Gravity Problem Help - Find Acceleration of Gravity 3000 Miles Above Earth

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SUMMARY

The discussion focuses on calculating the acceleration due to gravity at an altitude of 3000 miles above the Earth's surface. The initial calculation using the formula g = GM/r² yielded an incorrect value of 17.1 m/s². After clarification, the correct approach involves adding the Earth's radius (approximately 6,378 km or 11.2 x 10^6 meters) to the altitude, resulting in an accurate acceleration of approximately 3.17 m/s².

PREREQUISITES
  • Understanding of gravitational force equations, specifically g = GM/r²
  • Knowledge of Earth's mass, approximately 5.98 x 10^24 kg
  • Familiarity with unit conversions, particularly miles to meters
  • Basic understanding of Earth's radius, approximately 6,378 km
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  • Research gravitational force calculations at various altitudes
  • Learn about the implications of altitude on gravitational acceleration
  • Explore the concept of gravitational potential energy
  • Study the effects of altitude on satellite orbits and trajectories
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Students in physics, aerospace engineers, and anyone interested in gravitational physics and its applications in space exploration.

bengaltiger14
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I need to find the value of acceleration of gravity 3000 miles above the Earth's surface. I figured 3000 miles = 4.83 x 10^3 meters. Mass of the Earth equals 5.98 x 10^24kg. I used the equation:

g= GM/r^2 and get 17.1 m/s^2. This is too high a value. I don't think I am using the correct distance in my calculation. Any help Please?
 
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Ok, I think I got is. I need to add the radius of the Earth which is 11.2 x 10^6 meters to the 3000 miles and the acceleration = 3.17 m/s^2 correct??
 
bengaltiger14 said:
Ok, I think I got is. I need to add the radius of the Earth which is 11.2 x 10^6 meters to the 3000 miles and the acceleration = 3.17 m/s^2 correct??
Yes, it is correct (although I think you meant 3000 miles {4828 km} plus the radius of Earth {6378 km} equaled 11.2 x 10^6 meters.)
 

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