Calculating Orbital Radius of Ganymede using Orbital Period

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SUMMARY

The discussion focuses on calculating the orbital radius of Ganymede, one of Jupiter's moons, using its orbital period of 7.15 days. The participants apply Kepler's Third Law, represented by the equation (Ta/Tb)^2 = (Ra/Rb)^3, to derive the orbital radius. The user initially miscalculates the radius, obtaining 10.5342 units, while the textbook states the answer is 4 units. The conversation emphasizes the importance of performing algebraic manipulations before inputting values into a calculator to minimize errors.

PREREQUISITES
  • Understanding of Kepler's Third Law of planetary motion
  • Basic algebraic manipulation skills
  • Familiarity with orbital mechanics terminology
  • Experience with unit conversions in astronomy
NEXT STEPS
  • Study the derivation and applications of Kepler's Laws of planetary motion
  • Learn about the significance of orbital periods and radii in celestial mechanics
  • Practice solving problems involving orbital calculations using different celestial bodies
  • Explore common errors in algebraic calculations and strategies to avoid them
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Astronomy students, educators teaching celestial mechanics, and anyone interested in understanding the calculations behind orbital dynamics.

trekkie1701c
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Homework Statement


"If Ganymede, one of Jupiter's moons, has a pariod of 7.15 days, how many uits are there in it's orbital radius? Use the information given in Example Problem 1."

Example Problem:

"Galileo measured the orbital sizes of Jupiter's moons using the diameter of Jupiter as a unit of measure. He found that Io, the closest moon to Jupiter, had a period of 1.8 days and was 4.2 units from the center of Jupiter. Callisto, the fourth moon from Jupiter, had a period of 16.7 days. Using the same units that Galileo used, predict Callisto's distance from Jupiter."


Homework Equations



(Ta/Tb)^2=(Ra/Rb)^3

The Attempt at a Solution



My solution to the example problem:

Ta = 1.8
Tb = 16.7
Ra = 4.2

(1.8/16.7)^2 = (4.2/Rb)^3

.0116 = 74.088(Rb^3) (Dividing 1.8 by 16.2 and squaring, and cubing 4.2 and Rb)
.2264Rb=4.2 (Multiplying by Rb^3 and then taking the cubed root of everything)
Rb=18.5512 (Dividing by .2264)

I considered that since the book gave an answer of 19, that I was close enough, considering the book's want to round everything.

For the problem I'm having difficulty with, I did this:

Ta = 1.8
Tb = 7.15
Ra = 4.2

(1.8/7.15)^2 = (4.2/Rb)^3
.2517^2=4.2^3/Rb^3 (1.8 divided by 7.15)
.0634=4.2^3/Rb^3 (Squaring previous answer)
.0634(Rb^3)=4.2^3 (Multiplying by Rb^3)
.3987Rb=4.2 (Taking the cubed root of both sides)
Rb=10.5342 (Dividing by .3987)

The book, however, lists an answer of exactly 4. I can't figure out where I messed up.
 
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Well, if Tb > Ta then Rb > Ra. In other words, the text must have the wrong answer. (My result is the same as yours).

Next time, try to do all of the algebric manipulations before punching the numbers into the calculator. There are a couple of good reasons for doing this: you are less likely to make an error with the calculator, mistakes are a lot easier to track down and it makes it easier for the marker/fellow students to follow your steps (symbols are easier to follow than numbers).

Hope it helps.
 
Are you sure you are reading the problem correctly? A body with an orbital radius of 4 units would obviously have a shorter period than a body with a radius of 4.2 units.
 
I read it correctly. The problems I listed above are word for word what's in the textbook.

And, link2001. Thanks; I'll start doing that from now on.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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