# Physics: Circular Motion?

What is the acceleration due to gravity at an altitude of 1.00 × 106 m above the earth’s surface? Note: the radius of the earth is 6.38 × 106 m.

Relevance

Basic formula

F = m a

this is 2nd Law of motion

a = F/m

where F is the force of gravity

F = G M m / r²

a = F/m = G M / r² ......... (1)

G is gravitational constant

M is the mass of the Earth

r is the distance from the center of the Earth

We are not given G and M, but remember that the acceleration due to gravity on the surface of the Earth is g=9.8 m/s². As in the previous case

g = G M / R² ............ (2)

where distance from the center of the Earth is actually the radius of the earth R

Divide (1) by (2) to rid of GM

a / g = R² / r²

a = g R² / r²

r = R + h

where h is the altitude above the Earth's surface. Substitute this in the previous equation

a = g R² / (R+h)²

a = g (R/(R+h))²

a = 9.8 * (6.38*10⁶/(6.38*10⁶+1*10⁶))²

10⁶ cancels out

= 9.8 * (6.38 / 7.38)²

= 9.8 * 0.864²

= 7.32 m/s²