rotation.


This is, after all, why two bodies can exist in stable orbit - i.e. without smashing into each other. The orbital distance will be determined by the amount of angular momentum each body has; more angular momentum allows for a shorter orbital distance (where the strength of attraction is greater).

Your calculations suggest a constantly non-zero force of attraction between the Earth and Moon, which would imply they are accelerating towards each other, and would eventually collide. This is not the case... If you take into account the rotation, this attractive force disappears entirely.

The tidal forces arise only due to the radius of the Earth, and would not be visible if the Earth and Moon were modeled as particles. This is the reason why your estimate is many orders of magnitude higher than the true value.

Saying that, I can't find the quoted value of 999.9998g anywhere on the page.

The Wikipedia page quotes 2 µm/s² (i.e. 0.000002m/s²) as the magnitude of the tidal forces. The typical Earth mavity is 9.8 m/s². From this you can compute the different in weight due to tidal forces, as shown above.