#1
okay, i have a question about physics.
the gravitational field strength at the surface of the Earth is 9.8 N/kg right where the radius from the earth's centre = 1 right?
So if you were trying to find the gravitational field strength at a certain radius from the earth's centre you would divide 9.8N/kg by the square of that radius right?

For example, if you were two radii from the Earth's centre, then the gravitational field strength would be 9.8N/kg divided by (2)^2, right?
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#2
You can certainly work out how far objects are apart from their masses and the gravity between them. There were a few questions of this type in my first year exams. The law in question is Newton's law of gravitation.
#3
Gravity at a certain radius is given by

a = (MG)/r^2

M = mass of big object (earth)
G = Gravitational constant, NOT g, 6.67 x 10^-11, I can't remember, look it up
r = Radius of orbit
a = Acceleration due to gravity.

I'd normally explain it in depth, but I'm tired, so if you need any help, PM me, and I'll be happy to explain in depth in a day or so.

EDIT:

Yes, you're absolutely right, it's an inverse square law. Sorry, half asleep.
Last edited by LordBishek at Sep 21, 2008,
#4
the radius from the earths core, for all intents and purposes, is generally a constant, as regardless of how far you dig in, you are still pretty much the same distance as you were before.

(that is a direct quote from my physics professor)
I'm pretty sure mass is the only variable. or acceleration or something like that.
Last edited by tona_107 at Sep 21, 2008,
#5
Quote by tona_107
the radius from the earths core, for all intents and purposes, is generally a constant, as regardless of how far you dig in, you are still pretty much the same distance as you were before.

(that is a direct quote from my physics professor)
I'm pretty sure mass is the only variable.


Yeah, but I think he's talking about much larger differences - two radii or more.
#6
Quote by LordBishek

Yeah, but I think he's talking about much larger differences - two radii or more.



Oooh sorry bud, I guess I didn't really read it right.
#7
Okay thanks, because i had a bunch of questions about this kind of stuff and i tried using the equation g = G(m/r^2) but it didn't work cuz i realized if G = 6.67 x 10^-1 etc etc then i would need the actual distance, not just the radii. Thanks for clearing things up for me guys
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#8
Quote by hell_monkey
Okay thanks, because i had a bunch of questions about this kind of stuff and i tried using the equation g = G(m/r^2) but it didn't work cuz i realized if G = 6.67 x 10^-1 etc etc then i would need the actual distance, not just the radii. Thanks for clearing things up for me guys


No worries man. GM is also called the gravitational parameter, and it's used a lot in these kind of calcs, so use this, it might help:

http://en.wikipedia.org/wiki/Standard_gravitational_parameter

May be worth learning a couple of the ones that crop uo in exams like Earth's, Jupiter's and the Sun's.