Orbital problem giving answer much smaller than choices

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Homework Statement


see problem 9 attachment


Homework Equations


see problem 9 attempt attachment


The Attempt at a Solution


I come up with a solution that is much smaller than any of the choices. I do not know if I am to assume the orbital is circular. I do not know if there is another error besides that in my work.
 
Attachments
  • Printed table titled Situation 12.1 listing radius, surface gravitational acceleration, orbital radius and orbital period for Moons I, II and III, followed by question 9 with multiple-choice answers A to E in powers of ten metres.
    1031-10871-problem 9.jpg
    23.2 KB · Views: 479
  • Handwritten derivation: g = Gm/R^2, m = gR^2/G, then T = 2πr^(3/2)/√(Gm) substituted to T = 2πr^(3/2)/√(gR^2), rearranged to r = [T(gR^2)^(1/2)/2π]^(2/3).
    2899-33757-problem 9 attempt.jpg
    16.1 KB · Views: 521
Hello,

You need not assume that the orbit is circular. Use one of Kepler's Laws (which applies for both circular and elliptical orbits).
 
ok if I use kepler's third law would I start by using moon II orbital period and radius to solve for the mass of the planet?
 
You don't need to solve for the planet's mass. Remember that R^3 is proportional to T^2, and that must hold for all objects orbiting a body. Hence R^3/T^2 = constant.
 
ok so I can find that ratio from moon II data and then multiply that ratio by T^2 of moon III?