Strength of gravity question.

1. Feb 16, 2013

jaydnul

So is the strength of gravity proportional to the speed of acceleration of gravity? So would an object with 2g of gravity have a 19.62 m/s^2 acceleration? Im sure this sounds like a stupid question

2. Feb 16, 2013

phinds

I wouldn't say it is a stupid question, I'd say it is an incomprehensible question. What is the "speed of acceleration of gravity" ???

3. Feb 16, 2013

Bandersnatch

Eh, come on phinds, that looks like it's meant to be "magnitude of acceleration due to gravity".
Here's your belated award:

And to answer the OP, when you say something "has X gravity", you're saying the same thing as "it produces a field of strength X", where X has units of acceleration.
A "g" is such an unit, just as 1 m/s^2 is. Whichever you use doesn't matter.
So, then, saying "it's got 2g of gravity" is exactly equivalent to saying "it's got 2*9,81m/s^2 of gravity" to "it's got a 2g gravitational field potential" to "it's got a 2*9,81m/s^2 gravitational field potential".
If you put any kind of mass in such a field, it'll feel a force of gravity F=m*a, where a is the potential.

It's improtant to remember, however, that whenever we talk about gravitational potential, we must specify a certain point in space in which we measure it. In most cases in everyday use it's obvious, but to be truly clear, one should say something like "it's got 2g on the surface", as the acceleration measured anywhere further from or closer to the centre of the field would be different.

4. Feb 16, 2013

CWatters

It's a badly worded question but...

If you are in space (eg a long way from a planet) in a rocket that is accelerating at 19.62 m/s^2 then it will feel like you are experiencing 2g. Your cup of coffee will feel twice as heavy (actually more so because your arm will also feel twice as heavy).

5. Feb 16, 2013

Drakkith

Staff Emeritus
Hey, you found Phind's missing dogtag! I've been looking for that everywhere!

6. Feb 16, 2013

jaydnul

Haha sorry. Isn't earth gravity acceleration something like 9.81 m/s^2? We also describe earths gravity as being 1g (I assume the g stands for gravity, if not, please tell me). So would an object with a gravity of 2g have an acceleration of 2*9.81 (19.62) or is the 9.81 a universal rate of acceleration? (Sorry meant to say rate, not speed of acceleration)

7. Feb 16, 2013

Staff: Mentor

Sure, you can say that the strength of earth's gravitational field (near the surface) is 1g.
Still not quite sure what you mean. Are you asking: If the strength of the gravitational field were twice as much, would a falling object have twice the acceleration? If so, then yes.

If not, then ask again. (The phrase "an object with a gravity of 2g" is not clear.)

8. Feb 16, 2013

jaydnul

Ahh i see. So theoretically, something falling towards the earth will have an acceleration slightly less than 9.81 (proportional to its distance from the surface) until the instant it hits the surface?

9. Feb 16, 2013

Staff: Mentor

Something like that. The acceleration due to gravity is inversely proportional to the square of the distance from the earth's center. Since the radius of the earth is about 6.4 x 106 kilometers, if you were that distance above the earth's surface (which is quite far) the acceleration would be 1/4 as much as it is on the earth's surface.

But as long as you stay not too far from the earth's surface, you can consider the acceleration due to gravity to be a constant equal to g.

10. Feb 16, 2013

xodin

An object doesn't have a fixed gravity, it has a fixed mass. The force felt between two masses is described by the law of universal gravitation:

F=G(m1*m2)/r^2

Where G is the gravitational constant, more or less that "universal rate" called a constant that you were asking about:

G=6.67398 × 10^-11 m^3 kg^-1 s^-2

m1 is the mass of the object (in kilograms)
m2 is the mass of some other object (in kilograms)
r is the distance between the centers of mass of the two objects (in meters)
F is then the mutual force felt by each mass from the other mass (in Newtons)

So m1 feels that force by m2 and m2 feels that force by m1, due to Newton's Third Law (equal and opposite reactions).

Subsequently, the acceleration resulting from a single object (as opposed to the force on a second object), can be found by eliminating m2 from the equation--yet it nevertheless still depends on a distance from the mass, namely r!

a=G*m1/r^2

a is the acceleration (in m/s^2)

This is a result of taking the force equation and using Newton's second law for a hypothetical second object F=m2*a:

F=G(m1*m2)/r^2=m2*a -> a = G*m1/r^2

Edit:

So, using the mass and radius of the Earth we have:

a = G*m1/r^2 = (6.67398 × 10^-11 m^3 kg^-1 s^-2)*(5.972 × 10^24 kg)/(6.373 × 10^6 m)^2 = 9.81 m/s^2

The actual acceleration depends on a person's specific location on the surface of the Earth. It can vary from around 9.78 m/s^2 to 9.83 m/s^2 on Earth's surface, and possibly more at the extremes, depending on latitude and height above/below sea level, among other things.

Last edited: Feb 16, 2013
11. Feb 16, 2013

phinds

Damn ! I gotta get me one of those.

12. Feb 18, 2013

Lsos

The gravity on Jupiter is about 2.5 g...so an object falling at Jupiter's surface would accelerate at 24.79 m/s^2.

On the moon, you would accelerate at 1.62 m/s^2.

Nothing universal about 9.81 m/s^2. It's just something we happen deal with every day.