# Magnitude of acceleration due to gravity

1. Mar 3, 2014

1. The problem statement, all variables and given/known data
Satellites are placed in a circular orbit that is 1.20 × 106 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?

2. Relevant equations
My book is saying that the relevant equations are:

W = G((ME m)/r2)

W = mg

3. The attempt at a solution
After looking at this post, I tried the formula g = MG/r^2. (M being the mass of the Earth)

g = ((6.67 x 10^-11)(5.97 x 10^24))/((1.20 x 10^6)^2)

This^ yielded 276.53 m/s^2, which is apparently wrong.

2. Mar 3, 2014

### Ronaldo95163

g' = (Re^2/r^2)*g

r = Re + h where Re is the radius of the earth and h is the height above it.

g = acceleration due to gravity on the earth

substitute the values given in the equation in that formula and you should get the answer for g'

3. Mar 3, 2014

### DataGG

As Ronaldo said, $r$ is the distance of the object to the CM of earth. That means, as Ronaldo stated: $$r=1,20*10^6+r_{earth}$$

4. Mar 3, 2014