Homework Help: Period of Planets orbiting a Star

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1. May 4, 2015

Soniteflash

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

3. 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 .....

2. May 4, 2015

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 !

3. May 4, 2015

Soniteflash

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. May 4, 2015

BvU

How do you deduce that from your equations ?

5. May 4, 2015

Soniteflash

I cancelled m1 out.

6. May 4, 2015

BvU

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

7. May 4, 2015

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.

8. May 4, 2015

BvU

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

9. May 4, 2015

Soniteflash

How would your approach look like? I am wondering how I would arrive at T2

10. May 5, 2015

BvU

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)

11. Jan 7, 2018

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 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. Jan 7, 2018

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 !

13. Jan 9, 2018

MiaPow

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

14. Jan 10, 2018

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: