How much more gravitational pull does the moon have then an ear ring on an ear?

The ear ring weighs 1 gram and is .25" away from the ear.

1. More then 125,000 times the influence then the ear ring.

2. More then 10,000 times the influence then the ear ring.

3. More then 100 times the influence.

4. About twice or more time the influence of the ear ring.

5. The ear ring's influence is equal to or more then the moon's

 

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Yes, it's really very simple. But thanks for taking an interest in my math skills. Fortunately I might be smart enough to know the answer before I do the math - which is an incredible advantage in checking the math. Are you going to vote or not James?

 

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Woops. I made assumptions only to find specifications in the first post.
Having said that, I had pretty good assumptions (my factor was about 50,000)

 

I got 19,400 after calculation.

 

Calculating is wrong. Right?

I flipped my coin ( luckily, I found a penny laying under my Rolls ) but it rolled into the gutter.

So, I don't know any more about this then any other guesser.

Maybe vote later.

 

I saw a program on tv that claimed that the gravitational pull from the moon was actually many many times less then stated in answer 5. (an object a gram about 20" away) I guess it and I were both wrong. James and I both got very similar results from very different ways of calculation, so I have to conclude that the program was probably wrong. I'm very surprised. It actually airs again today at 3pm pt. I think the anology they used was a pea about 20" away.

Wednesday, April 12, 2006, at 3P

Naked Science
Moon Mysteries [] National Geographic Channel

 

Worked out a ratio of about 19,990, but I suppose another way of looking at it is that since we're in a free-fall orbit around the moon, we don't feel any of the moon's gravity.

 

Is that entirely true? We have inertia right? I mean if we felt no pull from the moon would we have tides? When the moon is directly above us, don't we experience less gravity then when it is on the other side of the earth? Isn't this feeling the moon's gravity? I'm prob. missing your point. You prob mean that we don't feel the acceleration of the gravity like we would in a car because in a car the energy is transfered to us through the transfer of force through our bodies from one atom pressing against another, where in gravity every atom is accelerated at the same time. Like the difference between 1. accelerating in a gravitational field - like a space ship gaining speed as it gets closer to a planet - might not feel any acceleration and 2. The rocket accelerating by way of chemical rockets.

 

This is true. You wouldn't feel the moon's gravitational pull at the Earth's centre of gravity, but since the moon's gravitational field varies with distance, you'll feel the difference between the field strength at the centre of the Earth and at other points. This is what causes tides. We still only feel a small fraction of the moon's gravitational field.

Exactly.

 

Wednesday, April 12, 2006, at 3P
Naked Science
Moon Mysteries [] National Geographic Channel

40 min. and 56 sec. into the program: " The moon's gravity is now so weak (gravity felt on earth) it is like a pea held above your head"

I'm really surprised, NG is usually very accurate about it's data.

 

The original question was poorly worded. I doubt the program was wrong.

Pryzk almost got it right:

even closer,

It's not just a small fraction. Tidal forces (or third-body effects) vary with the inverse <i>cube</i> of distance. The tides would be (roughly) eight times greater than present if the Moon was orbiting the Earth at half its current distance. Because the moon is so far away, the <i>net</i> effect of the moon on a person is very tiny.

 

Well, I'm very confused now, most everyone calculated that the moon was about 20,000 times more influencial gravitationally then the ear ring - and I know that James and I took into account the inverse square law. What do you calculate it to be DH?

 

Tortise: you work out the difference between the moon's gravitational field at the center of the Earth and at a point on the surface.

If I haven't made any errors, I get a factor of up to about 680 times, depending on where you are on Earth. That's the Earth's acceleration 'hiding' nearly 97% of the moon's gravitational attraction if the moon is directly overhead.

 

Ok - thanks for your comments

DH are you sure about this?

Last time I checked if you half the distance, it's 4x the intensity. This is called the inverse square law. 2^2=4 or 2x2=4

p wrote:

I'm not sure that this is consistent with GR or Newton. It really doesn't matter how intense earth's gravitational field is, the moon's influence is simply a matter of distance and mass.

 

The tidal force is the difference between 1/R^2 and 1/(R-r)^2 (leaving out G and the 2 masses), where R is the Earth-moon distance and r is the Earth's radius. (R-r) just gives you the distance from the moon to the point on the Earth's surface closest to the moon. You can do a bit of algebra on the subtraction to get:

r * (2R - r) / (R^2 * (R - r)^2)

This simplifies to 2r/R^3 if you assume r is small compared to R, so that's where the cube comes from. DH's post prompted me to actually work this out.

 

The magnitude of the gravitational field is simply GmM/R^2 as you used, but given that we're accelerating toward the moon its the amount that isn't 'masked' by the acceleration that you'd measure, feel, and is responsible for tidal effects.

It's for the same reason that astronauts in orbit feel weightless in their shuttle, despite the significant gravitational force exerted on them by the Earth.

 

Gravity follows an inverse square law. Tidal effects do not. They follow an inverse cube law.

Which has a greater gravitational effect on the Earth, the Sun or the Moon? The Sun, obviously, since M_s/r_s^2 >> M_m/r_m^2. Which has a greater tidal effect on the Earth, the Sun or the Moon? The Moon, not so obviously, since M_m/r_m^3 > M_s/r_s^3.

If you don't believe me, try this: http://en.wikipedia.org/wiki/Tide#Tidal_physics: