Calculating Kepler's Constant for Earth Satellites

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SUMMARY

The discussion centers on calculating Kepler's constant for Earth satellites using the formula K=R³/t². The user attempted to calculate K using the moon's distance from Earth (384,403 km) and its orbital period (27.3 days), resulting in an answer of 1.02x10¹³. However, the expected value is 9.85x10¹², which can be attributed to differences in significant figures used in the calculations. The discrepancy highlights the importance of precision in scientific calculations.

PREREQUISITES
  • Understanding of Kepler's laws of planetary motion
  • Familiarity with the formula K=R³/t²
  • Knowledge of significant figures in scientific calculations
  • Basic understanding of orbital mechanics
NEXT STEPS
  • Review the concept of significant figures in scientific calculations
  • Learn more about Kepler's laws and their applications in celestial mechanics
  • Explore the calculation of orbital periods for different celestial bodies
  • Investigate the impact of measurement precision on scientific results
USEFUL FOR

Students studying planetary mechanics, educators teaching orbital dynamics, and anyone interested in the precision of astronomical calculations.

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Homework Statement


Determine Kepler's constant for all Earth Satellites. No information is given, only the question.

Homework Equations


K=R³/t²

The Attempt at a Solution


I decided to use the moon as a satellite. So I went K=(384,403,000)³/(2,360,594.88)²

For R I used the distance from the moon to Earth in meters (384,403km) and for t I used the time it takes the moon to revolve around Earth in seconds (27.3 days).

The answer I get is 1.02x10¹³.
The answer is supposed to be 9.85x10¹²

Thanks for any help!
 
Last edited:
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Is there any other information I need to provide? Please help, I have a test on Planetary Mechanics tomorrow and my teacher refuses to help me with this question. He told me what the answer should be and said to work it out on my own. I asked some of my classmates and they could not get it either...
 
I would do it like you did.
These values are very close.
10.2 * 10^12
9.85 * 10^12
So why do you think that is? You and your teacher used different number of significant figures.
3.8 * 10^8
and
2.36 * 10^6
Only two or three significant figures.
Use these values for your calculation and you get your teacher's results. Silly huh? :smile:
It's weird how your teacher would use two significant figures for one and three significant figures for the other.
 
Last edited:
Right on, thank you very much for your help. It seemed strange because I was so close.
 

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