What is the equation for calculating orbital period?

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Homework Help Overview

The discussion revolves around calculating the orbital period of an asteroid around the sun, specifically in relation to Earth's orbital period. The original poster attempts to find the equation for the orbital period after determining the radius of the asteroid's orbit.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss Kepler's Third Law and its application to the problem. The original poster questions the relevance of the constant in the equation and explores the implications of the asteroid's orbital period being four times that of Earth.

Discussion Status

Some participants have provided guidance on using Kepler's Third Law, while others are exploring the implications of the ratio of orbital periods. There is an ongoing examination of the relationship between the asteroid's and Earth's orbits, but no consensus has been reached regarding the specifics of the constant or the units of measurement.

Contextual Notes

There is a mention of multiple choice answers, indicating that the original poster may be working within a structured assignment or assessment context. The discussion also highlights the importance of specifying units when calculating the orbital period.

ghostbuster25
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Ok just a quick question
trying to work out the orbital period of an asteroid around the sun which has an orbital period 4times that of the Earth which is 1.5*10^11m
i worked out the radius of the asteroid to be 6*10^11 but can't find the equation for the orbital period.
Does anyone know it?

thanks
 
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Use 3rd Kepler's Law:

<br /> \frac{R^{3}}{T^{2}} = \textup{const}<br />

for all bodies orbiting the Sun. You do not even need the value of the radius of the Earth's orbit, but just their ratio, which is already given.
 
haha thanks, your helping me a lot today!

so I am guessing as its 4 times that of Earth its 4AU?

what is the constant?
 
ghostbuster25 said:
haha thanks, your helping me a lot today!

so I am guessing as its 4 times that of Earth its 4AU?

what is the constant?

It's irrelevant. The ratio on the lhs has the same value for the Earth and for the asteroid. Equate the left hand sides for both the bodies and you will have an equation which you will need to solve for the orbital period of the asteroid.
 
ok I am getting...T=\sqrt{4^3} = 8...which is one of the answers on my multiple choice :)
 
this would be true if you specified the units in which the orbital period is measured.
 

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