Help with Calculating Planet's Orbital Period

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Homework Help Overview

The discussion revolves around calculating the orbital period of a newly discovered planet that is eight times further from the sun than Earth, using principles from celestial mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Kepler's laws, particularly the relationship between the orbital period and the distance from the sun. There is an exploration of deriving the orbital velocity and period using gravitational principles.

Discussion Status

Some participants have provided hints and guidance on using Kepler's third law, while others have attempted calculations based on the provided information. There appears to be a confirmation of correctness regarding the calculations presented, but no explicit consensus on the final interpretation of results.

Contextual Notes

Assumptions include circular orbits and the application of Kepler's laws, with a note that actual orbits are elliptical but can be approximated as circular for this problem.

tinksy
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hi, could someone please help me with this problem??

If a small planet were discovered with a distance from the sun eight times that of the Earth, what would you predict for its period in (Earth) years. (i.e. how many times longer would it take to go round the sun than the Earth does.)
 
Last edited:
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i will not give you the answer straight away, but here's the hint---
get orbital velocity v as a function of radius r

dynamics of the system(*) :

from Newton's second law and law of gravitation,
(m*v^2)/R = (G*M*m)/R^2


get v(r) and substitute in the kinemetical relation which you got correct,that is, T=2(pi)r/v

this will give you T(r) which is usually called "kepler's third law".


*assumption: all orbits are circular

justification : though the orbits that actually elliptical ,the eccentricity is very small.(dont worry if you don't understand this,take this as a side remark).

cheers :smile:
 
Last edited:
thanks teddy...

could u tell me if I'm right?

i've used kepler's law T^2 (directly proportional to) R^3
so if R = 8R (as for Earth, R = 1AU so for the planet, R = 8AU)
kepler's law: R^3/T^2 = 1
therefore, T^2 = 8^3/1
so T = (root)512 = 16(root)2 ...? is that correct?
 
yup,its right.
bye :smile:
 
yaaaaaay! thanks
 

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