Calculating Kepler's Constant for Earth Satellites

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Homework Help Overview

The original poster attempts to determine Kepler's constant for Earth satellites, specifically using the moon as a reference. The problem involves calculating the constant using the formula K=R³/t², where R is the distance from the moon to Earth and t is the orbital period.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster calculates K using the moon's distance and orbital period but finds a discrepancy between their result and the expected answer. They question the accuracy of their inputs and seek clarification on the values used.

Discussion Status

Some participants have engaged by suggesting that the difference in results may stem from the number of significant figures used in the calculations. There is an acknowledgment of the closeness of the values, but no consensus has been reached on the exact cause of the discrepancy.

Contextual Notes

The original poster mentions an impending test on Planetary Mechanics and expresses frustration over a lack of support from their teacher, who has provided the expected answer but not guidance on the calculation process.

NDiggity
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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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