# Size and Gravity question **URL:** \ **Category:** Factual Questions\ **Created:** [November 6, 2013, 12:40pm UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235 "2013-11-06T12:40:31Z")\ **Posts on this page:** 11\ **Page:** 2
**Author:** ![Frodo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/frodo/32/302_2.png) [@Frodo](https://boards.straightdope.com/u/Frodo)\ **Post date:** [November 19, 2013, 6:17pm UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/21 "2013-11-19T18:17:27Z")
> [@kaylasdad99](#): > > 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. Yeah, but did she have her own event horizon? ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [November 19, 2013, 6:27pm UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/22 "2013-11-19T18:27:30Z")
It’s thought that gravity starts to work differently round about the Planck scale. Of course 1 Planck mass is actually a quite weighty ~22 micrograms which isn’t much lighter than the smallest insect. ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [November 19, 2013, 6:30pm UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/23 "2013-11-19T18:30:46Z")
> [@kaylasdad99](#): > > 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. To which you could’ve said “Why yes according to Newton’s law of universal gravitation all objects have their own gravitational field: how astute of you dear chum!”. At which point though your lunch money may’ve been stolen. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [November 19, 2013, 10:19pm UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/24 "2013-11-19T22:19:38Z")
On a similar note, when Chuck Norris does pushups, he really _does_ push the Earth down. ---
**Author:** ![Learjeff](https://avatars.discourse-cdn.com/v4/letter/l/94ad74/32.png) [@Learjeff](https://boards.straightdope.com/u/Learjeff)\ **Post date:** [November 19, 2013, 11:40pm UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/25 "2013-11-19T23:40:42Z")
> [@Frapp](#): > > 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. The planet under our feet doesn’t negate the effect. It does dwarf it, but even if the planet weren’t there, the gravitational force between you and say your very close friend is too small to be of any use to you. For example, in the movie Gravity, while Hot Lady and Macho Man are floating in space, even if they get really close, their gravity is so small that it’s negligible. For a worse example, in one of those silly Bruce Willis movies where he’s in space and lands on an asteroid and walks around … nuh uh. Even for a pretty darn big asteroid, he’d kinda bounce into it. My gosh, even on the moon (which is pretty damn big, WAY bigger than any asteroid) people walk in a floaty kind of way, with very heavy space suits on. > [@msmith537](#): > > 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 Excellent post, but nit pick: r^2 = distance between the objects in meters, squared ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [November 20, 2013, 12:00am UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/26 "2013-11-20T00:00:53Z")
If we’re going to be nitpicking, the formula, like any formula in science, doesn’t have any particular units associated with it. F is the force, not the force in newtons; m is the mass, not mass in kilograms; r is the distance, not the distance in meters; and so on. ---
**Author:** ![bob\_2](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bob_2/32/3341_2.png) [@bob\_2](https://boards.straightdope.com/u/bob_2)\ **Post date:** [November 20, 2013, 12:06am UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/27 "2013-11-20T00:06:33Z")
> [@Grey](#): > > 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. They used astronomy to measure the azimuth, and built stone buildings to house the pendulums. They did it on two opposite sides of the mountain. It think they were some way shorter than 60 feet. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [November 20, 2013, 12:18am UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/28 "2013-11-20T00:18:38Z")
And you’d be surprised at how precise scientific instruments can be, too. I’ve used analog calipers that could measure things to within a precision of 1/50 mm, just by sliding bits of metal against each other. ---
**Author:** ![Flander](https://avatars.discourse-cdn.com/v4/letter/f/8c91f0/32.png) [@Flander](https://boards.straightdope.com/u/Flander)\ **Post date:** [November 20, 2013, 3:18am UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/29 "2013-11-20T03:18:22Z")
> [@Chronos](#): > > If we’re going to be nitpicking, the formula, like any formula in science, doesn’t have any particular units associated with it. F is the force, not the force in newtons; m is the mass, not mass in kilograms; r is the distance, not the distance in meters; and so on. Yeah no shit! That’s how they kill us physics students on exam days. :smack: ---
**Author:** ![newme](https://avatars.discourse-cdn.com/v4/letter/n/e0b2c6/32.png) [@newme](https://boards.straightdope.com/u/newme)\ **Post date:** [November 20, 2013, 5:35am UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/30 "2013-11-20T05:35:18Z")
> [@Grey](#): > > 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. The instrument used at Schiehallion was a 10 foot (3 meter) [zenith sector](http://www.mhs.ox.ac.uk/students/96to97/zenith.htm). The _combined_ deflections from readings on the north and south sides of the mountain was 11.6". That means on each side, the plumb line was deflected about 5.8", which translates to ~.08 mm deflection at 3 meters. A US dollar bill is about .1 mm thick. ---
**Author:** ![Learjeff](https://avatars.discourse-cdn.com/v4/letter/l/94ad74/32.png) [@Learjeff](https://boards.straightdope.com/u/Learjeff)\ **Post date:** [November 20, 2013, 4:05pm UTC](https://boards.straightdope.com/t/size-and-gravity-question/673235/31 "2013-11-20T16:05:17Z")
> [@Chronos](#): > > If we’re going to be nitpicking, the formula, like any formula in science, doesn’t have any particular units associated with it. F is the force, not the force in newtons; m is the mass, not mass in kilograms; r is the distance, not the distance in meters; and so on. Since he expressed G in specific units, the other units are appropriate, for consistency. Since we’re nitpicking! 😉 [Previous page](https://boards.straightdope.com/t/size-and-gravity-question/673235.md?page=1)