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Author Topic: When the moon is Directly Overhead, How Much Less do I Weigh?  (Read 14393 times)

Offline chris

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Can someone good at physics suggest how much lighter I become when the moon is overhead, compared with on the opposite side of the planet? And how might one measure such an effect?

Chris
« Last Edit: 09/01/2008 23:05:01 by chris »


 

Offline syhprum

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I will start off the discussion, the Moon and the Earth rotate about a common C of G which is in fact below the surface of the earth.
Experiment and theory has shown that it is not possible to shield against gravity so my contention is that there is no effect. 
 

Offline lightarrow

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Can someone good at physics suggest how much lighter I become when the moon is overhead, compared with on the opposite side of the planet? And how might one measure such an effect?

Chris
For this you should give your mass. To my computations, gravity acceleration of Moon at 384,000 km (but there is about 8,000 km of difference between perigee and apogee, furthermore the distance is from Earth's centre and not Earth's surface) is about 3.4*10-5 m/s2 (I took 7.3*1022 kg as Moon's mass), so the composed gravity of Earth and Moon should be about 9.81 - 3.4*10-5 m/s2 when the Moon is overhead and 9.81 + 3.4*10-5 m/s2 when at the opposite side, so the difference is 6.8*10-5 m/s2. This means that if your mass is 70 kg, your weight difference would be 70*6.8*10-5/9.81 = 4.85*10-4 kg = 0.48 g.

Measuring the effect shouldn't be difficult with mothern scales: electronic analythic scales should have a sensitivity of 10-5g/10g = 10-6g/g;  0.48 g/70 kg ~ 7*10-6g/g.
Certainly, your question is very interesting. I had never thought about it and I'm wondering if I have made any mistakes in my reasoning/computation: at least in analythical chemistry the difference does not seem negligible.  ???
« Last Edit: 04/01/2008 19:06:11 by lightarrow »
 

Offline chris

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So how would modern scales measure the effect? Wouldn't the scales be pulled towards the moon too and because you are in the way (standing on them), they'd record the same weight?

Chris
 

Offline lightarrow

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So how would modern scales measure the effect? Wouldn't the scales be pulled towards the moon too and because you are in the way (standing on them), they'd record the same weight?

Chris
With a 2-plates scale, yes; but some electronics scales works in a different way: you put the sample on the plate and a magnetic field (which act repelling another constant magnetic field fix with the plate) is increased as to bring again the plate in the initial position, so there isn't another weight which equilibrate the sample, but a magnetic field (which is gravity-independent).
« Last Edit: 04/01/2008 19:25:08 by lightarrow »
 

Offline Soul Surfer

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The sorts of gravity meter they use for prospecting for minerals should allow you to meaure this quite well.
 

Offline chris

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So does everyone agree with lightarrow's suggested 0.48g weight loss?

Chris
 

Offline syhprum

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should we not also take into effect the orbital wobble of the Earth/Moon system as it rotates about the Sun.
 

Offline lightarrow

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should we not also take into effect the orbital wobble of the Earth/Moon system as it rotates about the Sun.

That's correct and the effect of the sun is about 180 times that of the moon! Your weight difference between day and night due to sun's gravity should be about 86 g! That's incredible! Are these things real?  ??? ??? Please tell me I'm wrong, I can't believe it!


Sun's mass ~ 2*1030 kg
Average Sun-Earth distance ~ 1.5*1011 m

Sun's gravity at that distance: GM/R2 = 6.7*10-11*2*1030/(1.5*1011)2 = 6.0*10-3 m/s2
Difference between night and day: 2*6.0*10-3 m/s2 = 1.2*10-2 m/s2
Weight difference for a 70 kg man: 70*1.2*10-2/9.81 = 8.6*10-2 kg = 86 g.
 

Offline syhprum

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In addition to the night-day variation there is a 28 day cycle due to the wobble of the C of G of the earth-moon system in its orbit about the Sun, this C of G is approximately 2565 Km below the Earths surface so there will be a 5130 Km variation in orbital distance each 14 days.
Numerous simplifying assumption have been made, ie that the orbital plane of the earth-moon coincides with that of the the orbit around the sun and is circular and no gravitational shielding exists.
A precise calculation taking all the factors would be rather challenging!

Are there many 70 Kg men ?, after xmas I am at least 85 
« Last Edit: 05/01/2008 13:37:28 by syhprum »
 

Offline lightarrow

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In addition to the night-day variation there is a 28 day cycle due to the wobble of the C of G of the earth-moon system in its orbit about the Sun, this C of G is approximately 2565 Km below the Earths surface so there will be a 5130 Km variation in orbital distance each 14 days.
Numerous simplifying assumption have been made, ie that the orbital plane of the earth-moon coincides with that of the the orbit around the sun and is circular and no gravitational shielding exists.
A precise calculation taking all the factors would be rather challenging!
Agree completely.
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Are there many 70 Kg men ?, after xmas I am at least 85 
:)
However I think Chris stayed in shape!
 

Offline chris

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I cetainly did!

So can anyone corroborate lightarrow's claimed 86g weight difference between day and night? That sounds like a lot.

Chris
 

Offline syhprum

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Take a trip down to the coast Chris and watch the sea sloshing about you will note that there is a pretty large gravity anomaly causing it
 

lyner

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Sun's gravitational effect: Lightarrow, in addition to the acceleration to the Sun's gravity - which gives you factor of +/- 40g effective weight, there is the acceleration due to the circular motion of the Earth's orbit around the Sun. I remember I did some sums a long while ago. The effect of, shall we call it centrifugal force, is in the opposite direction to that of the gravity and more or less cancels out. In the same way that an astronaut feels weightless when in orbit around the Earth, the kg mass and the Earth are both in 'free fall' around the Sun -so there is much less difference than the 40g. I calculated the difference to be about 1/1000 of that, taking both accelerations into account.
I was, originally, trying to work out the effect on tides but gave up when I realised that the tides actually consist of a wave which sloshes around the Earth once a day (ish) and the actual Q factor of what is, in effect, a resonator, would be a major factor in determining the actual heights of tides.

I calculated the sums of the rotational and gravitational effects due to the Solar orbit and the Lunar 'wobble'. This suggested that the difference due to the Moon's effect would be about 4e-5 N/kg whilst the effect of the Sun would be about 2.6e-6N/kg. It would mean a variation in measured weight of  4e-4% for the moon and 2.6e-5% for the Sun.
This isn't far from the ratio which they quote for the relative effect of Sun and Moon on the tides.
« Last Edit: 06/01/2008 16:02:09 by sophiecentaur »
 

Offline syhprum

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I have scoured Google for information and many graphs are available showing the daily-monthly variation etc, all agree that the contribution of the Sun and the Moon are in a ratio 2 to 1 approximately but none seem to give the actual amplitude of the effect.
Many complex second and third effects are disused but no details of actual amplitudes.
 

lyner

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Calculating the first order effects is easy.
Just apply the inverse square law to find the gravitational accelerations due to Sun and `Moon on the near and far faces of the Earth.
To find the accelerations due to circular motion (near enough for Jazz) you just use the formula "omega squared r" for the different distances. omega (angular frequency) is very small but the radius is big, so you get an answer which is more or less the same as the gravitational attraction.
If we were dealing with point masses, there would be no difference.
It would be good if someone else could do the same sums and reveal any arithmetical errors on my part.
I think my figures should be reasonable correct because, in both cases, the two accelerations were nearly the same and the differences were more or less the same on each face - implying that you should expect more or less equal heights of tide during the day - which is what you usually get.
At first I was amazed at the 'coincidental' almost total cancellation - then I reaslised about micro gravity in satellites and it made sense.

Actual tidal predictions don't seem to use this sort of calculation but rely on harmonic analysis of actual measured tides. They are based on seven harmonic constants which are different for each location.
 

lyner

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And yes - lightarrow must be in very good shape if he reckons on 70kg. I've been 85 kg for years and years.
 

Offline lightarrow

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Sun's gravitational effect: Lightarrow, in addition to the acceleration to the Sun's gravity - which gives you factor of +/- 40g effective weight, there is the acceleration due to the circular motion of the Earth's orbit around the Sun.
Right. Mistery explained!  :)
(How could I forget it?  ???)

Quote
I remember I did some sums a long while ago. The effect of, shall we call it centrifugal force, is in the opposite direction to that of the gravity and more or less cancels out. In the same way that an astronaut feels weightless when in orbit around the Earth, the kg mass and the Earth are both in 'free fall' around the Sun -so there is much less difference than the 40g. I calculated the difference to be about 1/1000 of that, taking both accelerations into account.
I was, originally, trying to work out the effect on tides but gave up when I realised that the tides actually consist of a wave which sloshes around the Earth once a day (ish) and the actual Q factor of what is, in effect, a resonator, would be a major factor in determining the actual heights of tides.

I calculated the sums of the rotational and gravitational effects due to the Solar orbit and the Lunar 'wobble'. This suggested that the difference due to the Moon's effect would be about 4e-5 N/kg whilst the effect of the Sun would be about 2.6e-6N/kg. It would mean a variation in measured weight of  4e-4% for the moon and 2.6e-5% for the Sun.
This isn't far from the ratio which they quote for the relative effect of Sun and Moon on the tides.

« Last Edit: 06/01/2008 20:51:01 by lightarrow »
 

Offline chris

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Okay, so to summarise, can we agree that, ignoring the wobble of the Earth / Moon system as it orbits the sun, every day the weight of a 70 kg man (assuming he exists) would change decrease by 0.48g when the Earth-Moon distance is at a minimum for the patch of Earth's surface upon which that man is standing?

(I just want to make sure that the physics brains on the forum are happy to sign off this answer before some random newspaper publishes this "fact" we've written here, attributes it to us and then makes us look stupid when it turns out we forgot to factor in something).

Cheers

Chris
 

Offline lightarrow

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When the moon is Directly Overhead, How Much Less do I Weigh?
« Reply #19 on: 10/01/2008 08:16:04 »
And yes - lightarrow must be in very good shape if he reckons on 70kg. I've been 85 kg for years and years.
You haven't read well my posts, I was talking about Chris! (I know from a picture of him that he is in good shape  ;D)
 

Offline syhprum

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When the moon is Directly Overhead, How Much Less do I Weigh?
« Reply #20 on: 10/01/2008 08:27:49 »
Can I just confirm what we are signing off to, over a 24 hr cycle as the earth rotates there is a 0.48g difference in our attraction towards the centre of the Earth between when the moon is directly above us to when the moon is directly below us (the other side of the Earth).
Sounds reasonable.
There is also a 28 day cycle when the earth/moon system rotates relative to the sun that produces an approximately 50% modulation of this effect.
 

lyner

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When the moon is Directly Overhead, How Much Less do I Weigh?
« Reply #21 on: 10/01/2008 10:49:32 »
Chris, the 'wobble' is all part of the calculation; you can't ignore it. The Earth is orbiting around the centre of mass of the Earth-Moon system. This centre just happens to be beneath the surface of the Earth; this is the 'wobble'. The radii of rotation of the bits which are facing towards and away from the Moon are different - so the accelerations due to circular motion are different. These two accelerations, added to the gravitational attraction of the Moon, produce the overall disturbance to the weight due to the Earth's mass. Without the wobble, the difference would be enormous - as in lightarrow's earlier post.
The Earth is rotating with a 24hour period, so, on one spot (on the equator) you would expect a total spread of weight of 0.00004N for every kg, or, on a spring balance, 0.004g  spread of measurements for a kg. With a 'beam' balance, there would be no difference, of course.
I was hoping that someone else could repeat the sums to get agreement on the actual value before we could commit ourselves.
A similar calculation can be applied to the parts of an orbiting space station; you would not be exactly 'weightless' because the net accelerations of the nearest and furthest parts of the ship would be different.
 

Offline turnipsock

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When the moon is Directly Overhead, How Much Less do I Weigh?
« Reply #22 on: 10/01/2008 13:16:21 »
what about when the sun and moon are directly overhead?
 

lyner

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When the moon is Directly Overhead, How Much Less do I Weigh?
« Reply #23 on: 10/01/2008 14:14:03 »
You should get the two effects adding up, vectorialy, wherever they are, relative to each other. Hence spring and neap tides! Springs are when Sun, Moon and Earth are aligned and neaps are when Sun and Moon are at right angles. I have to dip out on the quantitative relationship where tides are concerned - too hard guv.
 

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When the moon is Directly Overhead, How Much Less do I Weigh?
« Reply #23 on: 10/01/2008 14:14:03 »

 

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