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The effect of the Earth's Moon's gravitational pull on our weight

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Mike13294


Registered: Feb 04
Posts: 2
Cool

Hello there--if someone could please answer the following question and provide legitimate proof/fact I would be very grateful.

I was wondering if during Spring Tide (When the Earth, Sun and Moon are all aligned), if you were to stand within the hemisphere directly being affected by the gravitational pull--would your weight actually decrease for that given period of time? And by how much?

Thanks.

-Mike D.

02/11/04 01:31
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Shambolic


Registered: Nov 03
Posts: 369

I'll assume the order of the bodies is Sun-Moon-Earth.

Rough Answer
No, your weight won't change. Remember that the Sun/Moon will pull on Earth as well. Even though your body will be dragged towards them, the Earth will be dragged up behind you, and so still press against your feet equally hard. You will feel no lighter, and no experiment will show you to be any lighter.

Exact Answer
Because your centre of mass will be slightly closer to the Sun/Moon than the Earth's centre of mass, you will be pulled slightly more strongly. This is an example of a tidal force, and will make your effective weight decrease slightly.

A quick calculation, using the average distances for Earth-Moon and Earth-Sun show that your weight will only be smaller by one part in 6 million. That's a neglible effect!!

If you want me to post details of the calculation, just ask.
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Shambolic - keeping it complex analytic

02/11/04 02:36
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alfnapper


Registered: Nov 01
Posts: 352

Hi Mike12394,i hope your good at math,check out the following site it has all the different senerios for tidal waves.I hope it helps and welcome to space talk and maybe Shambolic can verify the equations on this site i'm not sure if there all still valid. http://www.cepo.interacesso.pt/Artigos/Astrol/GcalcEhtm
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02/11/04 03:15
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Shambolic


Registered: Nov 03
Posts: 369

That site is using slightly different assumptions to get its values for weight variation. That's why their value is a fair bit bigger than my one part in six million.

I ignored centrifugal effects, and just calculated the difference between...
1. The earth being all alone in the universe
2. The sun and moon pulling on the earth and a person standing on it.
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Shambolic - keeping it complex analytic

02/11/04 03:40
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Mike13294


Registered: Feb 04
Posts: 2

Thanks--for the input guys.

And Sham--I knew that it wouldn't effect it that great--but since during Spring Tide--the oceans on Earth do rise a significant amount--I figured that maybe we too could loose some weight.

02/11/04 17:58
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Shambolic


Registered: Nov 03
Posts: 369

The reason tides can occur fairly easily is simply because the water is free to flow. Even a small acceleration can result in a fairly reasonable displacement over time.

For a reason the effect on our weight is so small, look at these figures:

- Acceleration due to sun's gravity (measured at average earth-sun distance): .0059 m.s^-2

- Acceleration due to moon's gravity (measured at average earth-moon distance): .000033 m.s^-2

- Acceleration due to earth's gravity (at surface of earth): 9.8 m.s^-2!!

As usual, if you want me to justify my figures, just ask, and I'll post the calculations.
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02/12/04 01:43
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Coppernicus2


Registered: Jun 03
Posts: 158

OPPOSITE FACE MOON TIDAL BULGES ARE ELECTROSTATIC?

I've never found a truly satisfactory reason in all my readings why on the opposite side of the globe facing away from the moon or sun, are nearly the same as the tidal bulge facing them. The idea that the gravitational pull on the opposite side of the planet is canceled by the planet does not satisfy me. 'Push' gravity does not explain it effectively either. So I would like to present an idea which I suspect is original, if not crazy, though another may have thought of it first: the risen tide on the planet's opposite face away from the moon or sun is due to electrostatic charge.

As outlandish as this may appear at first blush, consider the following suggestions:

1. There is a magnetic field from pole to pole, if off axis, that permeates the globe. This magnetic field gives a negative charge value to the northern hemisphere and a counter positive charge to the southern hemisphere, with a more neutral reading around the equator, where the positive and negative meet.

2. Tides appear to be greater towards the poles and lesser near the equator, which cannot be accounted by for planetary spin, since that would yield the opposite effect. So another cause must be found, which leads to the idea that like charges repel, even in the planet's ocean mass.

3. If like charges for large water mass, such as the oceans, repel, then the waters on the northern hemisphere, as they rise in response to the moon's or sun's gravity, will create a bulge towards the gravitational attraction and one opposite to it. This bulge creates a large negatively charged mass which, being of like charge, will be repelled on the opposite side of the planet, so that there too the water will bulge outwards. So due to like charge repelling like charge, the opposite side of the planet's mass is now counter-bulging in response to the mass bulge on the side facing the moon or sun. On the equator, where the positive-negative charge more or less cancels, there is less tidal opposite face response, so that there is less bulge than where the electrostatic charge is greater. On the southern hemisphere, the positively charged water mass likewise creates its own opposite bulge, same as in the northern.

4. Why does the equatorial water not bulge as much as towards the poles? My best guess is that it is already bulging, because of the planet's spin, so that the additional gravitational pull of the moon or sun does not displace this water as much as in either hemisphere because the centrifugal force already displaces it. Thus, equatorial waters should experience lower tides. I would expect, by this reasoning, that the waters near the artic or antartic circles should rise the most, and then taper off gradually towards the poles, or flatten out towards the equator. Of course, this basic principle is further modified by ocean currents and land mass interferences.

So by this reasoning, the tidal bulges on the opposite side of the planet are actually electrostatic phenomena, where like charge repels like charge, even in the ocean waters. The same would be expected in the land mass, though to a much lower degree. In effect, the planet is pulsating with a gravitationally activated bulge which then triggers a like response on the opposite side of the planet through like electrostatic force repulsion. If it did not do so, we would wobble chaotically, which we do not.

Pardon my ignorance, but I am not familiar with the above mentioned electrostatic theory of tides from any known sources, but would love to find out this it is not so. To my thinking, this much better explains why the tides create a like bulge on the opposite side of the planet, than any theory now found in existing textbooks. Blame it on the hot sun, but I got this idea while hiking up and down the rocky peaks near Pena Springs, Anza Borrego desert, last weekend, watching out for snakes and other prickly things. New barbs won't hurt me, so all challenges are welcome!


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I have formally 'resigned' (tactical withdraw) from the Space-Talk boards; mine were many questions, ideas, but no real answers. Thanks. 04/10/04.

Disclaimer: Please note the ideas expressed here by me are cutting edge theory, very speculative in nature, and not physics as it is being currently taught. Caveat lector.

03/17/04 03:07
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Shambolic


Registered: Nov 03
Posts: 369

Sorry, cop2, but gravity is all that is needed to explain a bulge on the opposite side of the earth from the moon.

Without getting too technical:

Think about placing three particles in the moon's gravitational field, next to each other, in a line pointing at the moon.

Diagramatically: (MOON) (P1) (P2) (P3)

The moon will pull on P1 more than on P2, so they will separate. This is equivalent to the bulge facing the moon. But it will also pull on P2 more than on P3, causing those two particles to separate as well. If you consider P2 to be at the centre of the earth, this explains the tidal bulge facing away from the moon.


(By the way...

quote:
This magnetic field gives a negative charge value to the northern hemisphere and a counter positive charge to the southern hemisphere
This is just wrong. A magnetic field does not lead to any charge imbalance.
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03/17/04 08:10
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wintermute11


Registered: Mar 03
Posts: 485

In the above anology, P1 experiences a larger force due to gravity. With regards to tides, the water experiences a larger pull and bulges toward the moon. P3 is further away so it experiences the smallest force due to gravity. With regards to tides, the water experiences the smallest pull which allows it to relax and "bulge" outward. P2 is stuck between the two. Take a water balloon and pull it from opposite sides. The middle part, representing P2 will contract correspondingly. Water is slightly compressable, so what you're probably seeing is water density fluctations. Water also is fluid, but this causes another effect.

quote:
Pardon my ignorance, but I am not familiar with the above mentioned electrostatic theory of tides from any known sources, but would love to find out this it is not so. To my thinking, this much better explains why the tides create a like bulge on the opposite side of the planet, than any theory now found in existing textbooks.

Before you knock the text books, you should read them first. Study up on classical electrostatics and electrodynamics (magnetism). You have serious issues with relativity so stay away from modern electrodynamics. Classical electrodynamics is good enough at nonrelativistic speeds (regarding the velocity of the moving charge, not the wave propagation). Tides, however, is a phenomenon due entirely to gravity.

03/17/04 14:14
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tmorten


Registered: Sep 03
Posts: 191

There are, of course, many other factors involved in tidal motion, such a the coriolis effect, continental landmasses, ocean depth, ocean floor friction, coastal topography etc etc. Laplace developed a model with several hundred factors.

03/17/04 16:24
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