Kepler's Law and Non-terrestrial Orbits (Not Earth)

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Homework Help Overview

The discussion revolves around a satellite orbiting a planet, with a focus on applying Kepler's 3rd Law to determine the orbital period. The problem specifies the distance from the planet to the satellite and the mass of the planet, but there is confusion regarding the conversion of the period into Earth days.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Kepler's 3rd Law and question how the original poster derived their period value. There are inquiries about the units of the calculated period and the method used to convert it into Earth days.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the calculations and assumptions made regarding the orbital distance and the relationship to the planet's properties. Some participants have pointed out the need for more detailed work to understand the original poster's results.

Contextual Notes

There is ambiguity regarding the reference to the planet's rotation and how it affects the interpretation of the period. The original poster's mention of "days of this planet" raises questions about the definitions being used in the context of the problem.

Axoren
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Homework Statement


A satellite orbits a planet at a distance of 6.80 multiplied by 10^8 m. Assume that this distance is between the centers of the planet and the satellite and that the mass of the planet is 3.08 multiplied by 10^24 kg. Find the period for the moon's motion around the earth. Express the answers in Earth days

Homework Equations


Kepler's 3rd Law
Period² = (4(pi)radius³)/(Gravitational Constant * Mass of Massive Body)

The Attempt at a Solution



I have the answer in days of this planet, however, I don't know how to get the answer in Earth days.

7779136.049 = Period.

I'm incredibly sure this is a correct value, I've done multiple calculations and always got within 0.1% of this value

I saw this question in the archive, but I don't believe anyone realized the answer was in Earth days or they just did not show their work (which is what I was hoping to see).

Please help.

Found my problem, needed to divide by days represented in seconds (Period / (60 * 60 * 24));

Answer was 90 days. (decimal dropped out of laziness, forgive me significant digits.)
 
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HOW did you get it "in days of this planet" when nothing is said about the planet's rotation?
 
Axoren said:

Homework Equations


Kepler's 3rd Law
Period² = (4(pi)radius³)/(Gravitational Constant * Mass of Massive Body)

The Attempt at a Solution



I have the answer in days of this planet, however, I don't know how to get the answer in Earth days.

7779136.049 = Period.
How did you go from Kepler's third law to 7779136.049? What are the units for this number?

In other words, show your work. It's a bit difficult to find where your mistake is if you don't show us how you arrived at your result.
 
Axoren said:

Homework Statement


A satellite orbits a planet at a distance of 6.80 multiplied by 10^8 m. Assume that this distance is between the centers of the planet and the satellite and that the mass of the planet is 3.08 multiplied by 10^24 kg. Find the period for the moon's motion around the earth. Express the answers in Earth days

Homework Equations


Kepler's 3rd Law
Period² = (4(pi)radius³)/(Gravitational Constant * Mass of Massive Body)

The Attempt at a Solution



I have the answer in days of this planet, however, I don't know how to get the answer in Earth days.

7779136.049 = Period.
How did you get the radius of planet X if the question on,y mentions the satellite's orbital distance?
 
Mrrss24 said:
How did you get the radius of planet X if the question on,y mentions the satellite's orbital distance?

That's not the planet's radius, it's the satellite's orbital radius.
 

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