Kepler's 3rd law - which object in numerator?

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SUMMARY

This discussion focuses on the application of Kepler's 3rd Law to determine the orbital period of celestial bodies. The formula discussed is (TA/TB)2 = (RA/RB)3, where T represents the orbital period and R the radius. The confusion arises regarding which object should be designated as "A" and which as "B." It is established that the less massive object typically corresponds to "A," and the formula can be applied regardless of the order of the objects, as long as the correct values are used. The example of Ganymede and Io illustrates this application.

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  • Familiarity with orbital mechanics and gravitational interactions
  • Basic knowledge of mathematical operations involving exponents and square roots
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  • Study the derivation and implications of Kepler's 3rd Law in greater detail
  • Explore the gravitational equations, specifically T=2π√(a3/G(m1+m2))
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Astronomy students, astrophysicists, and anyone interested in celestial mechanics will benefit from this discussion, particularly those looking to apply Kepler's Laws to real-world scenarios involving orbital calculations.

aphysicsmanduh
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Hi folks! I have been searching and searching and cannot come up with a clear answer. When applying Kepler's 3rd law to 2 objects to solve for an unknown period or radius, which orbiting object (satellite) is in the numerator and which is in the denominator?

Here is the formula I am referring to:

( TA / TB )^2 = (RA / RB)^3

so if you're solving for orbiting period you would have TA = √(RA/RB)^3 * TB^2

I get different answers if I switch them so there must be some sort of "rule" that tells me which object is "A" and which is "B". From my searching, it looks like period and radius "A" are for the less massive object. I also wasn't sure if it had something to do with which object was closest to original object being orbited. An example is the moons of Jupiter. If I was trying to find the period of Ganymede given its radius and the info. for Io, which object would be the numerator and which would be the denominator in Kepler's 3rd law equation.

Any help would be great so I can sleep tonight! :)
 
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T=2\pi\sqrt{\frac{a^3}{G(m_1+m_2)}}

"a" is the sum of the semi major axis of the two bodies.
 
It doesn't matter which is which.

The erroneous results might be due to:

aphysicsmanduh said:
TA = √(RA/RB)^3 * TB^2

Perchance you've lost a square root there?

TA = √(RA/RB)^3 * √TB^2
 

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