As we saw earlier, when we studied the HR diagram,

hot stars come in basically two flavors; one group astonishingly

bright, much more massive than the Sun, and the other

the stellar graveyard of objects very much like our Sun.

Cooling down like a dying ember and being very small.

How do we know they are small?

Remember that the luminosity of any star can be given

by its surface area, 4 pi r squared,

times sigma, a constant, times T to the 4th power.

We know that the luminosity is very small for these objects.

The temperature is very high. The only way we can

have a small luminosity is if the size of the star, its radius is very small indeed.

And in fact, the radii of most white

dwarfs are about equal to that of the Earth.

Imagine that!

A mass about equal to the Sun in a size no bigger than a fairly

small planet like the Earth. So back to GK Per.

We have an object with a star similar to the Sun.

A bit cooler being a K star, and a hot white component.

We know it can't be very luminous in the visible part of the spectrum.

Because we know its distance. And at that distance, a hot O or B star

would make GK Per hundreds of thousands of times brighter than we observe it.

So the second component must be a white dwarf.

Is there any other clue that this is the correct picture?

Yes!

Again, using Doppler measurements, we can see the composite spectral

lines moving back and forth over a period of about two days.

You can see that in the data shown here. This cosmic dance, as we shall see,

is indicative of a double star system whose components orbit each other.

With a period of about two days. But what about our 352

second period, what can that be and which star is responsible.

It can't be an orbital period because that's already been spoken

for in the two day clock observed in the spectral lines.

But now, let's look at the next obvious choice.

Rotation, of one of the stars.

Let's start with the K star. We know from theoretical considerations,

that these are objects about 0.7 times the radius of

our own Sun. This will make this star a ball equal

to a radius of about 5 times 10 to the 8th meters.

We can also use the binary periodicity observations.

To show that the mass of the k star is about

one half a solar mass, or about 1 times 10

to the 13th kilograms. Now if

this star is spinning around in 352 seconds, This

means that the speed at the equator must be 2 pi R,

which is the distance traveled, divided

by the amount of time it takes for

it to travel once around 2 pi R over T.

This means it's speed is about 9 times 10 to

the 6 meters per second.

Boy, that's pretty fast, what can possibly prevent

the stuff from flying apart at the equator.

We would need some sort of force to hold it together.

Is gravity upto the task?

Let's see.

You might remember that objects moving in a circle of radius r are accelerated

to the centre of that circle through their centripetal acceleration.