Pulsar Radius from its rotational period

AI Thread Summary
To determine the upper limit of a pulsar's radius based on its rotational period, the discussion revolves around the balance of gravitational and centrifugal forces. The gravitational force can be expressed in terms of the star's density and radius, while the centrifugal force relates to the pulsar's rotation rate. By equating these forces, the relationship between density, radius, and period is established. The final equation derived shows that the density multiplied by a constant equals a function of the period, providing a method to calculate the maximum radius. This approach effectively links the pulsar's physical properties to its rotational characteristics.
TFM
Messages
1,016
Reaction score
0

Homework Statement



A pulsar emits bursts of radio waves with a period of 10 ms. Find an upper limit to the radius of the pulsar.

Homework Equations



Not Sure

The Attempt at a Solution



Can anyone help with this, I cannot see how the period will help tell you the upper limit to the radius. I know that pulsars are basically neutron stars, and they have high densities (10^15 kg/m^3), but I ams lightly unsure how to get the radius of the pulsar from its period.

Any helpful suggestions would be most helpful,

Thanks in advanced,

TFM
 
Physics news on Phys.org
What happens when the rotation rate is fast enough that centrifugal force on a point on the surface as is stronger than gravity?
 
Well, Gravity is pulling down, the centrifugal force is pushing outwards, so items on the surface would be "pushed" off of the surface.
 
Correct - so at a certain speed the surface would break off, doesn't this set a maximum radius for a given rotation rate?
 
Indeed it would,

So:

mg = m\omega^2r

g = \omega^2r


And since:

Omega = \frac{2\pi}{Period}

Thus:

g = \frac{4\pi^2}{Period^2}r

Does this look okay?
 
You will need to write 'g' for the star in terms of it's mass (or density) and radius.
 
True, but we aren't given a mass for the star? Would we use the density as being 10^15?
 
Okay, so if we use:

g = -\frac{MG}{r^2}

and

M = density*volume

M = density*(\frac{4}{3}\pi r^3)

g = -\frac{(density*(\frac{4}{3}\pi r^3))G}{r^2}


Thus:

-\frac{(density*(\frac{4}{3}\pi r^3))G}{r^2} = \frac{4\pi^2}{Period^2}r

Since we need the magnitude only for g:

\frac{(density*(\frac{4}{3}\pi r^3))G}{r^2} = \frac{4\pi^2}{Period^2}r

(density*(\frac{4}{3}\pi ))G = \frac{4\pi^2}{Period^2}

Does this look better?

TFM
 

Similar threads

Back
Top