Period of Revolutions of a Planet

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Homework Help Overview

The discussion revolves around a problem involving the period of revolution of a planet around a neutron star, specifically before and after the star collapses from a radius of 10,000 km to 3 km. The original poster expresses confusion regarding how to approach the problem, particularly in finding the necessary velocity to calculate angular velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevance of Kepler's Laws and the relationship between period and radius. There are questions about whether the problem pertains to the planet's new period or the star's period. Some participants suggest using conservation of angular momentum and inquire about the angular momentum of a uniform sphere.

Discussion Status

Participants are exploring various interpretations of the problem, with some suggesting that the collapse of the star does not affect the planet's orbit. There is a mix of confusion and attempts to clarify the concepts involved, particularly regarding angular momentum and rotational rates.

Contextual Notes

Some participants note that the original poster may be missing information or that certain aspects of the problem are not clearly stated. The discussion includes references to specific equations and principles relevant to rotational motion and angular momentum.

shanklove
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Homework Statement


A neutron star has a radius of 10,000 km, and takes a planet 30 days to complete one revolution around the star. When the star collapses, the new radius is 3 km. Find the new period of revolution of the newly formed neutron star.


Homework Equations


T=2(pi)/omega
omega = (v cross r)/r^2

The Attempt at a Solution


Not sure even how to approach this. I'm so confused and i can't find the velocity to find omega! this is my first post on PF, any help please?!
 
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shanklove said:

Homework Statement


A neutron star has a radius of 10,000 km, and takes a planet 30 days to complete one revolution around the star. When the star collapses, the new radius is 3 km. Find the new period of revolution of the newly formed neutron star.


Homework Equations


T=2(pi)/omega
omega = (v cross r)/r^2

The Attempt at a Solution


Not sure even how to approach this. I'm so confused and i can't find the velocity to find omega! this is my first post on PF, any help please?!

Are you doing Kepler's Laws? The relationship between period and radius, one squared and the other cubed. Or is your class studying some other subject? And are you asking about the new period of the planet or the neutron star... something seems missing or stated slightly skewed...
 
pgardn said:
Are you doing Kepler's Laws? The relationship between period and radius, one squared and the other cubed. Or is your class studying some other subject? And are you asking about the new period of the planet or the neutron star... something seems missing or stated slightly skewed...

We're doing oscillatory motion and rotational motion and angular momentum. Oh and I'm assuming that he was asking for the period of the planet, the period of the star wouldn't make sense.
 
The collapse has no effect on the orbit of the planet (that will probably be the next question)
For the star you just need conservation of angular momentum.
Whats the angular momentum of a uniform sphere?
 
Have you done the conservation of angular momentum?
 
mgb_phys said:
The collapse has no effect on the orbit of the planet (that will probably be the next question)
For the star you just need conservation of angular momentum.
Whats the angular momentum of a uniform sphere?

It's just 2pi/T. So would it be 2pi/2592000?
 
Angular momentum is = I * rotation rate
Where I depends on the shape but for a sphere it is proportional to r^2
(http://en.wikipedia.org/wiki/List_of_moments_of_inertia)

So if you make r half as much the moment of intertia becomes 4x smaller, since the angular momentum is the same then it must go 4x faster.

How much is r reduced in this case?
 
Not following...

From your link I= (2mr^2)/5 <----assuming you mean I= moment of inertia. But it would make more sense to me that I=r^2. So what do I do now with the rotational rate...? I'm so confused =X
 
Ok I think I got it!

I(initial)*Omega(initial)=I(final)*Omega(final)

So I(initial)*(2pi/T(initial)) = I(final)*(2pi/T(final))

If the distribution is symmetric then kMR(initial)^2*(I(initial))=kMR(final)^2*(I(final))

So the kMR would cancel out on each side, so it would have Ri^2*(2pi/Ti)=Rf^2*(2pi/Tf)

Divide 2pi/Ti on each side and then divide by Ri. Multiply Tf on the other side and you get:

Tf = (Rf/Ri)^2 * Ti

Plug in and solve.

I feel smart now =)
 

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