Calculating the Ratio of Planetary Orbital Periods

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SUMMARY

The discussion focuses on calculating the ratio of the orbital periods of two planets, A and B, where planet B has twice the mass of planet A and its semi-major axis is also twice that of planet A. Using Kepler's Third Law, which states that T^2/a^3 is constant for all planets orbiting the same star, the correct ratio of the orbital periods is derived. The final conclusion is that the ratio of the orbital period of planet B to that of planet A is 1, not 2, as the mass of the planets does not affect the calculation; only the mass of the Sun is relevant.

PREREQUISITES
  • Understanding of Kepler's Third Law of planetary motion
  • Familiarity with the formula T^2 = (4 π² / G M) r³
  • Basic knowledge of elliptical orbits
  • Concept of semi-major axis in orbital mechanics
NEXT STEPS
  • Study Kepler's Laws of planetary motion in detail
  • Learn about the implications of mass in gravitational systems
  • Explore the derivation and applications of T^2 = (4 π² / G M) r³
  • Investigate the characteristics of elliptical orbits and their parameters
USEFUL FOR

Students studying physics, particularly those focusing on celestial mechanics, astronomers, and educators teaching orbital dynamics.

lizzyb
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Homework Statement



Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical orbit of planet B is two times larger than the semi-major axis of the elliptical orbit of planet A.

What is the ratio of the orbital period of planet B to that of planet A?

Homework Equations



[tex]T^2 = (\frac{4 \pi^2}{G M}) r^3[/tex]


The Attempt at a Solution



[tex]M_B = 2 M_A[/tex]
[tex]a_B = 2 a_A[/tex]
[tex]\frac{T_B}{T_A} = \frac{\sqrt{\frac{4 \pi^2}{G 2 M_A} 8 a_A^3}}{\sqrt{\frac{4 \pi^2}{G M_A} a_A^3}} = \sqrt{\frac{8}{2}} = \sqrt{ 4 } = 2[/tex]
but that was wrong. ?
 
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lizzyb said:

Homework Equations



[tex]T^2 = (\frac{4 \pi^2}{G M}) r^3[/tex]


The Attempt at a Solution



[tex]M_B = 2 M_A[/tex]
[tex]a_B = 2 a_A[/tex]
[tex]\frac{T_B}{T_A} = \frac{\sqrt{\frac{4 \pi^2}{G 2 M_A} 8 a_A^3}}{\sqrt{\frac{4 \pi^2}{G M_A} a_A^3}} = \sqrt{\frac{8}{2}} = \sqrt{ 4 } = 2[/tex]
but that was wrong. ?
Planetary Mass is not a factor. Kepler's Third law states that [itex]T^2/a^3[/itex] is the same for all planets. The M in your equation is the mass of the sun.

AM
 
thank you.
 

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