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Calculate the Period of a Planet.

  • Thread starter destinee
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  • #1
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Consider a planet with a period of orbit to be 0.241 years and an average radius of 0.39 astronomical units. Calculate the period of a planet with an average radius of 1.4 astronomical units.



If someone can answer and explain this question to me, I would be very thankful. I know it has something to do with Kepler's 3rd law (T²/R³), but I do not know how to apply it.The teacher counted my answer wrong. See my work below:



P1 = 0.241 years with a radius of 0.39 AU
P2 = ? years with a radius of 1.4 AU
So set 0.241 over 0.39 equal to X (? years) over 1.4

You get
0.3374=0.39x than divide
x = 0.865 year orbit
 

Answers and Replies

  • #2
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0.865 years ISN'T correct, You have forgotten to square each part, use this:

Using Kepplers 3rd Law:

P12/P22 = R12/R22

However, this is a poor question, it assumes these planets will have the same density/mass which is pretty much never the case. may want to mention this to your teacher!
 
  • #3
lewando
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Kepler's 3rd law is a bit more than "T2/R3". For a planetary system with a big sun and relatively tiny planets, it is more like "T2/R3=k", where k is approximately the same for all tiny planets.

You used a simple proportion--that's what got you in trouble. Find k and try again using the big sun equation.
 
  • #4
6,054
390
"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."

So for one planet, [tex]T_1^2 \propto a_1^3[/tex] For another, [tex] T_2^2 \propto a_2^3[/tex] Thus [tex]\frac {T_1^2} {a_1^3} = \frac {T_2^2} {a_2^3}[/tex] Can you see how you could use this?

However, this is a poor question, it assumes these planets will have the same density/mass which is pretty much never the case.
No, the question is OK. The law works as long as the planet's mass is much less than the mass of the star, which is pretty much always the case.
 
  • #5
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No, the question is OK. The law works as long as the planet's mass is much less than the mass of the star, which is pretty much always the case.
Really? Surely if one planet is extremely light and the other is heavy by comparison but still less than the sun then the force on the heavier planet and therefore it's orbital velocity would be much larger?

I'm not disagreeing, it's just this was the impression I was always under: that if there is a large difference in the relative mass of the two planets then the law can become skewed? Or is this just one of those assumptions we make because it pretty much never occurs? :S
 
  • #6
6,054
390
Really? Surely if one planet is extremely light and the other is heavy by comparison but still less than the sun then the force on the heavier planet and therefore it's orbital velocity would be much larger?
The coefficient of proportionality has, as a factor, [itex]\frac 1 {M + m}[/itex], where M is the mass of the star, and m of the planet. Typically [itex]M >> m[/itex], so that factor is almost exactly [itex]\frac 1 M[/itex] no matter what planet you choose. Take our Sun, Mercury and Jupiter, and see how this factor differs.

I'm not disagreeing, it's just this was the impression I was always under: that if there is a large difference in the relative mass of the two planets then the law can become skewed? Or is this just one of those assumptions we make because it pretty much never occurs? :S
As you can see from the above, of importance is the ratio planet to star, not planet to planet.
 

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