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Gravity and distance?

Started by Tree Penguin Jun 3, 2004 at 5:44 AM 12 replies 1.4k views

Original Post

Jun 3, 2004 at 5:44 AM

Hi, i have a question but due my n00bness it might look a little stupid or something, anyway here it is: I was wondering how the strength of gravity (maybe not gravity, i mean the force planets and stars excersize at eachother, please tell me if that''s not gravity

) decreases when distance increaces, is it linearly or exponentially or something else (i expect that). Can anyone give me the basic formula or equation? Thanks in advance! Ow, and if i said something stupid please tell me so i can learn

. And also i was wondering if anyone knows an article (or a serie of articles) about the atom structure (i know the molecule basics already from secondary school, maybe not enough but i will figure things out as i read), just interested. I don''t mind a difficult to understand article, i will figure it out (i got the time, the discipline and the will

RavNaz

Jun 3, 2004 at 6:00 AM

One of newtons laws :

F =G mM/R^2

G gravitational constance, m & M mass of bodies, and R is the distance seperating them.

G is something like 6.6x10^8, cant quite remember....

Tree Penguin

Jun 3, 2004 at 6:05 AM

Thanks! Is R measured from the centre of the objects? i guess so but just to be sure.

Shadowdancer

Jun 3, 2004 at 6:23 AM

Yes, gravity is calculated from the center of gravity (doesn''t it sound obvious that way?)

zngb

Jun 3, 2004 at 6:33 AM

G = 6.67 * 10^-11

you don''t wanna live in a universe where G is "6.6x10^8"

RavNaz

Jun 3, 2004 at 6:34 AM

good call Lord Gzoo. Did say I couldn''t quite remember though ;-)

Tree Penguin

Jun 3, 2004 at 7:38 AM

*Auch* (for a very short time)

uber_n00b

Jun 3, 2004 at 1:31 PM

I would just like to note that the force of gravity is proportional to 1/R^2 only outside of the mass of the object. Inside it's proportional to R (assuming a pretty even density distribution). So the graph of the force of gravity looks something like this:


gravity
|      /|     /  \_
|    /     \          
|   /       \__
|  /           \____
| /                 \____________
|/_______________________________\________________-radius
0       |
     radius object

(you're going to have to assume that the second part of that graph is smooth and roughly resembles 1/R^2 :)

EDIT ok the graph doesn't want to turn out right but you get the idea.

[edited by - uber_n00b on June 3, 2004 2:34:35 PM]

shadow12345

Jun 3, 2004 at 4:00 PM

I''ve never heard that before, why is that so?

PyroSA

Jun 3, 2004 at 4:38 PM

Imagine you would dig a hole right through the earth. Now If you were to drop through that hole, gravity would get less and less, until at the center of the earth, you have no gravity. why? because if you were to calculate gravity separately for every single atom in the earth once you start falling ''into'' the earth some of them would actually be pulling away from the center of gravity, effectively decreasing the force of gravity.

The center of gravity of an object is basically just a summed value of all its parts. Once you are inside this object, things get more complicated though.

CombatWombat

Jun 3, 2004 at 6:19 PM

As an added problem to the ''hole through the center of the earth'' theory...
If you jumped into said hole, when you passed the middle you would begin to slow down and eventually go back the way you came.
I think you''d end up in a sort of simple harmonic motion swinging back and forth if friction was ignored.
If there IS friction, you''d get stuck in the middle of the earth at some point with no way to get home.

Doc

Jun 3, 2004 at 7:45 PM

Assuming the force dependence on R is linear within the object, yes you would execute simple harmonic motion, and eventually end up stuck at the centre of the Earth. At that point, it sucks to be you.

Zipster

Jun 4, 2004 at 1:08 AM

quote:
Original post by shadow12345
I''ve never heard that before, why is that so?

It''s kinda like Gauss''s Law for Electricity, only for gravity in this case. If you have a closed Gaussian surface, according to Gauss''s Law the electric flux though this surface is only proportional to the charge inside the surface. Now, replace charge with matter, and replace electric flux with gravitational flux (rough translation

). Assume that r is less than the radius of the object. The enclosed matter increases at a quadratic rate, but the distance falloff is inverse squared. They cancel out. But since surface area increases at a linear rate, the net increase in gravitational force is linear.

Once r is greater than the radius of the object, it behaves like a regular point mass, and the usual rules apply (inverse square falloff force).

Paradigm Shifter

Jun 4, 2004 at 4:31 AM

So what would it be like living on a world shaped like a giant doughnut? (Not a jam one).

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