Gravitational acceleration above the Earth's surface

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SUMMARY

The discussion centers on calculating the altitude above Earth's surface where gravitational acceleration equals 4.9 m/s². The initial approach utilized the formula g = GM/(R+h)², where R is Earth's radius and h is the altitude. The user incorrectly computed the altitude, arriving at 2.00e7 m. A correct method involves using the ratio of gravitational accelerations at two different altitudes, leading to the equation 2 = (R+h)²/R², which simplifies the calculation.

PREREQUISITES
  • Understanding of gravitational force and acceleration
  • Familiarity with the formula g = GM/r²
  • Knowledge of Earth's radius (approximately 6.37e6 m)
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of gravitational acceleration formulas
  • Learn about gravitational potential energy and its relation to altitude
  • Explore the concept of gravitational fields and their calculations
  • Investigate the effects of altitude on gravitational acceleration in different celestial bodies
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Students in physics, educators teaching gravitational concepts, and anyone interested in astrophysics or the effects of altitude on gravitational forces.

bearhug
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At what altitude above the Earth's surface would the gravitational acceleration be 4.9 m/s^2?

I thought this what very simple but apparently it is not since I got the wrong answer and can't figure out what I did wrong.
I used g= GM/r^2= GM/(R+h)^2 R being radius of Earth and h being the altitude.

g= 4.9m/s^2

4.9= (6.73e-11)(5.98e24)/(6.37e6+ h)^2
4.9= 3.99e14-(6.37e6+h)^2
solved for h and got 2.00e7m

Can someone tell me where I went wrong? Thanks
 
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The approach is OK, but the computation has an error. You really only need the radius of the Earth and g at the surface to do this, but your approach is OK too.

9.8m/s^2 = GM/R^2
4.9m/s^2 = GM/(R+h)^2
2 = (R+h)^2/R^2 <== dividing 1st eq by 2nd eq
etc
 

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