Magnitude of acceleration due to gravity

[Shadow236]

Homework Statement

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?

Homework Equations

My book is saying that the relevant equations are:

W = G((M_E m)/r²)

W = mg

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.

 

[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'

 

[DataGG]

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

 

[Shadow236]

It looks like you guys are correct, because I got the answer correct and the book says the following...: http://twitpic.com/dx955g

Thank you!

 

[HalfLostAndSearching]

Your attempt at using the formula g = MG/r^2 was correct. However, the value you used for the radius (1.20 x 10^6 m) was incorrect. The radius in this case should be the distance from the center of the Earth to the satellite, which is 6.37 x 10^6 m (radius of the Earth + altitude of the satellite).

Correct calculation:

g = ((6.67 x 10^-11)(5.97 x 10^24))/((6.37 x 10^6 + 1.20 x 10^6)^2)

g = 9.81 m/s^2

The magnitude of the acceleration due to gravity at this distance is 9.81 m/s^2. This means that objects in this orbit will experience the same gravitational acceleration as objects on the surface of the Earth.