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:
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:
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:
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 / R2, 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
-
Rblack 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 Rblack hole .
Given that a black hole has a mass that's 10^6 that of the sun its
radius of no escape is:
-
Rblack 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 Rblack hole
, the above equation reads:
change in frequency / rest frequency = Rblack hole /
(10 x Rblack hole) = 1/10
or:
-
(1.55 x 1018 Hz - fobserved-theory) / 1.55
x 1018Hz = 1/10
Solving for fobserved , (which by now I know you
can do) you get:
fobserved-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 Rblack hole.
In any case, to within the margins of error of the observation the
recieved frequency IS 1.41 x 1018 Hz.