Help with Calculating Planet's Orbital Period

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SUMMARY

The discussion focuses on calculating the orbital period of a hypothetical planet located at a distance eight times that of Earth from the Sun. Utilizing Kepler's Third Law, participants confirm that the relationship between the radius (R) and the period (T) is expressed as T² ∝ R³. Given R = 8 AU, the calculation yields T = 16√2 Earth years, confirming the correctness of the approach. The discussion emphasizes the application of Newton's laws and the assumption of circular orbits for simplification.

PREREQUISITES
  • Understanding of Newton's laws of motion and gravitation
  • Familiarity with Kepler's laws of planetary motion
  • Basic knowledge of circular motion dynamics
  • Ability to manipulate algebraic equations and square roots
NEXT STEPS
  • Study Kepler's Third Law in detail, focusing on its mathematical derivation
  • Explore the implications of elliptical orbits compared to circular orbits
  • Investigate the gravitational constant (G) and its role in orbital mechanics
  • Learn about the calculation of orbital velocities for various celestial bodies
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Astronomy enthusiasts, physics students, and anyone interested in celestial mechanics and orbital dynamics will benefit from this discussion.

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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