Find the orbital periods of the two small satellites without using the mass of Pluto?

Question

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already was known to have a large satellite Charon, orbiting at 19,600 km with an orbital period of 6.39 days.

1) Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto.

Answer 1

You can do this easily enough IF you assume that the masses of all the satellites including Charon are negligible complared to Pluto's mass.

Just set the centripetal force equal to the gravitational force and solve for the period:

m omega^2 r = G M m / r^2

where M is Pluto's mass and m is the satellite mass

T = (2 pi / omega) = 2 pi sqrt (r^3 / GM)

We see that the period goes like r^(3/2)--a result that you also could have taken directly from Kepler's Laws.

So your answer for the period of the small sats is:

T2 = T1 * (r2/r1)^(3/2)

Use Charon's period and radius for T1 and r1.  And plug in each small sat's r2 to get it's period T2.  So far, so easy.

BUT!--- if Charon is of siginicant mass compared to pluto, you can't use it's mass in Newton's 2nd law and have it cancel--you are forced to use its reduced mass:

mred = Mm / (M+m)

So a = F / mred = G (M+m) / r^2 =  omega^2 r

T = 2 pi sqrt (r^3 / (M+m) )

Now this makes the problem impossible as far as I can tell because you don't know Charon's mass OR pluto's.  It turns out Charon has about 11-12% of pluto's mass.  So the error in your answer is going to be of that magnitude if you assume it is small (reduced mass = Charon's mass).

Bottom line: if you haven't learned yet about reduced mass (ie you are in high school or introductory physics), disregard everything I said after the simple solution and just chalk it up as an oversight of whomever wrote the problem.  If you are in an analytical mechanics class, tell your prof that you aren't given enough info in this problem to get a solution.

In case you are in high school and are curious what the fuss is about, the reason Kepler's Law doesn't work here is that to derive it you have to assume that the body being orbited is so massive that it doesn't move--and that the satellite does all the moving around the planet's center of mass.  In reality, both the satellite and the planet orbit around their common center of mass.  This fouls up the math somewhat.  You can read the wiki link and understand how.

http://en.wikipedia.org/wiki/Reduced_mas...

Answer 2

The formula for the time period of a satellite orbiting a planet is:-

        T   =   2 * pi * SQR( R^3/G*M)

where "T" is the time for one orbit

  "R" is the radius of the orbit

  "M" is the mass of the planet

  "G" is the gravitational constant.

  

If we square both sides, we get:-

       T^2    =    4 * pi^2 * R^3/G*M

  

"M" is the mass of p**o, which is constant here.

"G" and "pi" are also constants.

   Therefore     T^2  is proportional   to R^3 (one of Kepler's laws)

         or     T^2 / R^3  is a constant

So when we work this out for any satellite of Pluto, we get the same answer.

    Therefore for one satellite|:-

             6.39^2 / 19,600^3   =   Time^2 / 48000^3

   Time for one satellite  =  24.49 days at distance of 48000 Km

    And the other  6.39^2 / 19600^3   =   Time^2 / 64000^3

  Time for other satellite  =  37.70 days. at distance of 64000 Km

Since the above are ratios, you do not have to use MKS units, just as long as the units are the same on both sides of the equation.