Size and Gravity question

Factual Questions
1

Disclaimer: Not homework. Legitimately curious about the topic.

I remember reading somewhere that something (or someone) having mass meant that they had their own bit of gravity. It’s so miniscule that the planet under our feet negates any effect we would have on anything else, but it’s there.

At what size would something begin to affect the local gravity around itself while on Earth? How about in space?

I was thinking about it in the context of television programs and movies that involve giant robots, like Transformers, Power Rangers, Gigantor, etc. I would even be curious about it in the context of the Enterprise(s) from Star Trek.

2

Any object that has mass has gravity. The pull of gravity is stronger as the objects get larger and closer together.

It can be expressed by the following formula:
F = (G * m1 * m2) /r^2

Where
F = force of gravity in Newtons or kg·m/s2
G = Gravitational Constant = 6.67384 × 10-11 m3 kg-1 s-2
m1 = Mass of object 1 in kg
m2 = Mass of object 2 in kg
r^2 = distance between the objects, squared in meters
IOW, in order to significantly attract another object while on or near the Earth, the object in question would need to have a significant mass relative to the earth. Unfortunately that means an object that is either the size of a small moon or so small and dense that it forms a black hole (which has it’s own set of problems). And rather than affect other objects, it would probably just drop through the floor and sink to the center of the Earth.

3

Well, ‘surface gravity’, that is, the acceleration due to gravity when you’re touching an object, is proportional to the radius and density of the object, assuming a constant shape like a ball.

The radius of the Earth is about 6400 km. Assuming a constant density, a ball of mostly metal and a little rock with 64km radius, (about the size of 92 Undina, a main-belt metallic asteroid,) would have 1% of the surface gravity of the Earth. Of course, it’s a matter of opinion where you put the line of a ‘noticeable’ tug of gravity, but that seems like a good rule of thumb to me.

Obviously we don’t have any free-standing structures that large on the Earth’s surface.

I hope this helps.

4

Tried to add this in edit and ran out of time, bah:

Thanks to msmith537 for providing the full math. When you’re looking at gravity in terms of acceleration instead of force, you can cancel out one of the m terms, because acceleration equals force over the mass of the object being accelerated.

You get to the bit about ‘proportional to radius and diameter’ by observing that if you’re on the surface of object 1, m1 is proportional to r^3*density. r^2 cancels out, leaving only one r term, and G is a constant. :slight_smile:

5

It took a mountain ( Schiehallion experiment - Wikipedia ) to deflect a plum bob 11.6" from true.

6

This would be more accurate if “negates” were replaced with “overwhelms” - the effect you have on other objects is not changed by the nearby planet, but it certainly is made hard to notice.

7

It should be possible to notice significant small-scale gravitational effects from objects well short of black hole density.

A small fragment of a neutron star (say, 1 mm in diameter) would have a mass of thousands of tonnes, which should work nicely.

8

To avoid other people experiencing the same momentary confusion I did, 11.6" here means “11.6 arc minutes”, not “11.6 inches”. The latter is a slight overestimate. :slight_smile:

9

Yeah, that should’ve been made clearer. :slight_smile:

10

Doesn’t 11.6" mean “11.6 arc seconds”, not “minutes”? The latter is a slight overestimate. :slight_smile:

11

Yep arcseconds. :smack: An earlier french experiment had 8 arcseconds of deflection as the wikipedian explicitly states. And you’re right ’ is minutes, " is seconds.

12

D’oh, yes. Of course initially all I could think of is “inches”.

13

In college, I did the Cavendish Experiment to measure the gravitational constant G. The picture half way down the page is a mock up of how it is done. The small lead balls are attracted to the large lead balls due to gravity.

It’s a pretty cool experiment, but it is extremely touchy. We did it in the basement of the physics building, because small vibrations on any other floor would have set it off.

14

Would someone be willing to work this out in terms of linear distance for me? Let’s say you have a pendulum that is 10 meters long. How much many fractions of a millimeter would would it be off true with a deflection of 11.6 arc seconds? (At least, I’m guessing it would be fractions of a millimeter). I’m guessing this has something to do with that trigonometry stuff I didn’t get 30 or so years ago in high school! :wink:

15

For really small angles the arc length should be pretty close to the actual deflection.

Arc length is Radiusdegrees(radians) or Radiusdegrees*pi/180

So R*0.00005623838

Your 10m length would be a deflection of 0.5mm.

A more accurate estimate would require a less lazy poster. :slight_smile:

16

Thank you! So just to confirm I got the math right, a 20 m pendulum would be 1 mm off (seeing as there’s no squares or anything in there)? I’m amazed this effect could be set up so it would be discernible to the naked eye!

17

Or we could just use the deflection (S’) = Rsin(angle) = R*sin(0.00005623838)

18

Well assuming I’ve managed not to screw up my math in multiple ways … yes it would be deflected by 1 mm.

The problem is getting 60 odd feet of pendulum isolated to read a 1 mm deflection. They did it so obviously it’s possible.

19

When I was in high school, I was told by a not-particularly-cordial acquaintance that my mother had her own gravitational field.

Also, that when she sat around the house, she saw around the house. And that she had her own ZIP code.

20

I’m sure they were just making a mountain out of a mole hill. :slight_smile: