Using Kepler's third law, determine the period

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SUMMARY

This discussion focuses on applying Kepler's Third Law to determine the orbital period of a hypothetical planet located three times farther from the Sun than Earth. The correct calculation reveals that the orbital period is 5.2 years, contrary to the initial incorrect assumption of 3 years. The discussion also clarifies that the mass of the planet does not affect the orbital period, as Kepler's Laws assume the mass of planets is negligible compared to that of the Sun.

PREREQUISITES
  • Understanding of Kepler's Third Law of planetary motion
  • Basic algebra for solving equations
  • Familiarity with orbital mechanics
  • Concept of circular orbits in celestial mechanics
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  • Study the derivation and implications of Kepler's Third Law
  • Explore the differences between circular and elliptical orbits
  • Learn about the mass-independent nature of orbital periods
  • Investigate the gravitational effects of massive bodies in celestial mechanics
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Astronomy students, physics educators, and anyone interested in celestial mechanics and orbital dynamics will benefit from this discussion.

Calpalned
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Homework Statement


If there were a planet three times farther from the sun than the Earth is, how long would it have taken this hypothetical planet to orbit the Sun? Assume the orbit is a circle.

Homework Equations


Kepler's 3rd Law ##= (\frac{r_1}{r_2})^3 = (\frac{T_1}{T_2})^2 ##

The Attempt at a Solution


##= (\frac{3r_E}{r_E})^3 = (\frac{T}{1})^2 ##
T = 3 years Correct answer: T = 5.2 years
Where is my blunder? Thank you all so much.
 
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Calpalned said:

The Attempt at a Solution


##= (\frac{3r_E}{r_E})^3 = (\frac{T}{1})^2 ##
T = 3 years Correct answer: T = 5.2 years
Where is my blunder? Thank you all so much.
Show the math one step at a time.
 
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gneill said:
Show the math one step at a time.
I just found out my error. ##3^3 = 27##. I accidentally wrote 9 instead of 27.
Thanks for helping Gniell
 
I have one more question...
Part B of this question asks if I can use the data to solve for the mass of the hypothetical planet. The answer is no. "No mass data can be calculated from this relationship, because the relationship is mass- independent. Any object at the orbit radius of 3 times the Earth’s orbit radius would have a period of 5.2 years, regardless of its mass. "

The "...regardless of its mass" statement seems counter-intuitive to me. What if that planet turned out to be a super-giant red star much greater in mass than the Sun? Then the Sun would orbit the red star right? How can the red star still have an orbital period of 5.2 years? Thank you.
 
The "...regardless of its mass" statement needs to be taken in the context of the basic assumption of Kepler's Laws where the masses of all the planets are considered to be much less than that of the primary (the Sun).

So I guess one might say that the mass of the object doesn't matter as long as it is insignificant :smile:
 
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gneill said:
The "...regardless of its mass" statement needs to be taken in the context of the basic assumption of Kepler's Laws where the masses of all the planets are considered to be much less than that of the primary (the Sun).

So I guess one might say that the mass of the object doesn't matter as long as it is insignificant :smile:
Thank you
 

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