#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?

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?

#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.

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.

(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

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

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

#8

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.