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z0diac
2009-Aug-05, 02:24 AM
Does anyone know the % difference someone weights when the moon is directly overhead compared to when it is on the other side of the Earth? It has enough gravity to pull bodies of water around, so I'm guessing there is a measurable difference.

Anyone know the numbers for when both the moon and sun are overhead compared to being on the other side of the planet?

And, do labratories that deal with very precise measurements of weight, take this into account?

WayneFrancis
2009-Aug-05, 04:46 AM
You might think that it has a big effect because you can see the tides but it is pretty small when measuring small objects.

This is because the amount of gravity between 2 objects is dependant on the mass of the 2 objects and the distance squared. The ocean has a lot of mass thus is effected more by the moon then you, who have very little mass, is.

I started doing a big post with a lot of maths but I messed up in my calculation and in the end your question can be better addressed by how scientists measure something.

Most people think about weighing some and know you "weigh" less on the moon. When scientists "weigh" something they are really measuring its Mass. The mass of an object does not change no matter where you are (discounting Special Relativity effects)
So scientists would probably calibrate their scale right before doing the measurement if they where worried about that level of precision. They would do this by placing a weight of a known mass on the scale.

if you are interested in the maths then do the calculation and get back to us.
F = G(M1M2/r2)
where G = 6.67428x10-11m3kg-1s-2
The moon is 7.347 7 1022 kg and ~384,393 km away when overhead and ~384,405 km away when on the other side of the Earth.

Force when the moon is over head is the force of the Earth - force of the moon while overhead.
Force when the moon is on the other side of the Earth is the force of the earth + force of the moon while on the other side of the planet.

When you do the calculations you'll see that the force from the Earth is far larger then the force from the moon. Thus the moon's force being added or subtracted is very minor for small objects. The difference between being on one side of the Earth or other is only a difference of 0.99993756682302495775330753403284 because the distances are squared...ie
384,3932 / 384,4052
147,757,978,449/147,767,204,025 which =
0.99993756682302495775330753403284 difference.

z0diac
2009-Aug-05, 05:40 AM
But remember, one of your variables is incorrect :)

re:
The moon is 7.347 7 1022 kg and ~384,393 km away when overhead and ~384,405 km away when on the other side of the Earth.

The difference in distance would be the diamater of earth.

ie: put the moon over head, a person standing on earth, under the moon. To get the opposite, leave the moon overhead, just twist the earth 180 degrees so the person is sticking down instead of up. The change in distance is the diameter of earth, which you have calculated to be 12km across :)

I know the resulting figure will be very minimal though (and yes, I know the difference between mass and weight... believe it or not I took physics all through high school and up to 2nd year university, heh) but I was just wondering if anyone actually knew the % difference if someone was weighed while the moon was overhead, compared to when it was on the other side of the planet.

Silly question, I know. But interesting to see the answer :)

hhEb09'1
2009-Aug-05, 06:23 AM
And you don't use the squared terms, either, you're not comparing the size of the moon's gravity from one side to the other (they're in different directions, so their difference is just about twice one of them)--you're comparing the size of the difference between the moon's gravity and the earth's gravity from one side to the other. And the difference between squares is proportional to the cube.

WayneFrancis
2009-Aug-05, 06:42 AM
But remember, one of your variables is incorrect :)

re:
The moon is 7.347 7 1022 kg and ~384,393 km away when overhead and ~384,405 km away when on the other side of the Earth.

The difference in distance would be the diamater of earth.

ie: put the moon over head, a person standing on earth, under the moon. To get the opposite, leave the moon overhead, just twist the earth 180 degrees so the person is sticking down instead of up. The change in distance is the diameter of earth, which you have calculated to be 12km across :)

I know the resulting figure will be very minimal though (and yes, I know the difference between mass and weight... believe it or not I took physics all through high school and up to 2nd year university, heh) but I was just wondering if anyone actually knew the % difference if someone was weighed while the moon was overhead, compared to when it was on the other side of the planet.

Silly question, I know. But interesting to see the answer :)

6,378.1 x 2 is ~12km when you figure in that the distance of ~384,399km is the semi major axis of the moon
Are you really telling me you are worried about .75km at that range?
The orbit of the moon varies by over 45,000km. That along is going to be a bigger effect then if it is on the other side of the Earth by a factor of 3.

My point is basically the difference is swamped by local effects. IE you'll weigh more or less due to variations of gravity on the surface of the Earth.

The question you originally was very vague and could result in a wide range of numbers but all at a scale that is very insignificant. I'd be surprised if anyone has the answer at the top of their head if they where not some type of Savant.

But here ... the force difference is roughly
0.000033185622126304552636145808516139m3kg-1s-2 for a 125kg person

WayneFrancis
2009-Aug-05, 06:46 AM
And you don't use the squared terms, either, you're not comparing the size of the moon's gravity from one side to the other (they're in different directions, so their difference is just about twice one of them)--you're comparing the size of the difference between the moon's gravity and the earth's gravity from one side to the other. And the difference between squares is proportional to the cube.

Hense my


Force when the moon is over head is the force of the Earth - force of the moon while overhead.
Force when the moon is on the other side of the Earth is the force of the earth + force of the moon while on the other side of the planet.


Everything else being the same this equates to a difference of about 0.99993756682302495775330753403284 between the attractive force of the moon. Then its just a difference of adding or subtracting that force.

Sorry if I didn't make that clear. If there is something faulty with my logic please let me know.

MattTheTubaGuy
2009-Aug-05, 06:46 AM
Actually, the gravitational acceleration when the moon is on one side of the earth, compared when it is on the other side, is the same isn't it? That is why there is two high tides a day. It's because, compared to a place on earth at right angles to the moon, the moon pulls x amount more on the side that the moon is above, and x amount less on the opposite side, because the moon is further away, but because 'up' is the other way, there is x amount less gravitational acceleration on both sides compared to in the middle. Correct me if I'm wrong. Also, depending on the latitude, the rotation of the earth would have an effect as well.

EDIT:also

*0.000033185622126304552636145808516139m3kg-1s-2 for a 125kg person*

why do you have m3kg-1s-2 for a 125kg person?
shouldn't it be (the number)*125 ms-2
or (the number) mkg-1s-2 for any mass, and where did the 3 in m3 come from?

a force is measured in Newtons, or acceleration per mass, which multiplied by the mass, gives the acceleration.

I think that's right.

WayneFrancis
2009-Aug-05, 06:52 AM
Actually, the gravitational acceleration when the moon is on one side of the earth, compared when it is on the other side, is the same isn't it? That is why there is two high tides a day. It's because, compared to a place on earth at right angles to the moon, the moon pulls x amount more on the side that the moon is above, and x amount less on the opposite side, because the moon is further away, but because 'up' is the other way, there is x amount less gravitational acceleration on both sides compared to in the middle. Correct me if I'm wrong. Also, depending on the latitude, the rotation of the earth would have an effect as well.

I don't believe so. This is shown by the tide on one side of the earth being slightly higher then the other.

tony873004
2009-Aug-05, 07:00 AM
This is because the amount of gravity between 2 objects is dependant on the mass of the 2 objects and the distance squared. The ocean has a lot of mass thus is effected more by the moon then you, who have very little mass, is.
The amount of force is dependent on the mass of the 2 objects. But gravity is an acceleration, not a force, and is independent of the mass of the secondary body (in this case, the person, or the ocean). That is why a big rock and a small rock drop at the same rate. Acceleration only uses one M in the formula: GM/r^2.

When the Moon is at its average distance from the center of the Earth of 384,000 km, it is 377622 km away from you when overhead, and 390378 km away from you when it is directly below you.

6.672e-11*7.35e22/377622000^2=3.43897327300484E-05 m/s2
6.672e-11*7.35e22/390378000^2=3.21790121117883E-05m/s2
The difference between these is 2.2107206182601E-06 m/s2
which is about 2 ten millionths the strength of Earth's gravity of 9.8 m/s^2, therefore very insignificant.
And the majority of this difference is negated because when the moon is overhead, it is not only pulling you with an acceleration of 3.4e-5m/s2,but it is pulling the ground you are standing on with an equal acceleration.

MattTheTubaGuy
2009-Aug-05, 07:07 AM
^^^
wouldn't that be because the distance from the center of the earth to the surface is the same on both sides, but because of that the difference of force from the centre of the earth to the near side would be slightly more than from the centre to the far side? also would the barycentre of the earth and moon play any part? (centre spelled correctly, I am from NZ):)

astromark
2009-Aug-05, 08:11 AM
Some of my spelling gets a little imaginative at times... and I live in NZ also.:)I'm lazy...

Some of what has been said here is unfortunately not the whole story.

Sure there are two tides, per your comment... I agree with your thoughts. The actual weight calculation to the mass of a small body ( you ) is very small. The Earth and moon exert a lot of gravity force and our oceans respond as the tidal movement we all see. That force is not strong. The amount of force on the weight of a person is tiny. You would need some very precise equipment to measure it at all. Looking at what you have been told here. Its not much. Is it.

hhEb09'1
2009-Aug-05, 11:33 AM
6,378.1 x 2 is ~12km when you figure in that the distance of ~384,399km is the semi major axis of the moon
Are you really telling me you are worried about .75km at that range? :)

z0diac's point is that 6,378.1 x 2 is ~12 thousand km

JohnBStone
2009-Aug-05, 10:21 PM
Wont it be dwarfed by the centripetal force effects of say being on the equator vs being at the poles. (though after the recent rotation and relativity (http://www.bautforum.com/space-astronomy-questions-answers/88549-rotation-i-just-dont-get.html) thread I am less sure about this than before)

z0diac
2009-Aug-05, 11:00 PM
Wont it be dwarfed by the centripetal force effects of say being on the equator vs being at the poles. (though after the recent rotation and relativity (http://www.bautforum.com/space-astronomy-questions-answers/88549-rotation-i-just-dont-get.html) thread I am less sure about this than before)

That's something I never thought of before - is there also a measurable difference of someone standing on the equator compared to the north or south pole?

Tim Thompson
2009-Aug-06, 01:18 AM
The measured weight of anything is dependent on latitude, because of Earth's rotation, and on altitude because that sets the distance from the center of the Earth. Of course, it also depends on the motion of the Sun & Moon. Exploration geophysicists routinely measure the gravitational deflection caused by mountain ranges, individual mountains, hills, and probably even large buildings. Also note that the movement of underground water and ocean mass adds a time variable component to Earth's gravitational field, and therefore to measured weight. The measurement of this time varying component is the primary science goal of the Gravity Recovery and Climate Experiment (GRACE (http://www.csr.utexas.edu/grace/)) multiple satellite mission.

Precision experiments are able to measure the inverse square property of gravity over distances significantly less than 1 millimeter (Hoyle, et al., 2004 (http://adsabs.harvard.edu/abs/2004PhRvD..70d2004H) and citations thereto). I think that most of the effects we are talking about here are (or at least were) too small to be significant compared to uncertainties introduced by temperature gradients and the gravitational attraction of other laboratory apparatus, which are taken into account.

neilzero
2009-Aug-06, 01:25 AM
I don't know either, but my guess is you weigh a few grams more when the Sun and moon are both near the horizon. Both overhead, reduces the gravity and the Earth accelerates toward the moon which partially cancels the weight reduction, I think. The latitude variation and the Earth's crust density variations have little effect if you move less than a kilometer, between measurements. Neil

WayneFrancis
2009-Aug-06, 02:49 AM
:)

z0diac's point is that 6,378.1 x 2 is ~12 thousand km

Wooops I did 12km insteal of 12,000km. Got it. I didn't double check my math.

tony873004 said what I was trying to say in his post. Basically the change is far below the precision of the equipment doing the measuring.

cran
2009-Aug-06, 04:17 AM
The measured weight of anything is dependent on latitude, because of Earth's rotation, and on altitude because that sets the distance from the center of the Earth. Of course, it also depends on the motion of the Sun & Moon. Exploration geophysicists routinely measure the gravitational deflection caused by mountain ranges, individual mountains, hills, and probably even large buildings. Also note that the movement of underground water and ocean mass adds a time variable component to Earth's gravitational field, and therefore to measured weight. The measurement of this time varying component is the primary science goal of the Gravity Recovery and Climate Experiment (GRACE (http://www.csr.utexas.edu/grace/)) multiple satellite mission.

Precision experiments are able to measure the inverse square property of gravity over distances significantly less than 1 millimeter (Hoyle, et al., 2004 (http://adsabs.harvard.edu/abs/2004PhRvD..70d2004H) and citations thereto). I think that most of the effects we are talking about here are (or at least were) too small to be significant compared to uncertainties introduced by temperature gradients and the gravitational attraction of other laboratory apparatus, which are taken into account.

agreed ...

and if trying to measure the change in mass (or weight) of a person across ~14.5 days, [ETA: or even ~6 hours]
I think even the terrestrial variations could be swamped by variations in nutrient intake/digestion/discharge ...
even drinking water/sweating ...