What Altitude Above Earth Gives a Gravitational Acceleration of 3.53 m/s²?

AI Thread Summary
To find the altitude above Earth's surface where gravitational acceleration is 3.53 m/s², the equation a = GM/r² is used, with G being 6.67 x 10^-11 and M as 5.98 x 10^24. The calculated distance from the Earth's center is 1.06 x 10^7 m, but this value represents the total distance, not the altitude. To determine the altitude, the Earth's radius (approximately 6.37 x 10^6 m) must be subtracted from this total distance. The correct altitude can be found by calculating the difference between the total distance and the Earth's radius.
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At what altitude above Earth's surface would the gravitational acceleration be 3.53 m/s^2?

I used the equation
a= GM/r^2
G=6.67 x 10^-11 and M= 5.98 x 10^24
Solving for r gives 1.06 x 10^7 m which isn't right... can someone tell me what I'm doing wrong?
 
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Reread the question. You need the altitude above the Earth's surface, not the distance from the center of gravity of the earth.
 
How would I find that since all I know is the acceleration, mass and radius of the Earth?
 
You know r = 1.06 x 10^7 m, the problem isn't asking for 'r'.

The problem is asking how far from Earth's surface. You know 'r' of earth, how high above the Earth is r = 1.06 x 10^7m?
 
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