Magnitude of acceleration due to gravity

AI Thread Summary
The discussion focuses on calculating the acceleration due to gravity at a height of 1.20 × 10^6 m above Earth's surface. The relevant formula used is g' = (Re^2/r^2) * g, where Re is the Earth's radius and r is the total distance from the Earth's center. An initial calculation using g = G(ME m)/r^2 yielded an incorrect result of 276.53 m/s^2. Participants clarified that r must include the Earth's radius plus the height above the surface. The correct approach led to a successful calculation of the gravitational acceleration at that altitude.
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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((ME 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.
 
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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'
 
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As Ronaldo said, ##r## is the distance of the object to the CM of earth. That means, as Ronaldo stated: $$r=1,20*10^6+r_{earth}$$
 
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!
 

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