Calculating the Effective Value of g at 3330 m Above Earth's Surface

[strugglin-physics]

Calculate the effective value of g, the acceleration of gravity, at 3330 m above the Earth's surface.

Isn't the value of g always -9.8 m/s^2?

 

[robphy]

g has magnitude 9.8 m/s^2 near the surface of the earth

Generally, g=GM/R^2 where M=mass of the earth, R=distance from the Earth's center, and G is Newton's gravitational constant.

 

[strugglin-physics]

so if I wanted to Calculate the effective value of g, the acceleration of gravity, at 3330 km above the Earth's surface.

I would times 6.6742 × 10−11 (G) by Earth's mass and then divide it by 3330km^2? How do I know Earth's mass?

 

[arildno]

Write your "effective acceleration of gravity" as a fraction of "ordinary acceleration of gravity" (this bypasses the need to calculate the mass of the Earth and G, but you should know the radius of the Earth.)
And no, you should NOT divide with 3300 squared. Think again.