"The actual value of computed g, assuming no drag, also varies somewhat over the earth's surface due to the earth's spin (which causes a latitude-dependent centrifugal force to get subtracted), lack of a perfect spherical shape of the earth, and the proximity of very high mountain ranges. "

Thanks =]

wait...

ok I found the latitude the longitude and the elevation and used an online calculator which gave me

9.80272 m/s^2
+/- .000003 m/s^2

ok so what exaclty are we ignoring when we use the formula

Fg = (G m1 m2)/r^2

a "latitude-dependent centrifugal force " is ignored in that calculation

"lack of a perfect spherical shape of the earth"

Does this forumla assume that the Earth is a perfect sphere and we are ignoring that it is not???

"and the proximity of very high mountain ranges" How does very high mountain ranges effect the calculation???

And were assuming no drag force??? From the earth moving through space and spinning??? even though there would be one? Or would there not be one?

THANKS!!!

Yes, the fact that the Earth is an oblate spheroid varies the g vs over a sphere.

Centripetal force reduces apparent weight, not g, but many do not make that destinction.

Weight= mg -mv^2/r and v horizontally, the spin of the Earth, depends on latitude.

Now Mountains: g is a computed field vector dependent on Earth's Mass below it. If you go high on a mountain, you not only change elevation (reducing g) but you increase the mass pulling on you (increase g). Also, the material the mountain range is made of varies g (ie, the mountain range is granite or iron).

The drag is due to free fall, and should not have been mentioned. Actual acceleration to the Earth is g minus drag effects.

One last remark: Normally g is taught in HSchool as the acceleration due to gravity, in m/s^2. But you can do more with that, consider

Fgravity=mass*g or g= forcegravity/mass so g is in the units of N/kg, which is a force field. So consider g a force field intensity. This now puts a new light on the effects of the mountains...

When using the formula Fg = (G m1 m2)/r^2 to calculate the force of gravity, there are a few factors that are generally ignored or approximated.

Firstly, the formula assumes that the Earth is a perfect sphere, which it is not. The Earth is an oblate spheroid, meaning it is slightly flattened at the poles and bulges at the equator. This irregular shape introduces some minor variations in the gravitational acceleration (g) at different locations on the Earth's surface.

Secondly, the Earth's rotation causes a centrifugal force that is dependent on latitude. This force, known as the latitude-dependent centrifugal force, acts opposite to the force of gravity and slightly reduces the overall value of g. However, this effect is relatively small and often negligible for most calculations.

Thirdly, the presence of very high mountain ranges can also have a slight impact on the value of g. The gravitational force is inversely proportional to the square of the distance (r) between the objects, so if you are at a higher elevation near a tall mountain, the value of g would be slightly lower compared to being at sea level due to the increased distance between you and the center of the Earth.

Lastly, the formula does not take into account drag forces caused by the Earth's movement through space and its rotation. These forces are generally much smaller compared to the force of gravity and can be ignored in most scenarios.

So, when using the formula Fg = (G m1 m2)/r^2, it is important to understand that it provides an approximation of the gravitational force while ignoring factors like the Earth’s non-spherical shape, latitude-dependent centrifugal force, the presence of very high mountain ranges, and the drag forces. In practical calculations, these effects are usually considered only when dealing with highly precise measurements or specific scenarios where they become significant.