is acceleration to to gravity the same as equitorial surface gravity?

Answer 1

Yes, No, sort of.
You see the Earth is not completely the same all the way through. A value of 9.8 meters per second squared or 32 feet per second squared is the accepted value for gravity. It actually changes for altitude and composition of the Earth at every location. If you were at or below sea level over a large iron deposit, this force would be slightly greater. If you happen to be standing on a mile high pile of feathers, gravity would be slightly less.

If this is for your homework, they're probably just telling you to use the generally accepted value for gravity.

Answer 2

g = GM/R^2; where g is the acceleration due to gravity from a mass M (e.g., Earth), at a distance R from the center of the mass. G is a constant everywhere in the universe. So, for a given mass, g varies as the inverse of R^2.

Earth is not a perfect sphere; so R, the radius of the planet, is not constant from place to place on Earth's surface. In fact, at the equator R is larger than at the poles because Mother Earth is fat around the middle. So g at the equator would be less than g at the poles...even without the spin of the planet considered.

But there is the spin...360 degrees per 24 hours in fact. And on the equator that creates a centrifugal reaction that tends to act outward from the direction of gravity and, thereby, reduce the effective weights of objects from what they would be at the poles. That is, the effective net weight w = mg - mw^2R; where W = mg is the weight without the spin and mw^2R is the offsetting centrifugal force on a mass m spinning at w = 360 degrees/24 hours on the equatorial surface at radius R.

Bottom line, g is not fixed on Earth's surface. But for most problems g = 9.81 m/sec^2 or 32.2 ft/sec^2 will suffice because these values are averages over the entire surface of the planet.

Answer 3

No