Calculating the Ratio of Planetary Orbital Periods

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 3K views
lizzyb
Messages
167
Reaction score
0

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. ?
 
Physics news on Phys.org
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