# Gravity and value of g

1. Nov 16, 2007

### q01aq

Gravity and value of "g"

This could be a simple question, its just been a long time since I have done any physics.

Is there anyway to account for the constantly changing value of "g" as an object falls?
Example: A bowling ball (7.2kg) is dropped from the top of Mt. Everest (8848m), I would like to find the final velocity and final kinetic energy of the ball before it hits sea level (0m). I have found the value of g at the coordinates of Mt. Everest (27.988056 deg latitude) and 8848m, but the acceleration will change as the height changes.

So my question, is there anyway to account for this change in the acceleration due to earth's gravity as the object is falling? Will it even matter once it reaches terminal velocity? Thanks for the help!

-Bone

2. Nov 16, 2007

### Shooting Star

If we consider the earth to be a sphere, then the gravity of earth will decrease with height as the inverse square law. But this is the most basic approximation. The earth is also rotating about an axis, and the centrifugal force reduces the effective gravity more near the equator. Also, the earth is not a sphere but more of an oblate shape, due to which the g is more near the poles. Added to all this is the local variations of mass distribution, which has an effect on the value of g. After taking into all this, air resistance must be taken into account, the laws of which are mostly empirical and depends on the speed of the object and the shape of the object itself.

Depending on the problem and the degree of accuracy required, you have to choose what to consider and what to ignore.

3. Nov 16, 2007

### q01aq

So if I am assuming the earth is a sphere, then I can neglect the effect that the latitude has on the problem. Also, I am going to assume that the only force acting on the ball is that due to gravity (ie. pretend its in a vacuum). I think this eliminates eveything I am not worried about and will let me concentrate on the changing value of g based solely on the changing height.

4. Nov 16, 2007

### mgb_phys

The standard model is:
g = 9.78(1 + 0.0053 sin^2(lat) - 0.0000058 sin 2*lat ) - 3.086E-6 height

But geology ( density of surrounding rock ) effects it more than height.
If you are building telescopes the difference between 'UP' as in a line from the centee of the earth and 'UP from gravity can differ by 5arcsec because of local underground mass concentrations.

5. Nov 16, 2007

### q01aq

This is the formula I have actually used to find the value of g at the top of Mt. Everest. What I am thinking though, is that when the bowling ball is dropped, the acceleration it experiences initially will be smaller than the acceleration it experiences at some other point in its fall (say at the half way point). Maybe I am over thinking things here, but is there some sort of calculus function or something (integral pops into my head) that could account for this constant change in acceleration?

That tidbit about the geology is pretty interesting, thanks for all th input!

6. Nov 16, 2007

### katchum

g(h) = dv/dt = d(dh/dt) = 9.78(1 + 0.0053 sin^2(lat) - 0.0000058 sin 2*lat ) - 3.086E-6.h

v = dh/dt

Integrate:

int(int(1/(9.78(1 + 0.0053 sin^2(lat) - 0.0000058 sin 2*lat ) - 3.086E-6.h)))= 1/2.t^2

And h for the integral = height of the mountain.
So t = time till you reach the ground = ....

Then:

v = dh/dt
Fill in the equation for h(t) and fill in the time till you reach the ground and you'll have v.

I'm not sure of my answer, I'm just guessing around here! Anyone concur with my analysis?

Last edited: Nov 16, 2007
7. Nov 16, 2007

### mgb_phys

g is just inversely proportional to height.
So you equation is correct, or you could just work out 'g' at surface and 'g' at everest and assume a linear increase.

Of course if you are going to add this tiny correction should also allow for the bouyancy of the air being different at sea level and 30,000ft !

8. Nov 16, 2007

### Shooting Star

What exactly is the OP trying to find out would help a lot in making a model. I know that he wants to drop the ball, but is there more to it?

9. Nov 17, 2007

### dynamicsolo

As I read the problem, I wonder if this is overly fussy. Are you taking an introductory physics course or something at a higher level? You may not be required to concern yourself in this problem with either variations in acceleration or with air resistance. (It's a fair question as to whether the bowling ball would simply have reached terminal velocity by the time it fell almost 9 km.)

Having posed and worked on a related problem, I can tell you that g changes very little (some parts per thousand) over the height of Everest...

Last edited: Nov 17, 2007
10. Nov 17, 2007

### nrqed

It depends at what level your class is! If you are getting to a point where you worry about the effects of latitude, then you should worry even more about air friction!! And other effects.

On the other hand, at a lower level you may simply use conservation of energy!. For the gravitational potential energy, just use $- G m M /r$.

11. Nov 20, 2007

### q01aq

This isnt for any specific course, I am just trying to get a better understanding of potential energy. Specifically I am interested in electrical potential energy but I thought I could better grasp the concept if I considered this problem first.

In a hydrogen atom, considering the proton in a fixed position, and the electron held at a distance from the proton. What would be the electron's final velocity just before impacting the proton? I am trying to come up with the electrons beginning electrical potential energy and comparing it to the final kinetic energy. I am neglecting gravity obivously and I know that things do not work this way in a hydrogen atom. This is just where I was trying to go with the Everest/bowling ball question. Any comments?

I have been just using classical physics to determine the energies but am getting different values for potential and kinetic energies.