What Is the Magnitude of Satellite Acceleration Due to Gravity?

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SUMMARY

The magnitude of satellite acceleration due to gravity at a height of 3.59 x 107 m above the Earth's surface is calculated using Newton's law of gravity. The gravitational constant (G) is 6.67259 x 10-11 N*m2/kg2, and the mass of the Earth (ME) is 5.98 x 1024 kg. The correct acceleration is determined to be 0.233 m/s2, as the radius (r) must be measured from the center of the Earth, not the surface.

PREREQUISITES
  • Understanding of Newton's law of universal gravitation
  • Familiarity with gravitational constant (G)
  • Knowledge of mass of Earth (ME)
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of Newton's law of universal gravitation
  • Learn about the implications of gravitational acceleration in satellite motion
  • Explore the concept of circular orbits and their dynamics
  • Investigate the effects of altitude on gravitational force
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Students in physics, aerospace engineers, and anyone interested in satellite dynamics and gravitational effects at various altitudes.

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Homework Statement



Synchronus communications satellites are placed in a circular orbit that is 3.59 x 107 m above the Earth's surface. What is the magnitude of the acceleration due to gravity at this distance?

F=?
G=6.67259 x 10-11 N*m2/kg2
ME (Mass of Earth)= 5.98 x 1024 kg
r = 3.59 x107m

Homework Equations



F = (G*m1*m2)/(r2)
F = MA

The Attempt at a Solution



1) F = G * (ME2/r2) = 1.8514...N
2)F = MA --> (1.8514...N) = (5.98 x 1024)A
3) A = 0.309 m/s2

CORRECT ANSWER: 0.233 m/s2
 
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In Newton's law of gravity, r is the distance from the center of the earth, not the distance from the surface.
 
Doc Al said:
In Newton's law of gravity, r is the distance from the center of the earth, not the distance from the surface.

Ah, there's the problem. :P

Thanks. :D
 

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