Astronomy 102 Problem Set #6

Solutions

Problem #1: The pulsar in the crab nebula spins around its axis 30 times in one second. Using the dynamical equation of gravity, (Minimum Period² = 6 x 10¹¹ kg sec^2 / m^3 x Radius³ / Mass of star) show:
a. A pulsar cannot be a spinning white dwarf. A white dwarf has 0.8 x mass of the sun and the size of earth.
b. A pulsar can be a spinning neutron star. A neutron star has 1.4 x mass of the sun and the size of a mid-size town (less than 20 km radius).

Solution:
a. The MINIMUM POSSIBLE PERIOD for any object is given by the above dynamic equation of gravity. For example, if you substitute earth's mass and radius you would get:

P²= 6 x 10¹¹ kg sec²/m³ x (6.4 x 10⁶ meters)³ / 5.9 x 10²⁴ kg = 2.6 x 10⁷ s².

The period is the square root of this number:

P=5163 s, or ~90 minutes

That's the period of an artificial satellite in low earth orbit (just over the ~500km atmosphere). Earth, however, spins much slower. It completes one rotation around its axis in 24 hours. So the MINIMUM POSSIBLE PERIOD is the period in which if earth was spinning it would fly apart!

For white dwarf stars this number is:

P²= 6 x 10¹¹ kg sec²/m³ x (6.4 x 10⁶ meters)³ / x 1.6 x10³⁰ kg = 98 s².

The period is therefore only:

P= 9.9 seconds.

Now that's a fast period, but its also the fastest possible period! The white dwarf can spin faster, but not slower.
Lets compare this with the period of the pulsar:
Since the pulsar spins 30 times /seconds (that's the frequency here!) its period is:

P = 1/ 30 times/sec = 0.033 seconds.

Such a short period is not possible for the pulsar, because then the white dwarf flies apart. Since the pulsar is spinning faster than the faster possible speed for a white dwarf -- there is no way this pulsar is a white dwarf!

b. Same as in a., only for a neutron star the radius is much smaller now, which makes for a big difference in the period.

P²= 6 x 10¹¹ kg sec²/m³ x (2 x 10⁴ meters)³ / 2.8 x10³⁰ kg = 1.7 x 10^-6 s².

The period is again the square root of this number:

P = 0.0013 seconds

Now that's more like it! This number is a little smaller than the actual period. Remember the earth example? The period of the neutron star can be larger than this number. The pulsar in the crab number can therefore be a neutron star.

Problem #2: A black hole has a total mass that is 10⁶ times that of the sun.

a. What is the radius of that black hole (usually referred to as the black hole event horizon)? Size = 12 x 10^-11 m^3 /(kg x s^2) x Mass of black hole / c^2. c is the speed of light.
b. Assume that photons with frequency 1.55 x 10¹⁸ Hz are emitted from a distance that is 10 time the radius of the black hole. Using the gravitational Doppler effect show that the recieved frequency is 1.41 x 10¹⁸ Hz. (change in frequency / rest frequency = size of black hole / distance from black hole.

Solution:
a. The size of the black hole is obtained in a very interesting way: (through sheer luck here classical - non relativistic - physics works here, although its not quite clear why. The relativistic explanation yields the same result).
The energy of a particle trying to leave a gravitating body (earth, sun, black hole) is 1/2 m v2, where m is the mass of the particle and v is its velocity. The energy needed to leave that that body is 6.68 x 10^-11 M m / R^2, where M is the mass of that body and R its radius. By equating those two equations we get

R = 12 x 10^-11 m^3 /(kg x s^2) x Mass / v^2,

and by setting v=c the speed of light we get

R_{black hole}= 12 x 10^-11 m^3 /(kg x s^2) x Mass of black hole / c^2.

The meaning of all this is that the speed needed to leave the black hole is the speed of light. The escape velocity is the speed of light, and since in the special theory of relativity (the same one that gave us E=mc^2) the speed of light is the absolute limit, no one can escape from a distance SMALLER than R_{black hole .}
Given that a black hole has a mass that's 10^6 that of the sun its radius of no escape is:

R_{black hole}= 12 x 10^-11 m^3 /(kg x s^2) x 10⁶ x 2 x 10³⁰ kg / (3 x 10⁸ m/s) ²= 2.7 x 10⁹ meters

Now, this is just about 4 times the radius of the sun for a black hole a million times more massive than the sun!

b. The way in which the gravitational Doppler effect formula is derived is very similar to what we have done in class: By use of the equivalence principle. It is possible to find out that

change in frequency / rest frequency = size of black hole / distance from black hole.

(The idea here is similar to that in a: Using conservation of energy one finds that the energy of a particle near a black hole is E'+6.68 x 10^(-11) M/R^2, and away from the black hole it is E. If you than say that the energy of a photon is hf, as we did in the earlier stages of the course, you get the equation below.)
Anyways, since the frequency is emitted at 10 x R_{black hole}, the above equation reads:

change in frequency / rest frequency = R_{black hole} / (10 x R_{black hole}) = 1/10

or:

(1.55 x 10¹⁸ Hz - f_{observed-theory}) / 1.55 x 10¹⁸Hz = 1/10

Solving for f_observed , (which by now I know you can do) you get:
f_{observed-theory}= 1.395 x 10¹⁸ Hz.
In principle you could also get 1.705 x 10¹⁸ Hz, but that would be very different from the observed frequency. Even so, the observed frequency is somewhat lower than the theoretical predicted value. The reasons for the slight difference are:
1. The accuracy of our measurement of the redshift is still very low. Probably ~10% at best.
2. At this point in time there is no strong THEORETICAL argument that the line source should be at 10 x R_{black hole.}

In any case, to within the margins of error of the observation the recieved frequency IS 1.41 x 10¹⁸ Hz.