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

AI Thread Summary
The discussion revolves around calculating the orbital period of a satellite using Kepler's Third Law. The satellite orbits at a distance of 6.80 x 10^8 m from the center of a planet with a mass of 3.08 x 10^24 kg. The initial calculation yielded a period of approximately 7779136.049, but the user struggled to convert this into Earth days. Ultimately, the user identified that dividing by the number of seconds in a day provided the correct answer of 90 Earth days. Clarifications were made regarding the distinction between the satellite's orbital radius and the planet's radius, emphasizing the importance of accurately interpreting 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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