Satelite orbiting earth in a circular path

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SUMMARY

The discussion centers on calculating the orbital speed of a satellite orbiting Earth at an altitude of 1000 km. The correct formula used is v = sqrt(G*M/r), where G is the universal gravitational constant (6.67E-11), M is the mass of Earth (6E24 kg), and r is the radius from the center of Earth to the satellite. The user initially calculated the radius incorrectly as 7.67E6 m instead of the correct value of 7.37E6 m, leading to an incorrect orbital speed of 7369 m/s. The correct calculation yields an orbital speed of approximately 7900 m/s.

PREREQUISITES
  • Understanding of gravitational physics and orbital mechanics
  • Familiarity with the universal gravitational constant (6.67E-11)
  • Knowledge of Earth's mass (6E24 kg) and radius (6.37E6 m)
  • Ability to perform calculations involving square roots and scientific notation
NEXT STEPS
  • Learn how to derive orbital speed using Kepler's laws of planetary motion
  • Study the effects of altitude on satellite speed and orbital decay
  • Explore the differences between circular and elliptical orbits
  • Investigate the implications of orbital mechanics on satellite communication
USEFUL FOR

Students studying physics, aerospace engineers, and anyone interested in satellite dynamics and orbital calculations.

cortozld
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Homework Statement



What is the orbiting speed (m/s) of a satelite orbiting Earth in a circular path 1000 km above Earth's surface? (Use 4 sig figs)

Homework Equations



universal constant of 6.67E-11
r of Earth 6.37E6 m (given by teacher)
mass of Earth 6E24 kg (given by teacher)

derived v=sqrt(G*M/r)

this is what i have read will solve this equation for me


The Attempt at a Solution



First I converted the 1000 km above Earth to m and added that and the radiace of Earth together to get 7.67E6 m (r) then I solved for v: sqrt(6.67E-11*6E24/7.37E6)=7369 m/s

what am I doing wrong? There is also a possiblity that the teacher is wrong, its happened before on his website
 
Last edited:
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Welcome to PF!

Hi cortozld! Welcome to PF! :smile:

(have a square-root: √ :wink:)
cortozld said:
r of Earth 6.37E6 m (given by teacher)
…
First I converted the 1000 km above Earth to m and added that and the radiace of Earth together to get 7.67E6 m (r) then I solved for v: sqrt(6.67E-11*6E24/7.67E6)=7369 m/s

You've used 7.67E6 instead of 7.37E6 …

does that make a difference? :smile:​
 
that was an accidental typo, my bad. I believe its my teachers website, he's been having lots of problems with it since the year started. So unless anyone has an idea about what I'm doing wrong I'll just have to talk to him on Mon.
 

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