# My Own Personal Gravity

http://www.straightdope.com/mailbag/mgravityspeed.html

…got me to thinking. If, as in the example, two little beads
have an (admittedly small) gravity, don’t I have a personal
gravity? I know it would also be small, but surely it’s
there? How do I measure my gravitational force?
And are there any tiny objects or particles that would
be so small as to fall into my gravitational field and crash
to the surface?

I know that housecats seem to gravitate towards me,
but I think that’s just 'cause they know I’m allergic.

You can calculate the force of gravity between any two objects with the formula F = (G x m x m’) / r[sup]2[/sup], where G is the universal gravitational constant (6.672 x 10[sup]-11[/sup] Nm[sup]2[/sup]kg[sup]-2[/sup]), m and m’ are the masses of the objects and r is the distance between their centers.

As you can tell from QED’s formula, anything with mass exerts gravitational force. If you were in a frame that apparently neutralized the effect of Earth’s gravity (like an orbiting spacecraft or a falling elevator), then you might notice objects floating towards you–very slowly, per the formula (and you towards them, since they have gravity too). On Earth things fall too fast for you to notice the gravitational force between you and them, but it’s there

Of course. Everything with mass has a gravitational field (well, I’m not sure about the quantum level, since quantum gravity still hasn’t been worked out).

The size of the particles doesn’t really matter (other than the fact that, if it’s massive enough, you’ll move toward it faster than it’ll move toward you). The problem is that the earth’s gravitational field is so massive that your gravity can’t compete, and everything will fall to the ground rather than to you.

Also, because of your relatively small mass, even in the absence of other gravitational fields, objects would be attracted toward you quite slowly.

I.e., the force between you (say 80 kilograms) and your twin brother a meter apart is equal to:

F = (6.672 x 10^-11 Nm^2/kg^2)(80kg)^2/(1m)^2
= 4.27 x 10^-7 Newtons

Or about 1/20 the force exerted by a paperclip (1 milligram) laying on a table. (With the usual caveat, “If I did my maths correctly.”)

A paper clip weights much more than a miligrame, I´m quite sure. I don`t have a precision scale right here, but my WAG is in the order of 0.4 to 0.6 grams.

Bit of a nitpick, but the formula given is true for point particles. Due to some tricks of vector calculus, it also holds for spherical bodies. But for a more unusual shape, you’d have to do some work to get at the actual force.

None of which changes the basic principle, which is that you do have your own gravitational field, it’s utterly insignificant unless you’re way the heck out in the middle of nowhere, and that it causes a force which looks an awful lot like GMm/R[sup]2[/sup].

And a paperclip is indeed somewhat larger than a miligram; I’ve always heard it quoted as close to a gram, which sounds like the right order of magnitude,

Measuring the gravitational force between two big objects of known mass is (I believe) the most common way to get a value for the constant of gravitation G. This was done as early as 1798.

You (and g8rguy) are of course correct. I was misremembering an old elementary science text which used a paperclip as the example for a gram, not a milligram.

The corrected comparison would be that the force between you and your twin one meter away would be 1/20,000 the force of a paperclip laying on a table. Or, for something that actually weighs about 1/20 of a milligram, the force exerted by a grain of salt laying on the table.

Correct. Cavendish used a torsion balance to determine the gravitational constant. Two small balls of known mass were suspended on a fine fiber. Two large balls of known mass were placed a known distance from the small balls. The attraction of the small balls to the large balls produced a force which caused the fiber to twist, and this twist could be measured and the Universal Constant of Gravitation (G) determined. Incidentally, once this was known, the only unknown in the equation F=Gmm’/r^2 was the mass of the earth, which could then be solved for, so this experiment is often referred to as Cavendish “weighing the earth.” Google on “weighing the earth” and you’ll find lots of info about it. I believe that more sensitive versions of this experiment are still the usual method for obtaining more accurate values of G.

Isn’t the other problem with noticing personal gravity (say, you’re free-falling in orbit with dust particles surrounding you) that, with such small masses, pretty much every other kind of force is stronger and more noticeable than gravity? If I’m on the ISS, just hanging out in midair, isn’t it more likely that static electricity will draw dust particles to me than that gravity will? (assuming the ventilation system wasn’t blowing everything around…)

So far as I know, it’s the only way to get a value for G, which is why our measured value has such low precision compared to most physical constants: It’s really hard to measure such a tiny force. And torsion balances, in various forms, are still used for modern measurements of G, such as those at extremely close range (it’s theorized that gravity might not be inverse square at sufficiently close ranges, and it was also theorized that “sufficiently close” might be as large as a millimeter or so).

If the solar system were scaled down so the Earth was the size of a grape would the orbits remain the same?

No. If the Solar System were scaled down the planets would be much closer to the Sun, and therefore not in the orbits they are in at the moment, which are where they are, not where they would be. That is, the Earth cannot travel in an orbit 1.47 * 10^11 metres from the Sun if it is 1.2 * 10^2 metres from the Sun.

Like to rephrase that question?

Regards,
Agback

I admit to finding the question confusing.

If you kept the densities constant and shrunk the earth to the size of a grape and the rest of the objects in the solar system to scale, then of course the orbit changes.

If you kept the masses the same but changed the densities, then the orbit doesn’t.

I get the feeling that what you want to do is to make all the objects several orders of magnitude smaller, proportionally lighter, and shrink the orbits proportionally as well, then see if the shrunked orbits are correct. Is that the case?