Period of Planets orbiting a Star

  • #1

Homework Statement


Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1 is longer than the orbital period of Planet 2. What could explain this?

A) Star 1 has less mass than Star2.
B) Star 1 has more mass than Star 2
C) Planet 1 has less mass than Planet 2
D )Planet 1 has more mass than Planet 2.
E) The masses of the planet are much less than the masses of the stars.

Homework Equations



F=(G m1 x m2 ) / (r2)
ac = mv2 / r
(2π x r ) / T = V

The Attempt at a Solution


I think it is C.
I used F=(G m1 x m2 ) / (r2) and set it equal to mv2 / r
I didn't see anything related to Period so I remembered that circumference divided by period equals V.
I solved for V in the first equation and got : v = √[(Gm)/r]. Mass is the only variable that could cause the speed and therefore the Period to change. So I thought that increasing the mass of Planet would increase the speed and make the Period longer.
I probably messed up .....
 

Answers and Replies

  • #2
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You sure the mass in your expression is the planet mass and not the star mass ? In short: sort out which is which in m1, m2 and m and which one divides out !
 
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  • #3
You sure the mass in your expression is the planet mass and not the star mass ? In short: sort out which is which in m1, m2 and m and which one divides out !
Ohhhhh, dumb me...
I assigned m1 as the Planet and m2 as the Star.
Since the m in mv2 / r is referring to the orbiting mass, the mass of the planet cancels out and leaves me with the star so it should be B right?
 
  • #4
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How do you deduce that from your equations ?
 
  • #5
I cancelled m1 out.
 
  • #6
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I mean how do you deduce that it's B and not A
 
  • #7
So if V increases that means the period would increase. Oh wait. T is in the denominator of 2πr / T.. so that means if I increase V that would mean that Period would go down. I assume that's my mistake. So it has to be that T increases when V decreases meaning that the mass of the star has to be less. So A.
 
  • #8
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Something like that. If you work out T2 you get something with .../(GM..)
 
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  • #9
Something like that. If you work out T2 you get something with .../(GM..)
How would your approach look like? I am wondering how I would arrive at T2
 
  • #10
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  1. F=(G m1 x m2 ) / r2
  2. not ac but Fc= m2v2 / r
  3. (2π x r ) / T = v
3: ##\ \ \displaystyle {1\over T^2} = \left ({v\over 2\pi r} \right )^2\ \ ##. Now equate 1 and 2:
$${v^2\over r} = {Gm_1 \over r^2} \quad \Rightarrow \quad 1/T^2 = \left ({v\over 2\pi r} \right )^2 = {1\over (2\pi)^2} {Gm_1 \over r^3} \quad \Rightarrow \\ T = 2\pi\sqrt {r^3\over Gm_1} $$

as in wikipedia (here you can ignore the planet mass m wrt the M of the star)
 
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  • #11
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why did this get so complicated... [V][/orb]=√GM/r
And since the problem indicates that r is the same for both systems, it isn't going to affect the speed of the orbital, G is also a constant so it isn't gonna affect the system, however, a change in M can affect the speed of the orbital, for example: If you increase M the V orbital is gonna speed up, and if you decrease M its gonna slow down. Knowing that, we can see that the planet is taking longer to orbit because its slower than the other planet, so V orbital is slower, and since M is the only changing variable, we can see that M for planet 1 is less than planet 2.
 
  • #12
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Hello MiaPow, :welcome:

Don't hijack another thread -- Start a new thread and state the problem that YOU have when working this out. Where do you get stuck ?

What kind of assistnce can be brought to bear on a question like 'Why did this get so complicated ... ?'
I don't think it's all that complicated at all !
 
  • #13
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Oh I apologize for my poor wording. I had no problem following through your solution, just an alternative method of answering.
 
  • #14
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No problem. The OP probably needed some specific guidance and things got intricate before they became (almost) straightforward. Not everyone 'sees through' an exercise straightaway :rolleyes: (not even helpers...)
And, eh: :welcome:
 

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