# Why 9.8m/s^2?

1. Apr 7, 2004

### Imparcticle

Why is acceleration due to gravity always going to be 9.8/s^2??
(in the formula GE=grh)
Does it have any thing to do with the prominent gravity pull? Like if you were to calculate the acceleration due to gravity of an object falling towards the center of jupiter (this object is NOT in free fall), would you use 9.8/s^2 for g in the aforementioned formula?

2. Apr 7, 2004

### Chen

The gravity acceleration anywhere in the universe is:
$$g' = G\frac{M}{r^2}$$
Where G is the gravitational constant, M is the mass of the object that is pulling you and r is your distance from its center. It just so happens that on Earth, g is 9.8m/s2. On any other planet you would have to adjust the values of M and r and would probably get a different value for g'.

3. Apr 7, 2004

### math4me

Acceleration caused by gravity

Newtons 2nd law

Acceleration caused by gravity =

Force
-----
Mass

=

9.81N
-----
1Kg

= 9.81 ms^-2

(Force to accelerate 1kg by 1ms^-2)

4. Apr 7, 2004

### turin

It looks like Chen alluded to this, but failed to make it explicit. In some contexts, the sybol "g" is used to identically mean the number "9.81" with the attached units "m/s2", or the dimensional quantity with that value in some arbitrary unit system. Whereas, the acceleration due to the gravitational pull of some arbitrary massive body is represented by the symbol " g' " with the prime indicating that it is not a specific quantity assignment, but is more generally defined by the equation that Chen gave. For instance, you may hear/see that the gravity on the moon is only 1/6 gs. That means that the number for g' is (1/6)g = (1/6)(9.81 m/s2). In a different context, you may hear/see that the gravity on the moon is g = 1.6 m/s2. Beware of the context!

5. Apr 8, 2004

### Imparcticle

So g = 1.6 m/s2 is totally diffrent from (1/6)g = (1/6)(9.81 m/s2) because the first one is not specifying the numerical value of
g1?

But I still don't know WHY g=9.8m/s2 for earth.

6. Apr 8, 2004

### Chen

Because if you plug Earth's mass and radius into this equation:
$$g' = G\frac{M}{r^2}$$
You will get a value of 9.8m/s2.

Last edited: Apr 8, 2004
7. Apr 8, 2004

### Kurdt

Staff Emeritus
Also g is not constant for the Earth. As you can see the formula is inversely proportional to the radius squared so as one goes further up into the stmosphere you will experience a weakening g acceleration. We merely think it is constant because the heights from earths surface we encounter daily are not significant enough to be of any consequence to g.

8. Apr 8, 2004

Staff Emeritus
And the earth is not a perfect sphere, either, or perfectly symmetrical in density. Gravity is slightly stronger on the suface at some places than at others.

9. Apr 8, 2004

### Chi Meson

And just to split a few more hairs, the average value at the surface of the earth at sea level is 9.801 N/kg, not 9.81 N/kg.

10. Apr 8, 2004

### turin

One thing to keep in mind, though, is that the numerical value of 9.81 m/s2 is, I'm pretty sure, the value at sea level. Whereas, the value at the top of mount Everest would then be 9.78 m/s2 (if I did the calculation correctly). At any rate, this is only a 0.3 % difference, but it is noticible.

EDIT:
OK, apparently I was incorrect to assume that the value at sea level is 9.81. The percent difference is still valid, though, which was the issue I intended to exemplify.

Last edited: Apr 8, 2004
11. Apr 8, 2004

### Andre

It's even much more complicated. g is rather variable all over the globe, depending on local mass differences, corrections for the EArth shape and for Earth spinning forces.

Grace and GOCE (Links of Nereid) are exploring gravity very detailed nowadays.

You could find out here the gravity of the planets and bigger moons in the solar system.

If you put in 1, you get the relative values compared to 1g with Earth

Last edited: Apr 8, 2004
12. Apr 8, 2004

### Njorl

Gravity variation is one of the ways of finding oil and natural gas deposits.

Njorl

13. Apr 8, 2004

### KingNothing

I think it's more of a theory question. No offense, jsut how I interpret it. Well, it has to be acceleration. Think about it this way. If it was not acceleration...when skydiving, you would not fall because your velocity wouldn't be 'changing' (crude definition of acceleration).

The fact that it's 9.8 for earth is just basically something that happened to be true. 'Why' the earth was made like that is a question that is beyond physics. Is your question along the lines of "why not 10, or 15, or 500?"? Well, it just turns out that it's not. In some places in the universe, it is. Just not here. Why do I not have a TV in my house? I just don't. Just something that 'happens to be true'.

Think about it this way: What would happen if it weren't 9.8? What would happen if it wasn't acceleration?

14. Apr 8, 2004

### Chen

By the way, if you hear things like "The pilot is experiencing 3G's" it means that the acceleration of the pilot is 3 times the acceleration of a falling body on earth.

15. Apr 8, 2004

### turin

Shouldn't it be the acceleration experienced by a body not moving WRT the earth, since the experience implies something local, and a body that is falling does not necessarilly experience any local acceleration?

16. Apr 9, 2004

### Kurdt

Staff Emeritus
Not necessarily local. IF you were in a rocket and were accelerating at 9.8m/s^2 then you would feel as though you were under the influence of of the gravitational field of the Earth.

17. Apr 9, 2004

### Chen

Turn, look up Einstein's Equivalence Principle.

18. Apr 9, 2004

### turin

Kurdt and Chen,
I believe y'all misunderstood my slight objection. I am quite familiar with the equivalence principle, in several forms in fact. But, any effect that an object experiences is a local effect in classical physics. I will refer the two of you to Einstein's eventual objection to Mach's principle, or, more fundamentally and previous to this, the modern principle of relativity in the first place. Non-locality is a QM conjecture, and I would venture to say that it is not accepted in the general physics community (though I cannot be sure of this). At the least, I don't think that QM is appropriate in the consideration of the g's experienced by a pilot.

Regarding the accelerating rocket, wrt what is it accelerating? Mach would have said the fixed stars. Einstein began his GR under such a notion, but eventually abandoned it for what he considered the more appropriate notion of a dynamical geometry. The stars are in different places at different times depending on the local inertial frame from which they are considered. Furthermore, the light (and gravitational effects) take a long time to get to a rocket that is lyrs away from them. It is this finite velocity of propagation that gives modern relativity theory its local character (well, one issue/example anyway). Now, a person in the rocket experiences a tug, if you will, a tug that would not be experienced without the boosters engaged. More philosophically (Mach's principle), the tug is experienced because the distance relations to the fixed stars are non-stationary. In GR, this is revised so a nontrivial connection coefficient (Christoffel symbol) at the point of observation, a local effect, that happens to extend far outside of the rocket in this frame.

My objection was initiated by the notion that, an object not accelerating with the ship and allowed to coast away in an inertial rest frame would not be able to distinguish from free-fall in a uniform graviational field. Observers inside the rocket would not be able to distinguish from being at rest on earth. It is being at rest on earth that causes one to experience acceleration. It is free-fall that frees one from the experience of acceleration and allows a local inertial rest frame. I think that you have it the other way around. A pilot experiencing 3 g's is experiencing 3 times the tug that he/she would experience if he/she were not falling but sitting still on the surface of the earth.

Now that I have clarified, do y'all still disagree?

Last edited: Apr 9, 2004