Two satellites in orbit around Jupiter

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SUMMARY

The discussion centers on calculating the orbital period of a satellite around Jupiter using Kepler's laws. The first satellite has an orbital period of 16 hours and an orbital radius of r, while the second satellite has an orbital radius of 4.0r. By applying the relationship T^2 ∝ R^3, the correct orbital period for the second satellite is determined to be approximately 130 hours, aligning with answer choice A. The calculations involve using gravitational constants and algebraic manipulation to derive the periods.

PREREQUISITES
  • Understanding of Kepler's laws of planetary motion
  • Familiarity with gravitational constants (G = 6.673 x 10^-11)
  • Knowledge of the formula for orbital period (T = (2πr) / v)
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study Kepler's laws in detail, focusing on the relationship between orbital period and radius
  • Learn about gravitational forces and their application in celestial mechanics
  • Practice solving orbital period problems using different celestial bodies
  • Explore advanced topics in orbital dynamics, including perturbations and elliptical orbits
USEFUL FOR

Astronomy students, physics enthusiasts, and anyone interested in celestial mechanics and orbital calculations will benefit from this discussion.

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Homework Statement



Two satellites are in circular orbits around Jupiter. One, with orbital radius r, makes one revolution every 16 h. The other satellite has orbital radius 4.0r. How long does the second satellite take to make one revolution around Jupiter?

A. 130 hours
B. 120 h
C. 140 h
D. 110 h
E. 90 h

Homework Equations



v = sqrt( GMj / r )

T = (2*pi*r) / v

G = 6.673*10^-11

Mj = 1.90*10^27

The Attempt at a Solution



I'm assuming that this is just an algebra problem at this point, but I'm having trouble finding r.

T = ( 2*pi*r) / sqrt( GMj / r )

r = [(GMjT^2) / (4pi^2)]^(1/3)

When I solve for r and then plug r back into the equation for T I do not get 16, which means I'm not doing something right here.

Do I need to convert the 16 hours / revolution into another unit before solving for r?

Any pointers?
 
Last edited:
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According to Kepler's law the square of the period varies as the cube of the orbit radius.

T^2 \propto R^3

This implies that for any given set of orbits around a common central body, that

\frac{T^2}{R^3} = Constant

You should be able to use this fact to determine the "mystery" orbital period.
 
Thanks, that did help a lot.

It seems like the method I was trying to work through would also work but is much longer.

It also doesn't help that when I do the calculation with either method I still do not get a value the exactly matches one of the answer choices, but is very close to answer A ( which did turn out to be correct ).
 

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