View Full Version : Weight variation

Spiny Norman

12-01-2009, 07:14 PM

If one's weight is a function of the strength of the gravitational field at one's location combined with one's mass ... then does one's weight change throughout the day as the moon rotates around the earth?

I would assume so, seeing as the strength of the local gravitational field must vary (as evidenced by the changing tides).

But if I just stand of my bathroom scales, the mechanism is presumably also affected by the strenth of the same local gravitational field strength (?). So what methodology could I use to estimate the amount of change between high-, low- and mid-tide levels?

(what set me to thinking about this was the effect of today's King Tide, the highest tide for the year).

Rincewind

12-01-2009, 07:57 PM

If one's weight is a function of the strength of the gravitational field at one's location combined with one's mass ... then does one's weight change throughout the day as the moon rotates around the earth?

True but the variation is well below what can be measured by most scales including household bathroom variety.

The attractive force is proportional to the mass of body and inversely proportional to the square of the distance. Therefore a back of the envelope calculation using the values of

radius of earth 6.37E+06 m

mass of earth 5.97E+24 kg

distance to moon 3.63E+08 m

mass of moon 7.35E+22 kg

So the ratio of the gravitational attraction of the earth is roughly 264,000 times stronger than that of the moon (for someone on the surface of the earth)

Or in other words if you weighed 100 kg your weight would fluctuate by less than 400 milligrams due to the position of the moon.

Spiny Norman

13-01-2009, 06:28 AM

Thanks RW, that's the bit that had escaped me ... I knew that the gravitational pull of the earth compared to the moon, for a body on the earth as opposed to on the moon, was roughly 6x stronger ... but I didn't know how to calculate the pull of the moon for a body on the earth.

OK ... so ... 400mg ... not enough to make a whit of difference. Shame that. I'm on a weight loss regime, and was thinking that I might be able to skew the results slightly by only measuring on certain days (using tide charts)!

The things I think of, sometimes, it scares the hell out of me ...

Denis_Jessop

13-01-2009, 10:29 AM

Nevertheless, one's weight changes throughout the day for other reasons. For example, I normally weigh a kilo or more in the evening than in the morning regardless of anything else. I have been checking my weight daily for many years because of a combination of checking fitness and controlling cholesterol levels, among other things.

DJ

Capablanca-Fan

13-01-2009, 11:22 AM

If one's weight is a function of the strength of the gravitational field at one's location combined with one's mass ... then does one's weight change throughout the day as the moon rotates around the earth?

I would assume so, seeing as the strength of the local gravitational field must vary (as evidenced by the changing tides).

The reason the tides are strong is the difference in the fields over the distance of the earth's diameter. The tidal forces, in the first approximation, are inversely proportional to the cube of the distance to the tide-causer, and directly proportional to the diameter of the tide-containing body (the tidal bulges are approximately the gravitational analogue of an electric dipole):

F(t) ≈ 2GMmr/R^3, where r is the earth's radius and R is the distance from the earth's centre to the moon's centre.

Your height is minuscule compared to the earth's radius, so the tidal effect on you is minute. The inverse cube law explains why the moon's tidal effect is greater than the sun's.

Rincewind

13-01-2009, 02:13 PM

The reason the tides are strong is the difference in the fields over the distance of the earth's diameter. The tidal forces, in the first approximation, are inversely proportional to the cube of the distance to the tide-causer, and directly proportional to the diameter of the tide-containing body (the tidal bulges are approximately the gravitational analogue of an electric dipole):

F(t) ≈ 2GMmr/R^3, where r is the earth's radius and R is the distance from the earth's centre to the moon's centre.

The "tidal effect" is not really what Frosty was after as that would just make him bulge in some direction. To effect his weight you just need to calculate the simple gravitational force using Newton's famous formula

F(g) = G M m / r^2

The weight fluctuation can even be estimated independent of the value of G by dividing the Earth's force by the Moon's force (as I did above) or if you like you can measure the force in Newtons, e.g. for an 100 kg person as around F(g) ≈ 3.72E-03 N.

Using a human "radius" of 1 m the tidal force works out to a force of F(t) ≈ 2.05E-11 N which is much smaller than the attractive force which may effect the measurement of weight.

Likewise the attractive force of the Moon on the Earth (and vice versa) is F(g) ≈ 2.22E+20 N, which is an order or two larger than the tidal force which is F(t) ≈ 7.8E+18 N.

The reason the human tidal effect is so small is that the radius of a human (1 m) is minuscule compared to the distance to the moon. Or alternatively the reason the tidal effect of the Moon is (comparatively) so strong for the Earth is that the Earth's radius is not negligible compared with the distance to the Moon.

As far as I can tell the only analogy between the tidal force and the electric dipole is the scaling law and in that sense there is an analogy but otherwise it is misleading especially since more correctly analogous gravitational dipoles are (at least) theorised.

Capablanca-Fan

13-01-2009, 03:03 PM

The "tidal effect" is not really what Frosty was after as that would just make him bulge in some direction.

The tidal effect comes from the difference in graviational force over a large distance. So in effect, Frosty was after the magnitude of this in his own body.

As far as I can tell the only analogy between the tidal force and the electric dipole is the scaling law and in that sense there is an analogy but otherwise it is misleading especially since more correctly analogous gravitational dipoles are (at least) theorised.

Of course, there is no analogy to positive and negative gravitational charge. But there is an analogy that electrical and gravitational forces are central and conservative so obey the Inverse Square Law. So Newton's Law of Gravitation is an analogy to Coulomb's Law:

F = -GMm/R² and F = -kQq/R²

So it should not be surprising that the equation for tidal force is analogous to that of a force from a point charge on an electric dipole aligned with the direction to the charge, both derived from the binomial expansion, where r is the radius of the body on which the tidal force acts, and the distances between the charges on the dipole:

F(t) ≈ -2GMmr/R³ and F(d) ≈ −2kQqr/R³

Rincewind

13-01-2009, 03:17 PM

The tidal effect comes from the difference in graviational force over a large distance. So in effect, Frosty was after the magnitude of this in his own body.

No he was clearly after the variation in weight which is affected by F(g) not F(t).

The tidal force F(t) is an internal force which comes about due to a (significant) gradient in the gravitational potential across a body and will not affect his weight at all.

Spiny Norman

13-01-2009, 05:43 PM

Well ... speaking of my bulges ... they do seem to go in a certain direction ... but I don't think they're primarily tidal at the moment! <lol>

Kevin Bonham

13-01-2009, 05:50 PM

(what set me to thinking about this was the effect of today's King Tide, the highest tide for the year).

In Melbourne? Hobart had very low afternoon low tides at 0.05 and 0.07 metres on Sunday and Monday respectively.

Spiny Norman

14-01-2009, 07:27 AM

They were both very high and very low respectively. Something to do with the Moon being at perigee (? closest to earth due to elliptical orbit, and perhaps sun and/or planetary alignments ?). Someone told me that day that it would be the highest king tide for the year, about 50cm above normal.

Which reminds me ... if natural variation of tides can shift up to 50cm higher than normal ... how does this compare to what's predicted for global warming over the next century? Certainly, down here at Phillip Island, the very high tide made not one whit of difference to anyone, as far as I could tell ... although the very low tide prevented a few jet skiers from enjoying things, as they were picking up too much seaweed in their intakes!

Desmond

14-01-2009, 08:17 AM

Nevertheless, one's weight changes throughout the day for other reasons. For example, I normally weigh a kilo or more in the evening than in the morning regardless of anything else. I have been checking my weight daily for many years because of a combination of checking fitness and controlling cholesterol levels, among other things.

DJYes and for this reason it is usually recommended to weigh yourself at the same time each day. Actually I weigh myself once a week on the same day too, to remove other variations such as different eating habits and activities over the weekend.

Rincewind

14-01-2009, 08:35 AM

Which reminds me ... if natural variation of tides can shift up to 50cm higher than normal ... how does this compare to what's predicted for global warming over the next century? Certainly, down here at Phillip Island, the very high tide made not one whit of difference to anyone, as far as I could tell ... although the very low tide prevented a few jet skiers from enjoying things, as they were picking up too much seaweed in their intakes!

If sea levels rise so that the normal sea level is the present king tide level then area which do have occasional problems with large tides will be like that permanently and areas which don't normally have problems will have problems when king tides occur on top of the heightened sea level.

Bruce Oates

15-01-2009, 10:16 AM

Way beyond me... I thought high tides were caused by all the fat

people swimming on the sunny side. :hand:

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