Period of Planets orbiting a Star

[Soniteflash]

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)
a_c = 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 ...

 

[BvU]

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 !

 

[Soniteflash]

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?

 

[BvU]

How do you deduce that from your equations ?

 

[Soniteflash]

I canceled m1 out.

 

[BvU]

I mean how do you deduce that it's B and not A

 

[Soniteflash]

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.

 

[BvU]

Something like that. If you work out T² you get something with .../(GM..)

 

[Soniteflash]

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

 

[BvU]

1. F=(G m₁ x m₂ ) / r²
2. not a_c but F_c= 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)

 

[MiaPow]

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 going to 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 going to speed up, and if you decrease M its going to 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.

 

[BvU]

Hello MiaPow,

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 !

 

[MiaPow]

Oh I apologize for my poor wording. I had no problem following through your solution, just an alternative method of answering.

 

[BvU]

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 (not even helpers...)
And, eh: