physics
posted by physics .
"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 latitudedependent 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 "latitudedependent 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...