Download Astronomy 15 - Problem Set Number 7

Astronomy 15 - Problem Set Number 7 - Due Friday May 24
(1) The Schwarzschild radius of a non-rotating black hole of mass M is
Rs =
2GM
,
c2
which is numerically equal to
Rs = (2.95 km) × mass in solar masses
Thus the sun’s Schwarzschild radius would be about 3 km, while a 10-solar mass
object has about a 30 km Schwarzschild radius. If you compress an object of mass M
to a radius smaller than this, it will be a black hole; nothing can escape from in (in
the classical, non-quantum theory). Let’s see what this expression implies about the
densities of black holes.
(a) Suppose the volume of a black hole is given by the simple expression
V =
4 3
πR ,
3 s
which it won’t be, since the space is so distorted; however, we can use this expression
for rough purposes. Find an expression for the average density of a mass M which is
compressed to just its Schwarzschild radius; now evaluate it numerically to show that
the average density
1.8 × 1016 gm cm−3
ρ̄ =
mass in solar masses
If a mass is compressed beyond this density, it will surely be unable to resist further
collapse.
(b) Let’s suppose for a moment that no object can be compressed beyond neutron star
density, no matter what pressure you subject it to. This is completely unphysical,
but helps illustrate the point, which is that even if this were the case, black holes
would be inevitable if Einstein’s theory of gravity is correct. Let’s take the density of
a neutron star to be 1014 gm cm−3 and compute how large a mass of this (supposed
incompressible) you would have to assemble to make a black hole. How does your
number compare to the actual upper mass limit for neutron stars (about 2.5 M )?
Why any discrepancy?
(c) Suppose you were given an indefinitely large supply of magic, incompressible water
(density = 1 gm cm−3 ). Show that you could make a black hole, given enough of it!
How large a mass would be necessary? What would the radius of this mass be? How
does this compare to the earth’s orbit? (To make very large-mass black holes, then,
you don’t even need much compression!)
(2) This problem is designed to give you some idea of how information from binary star
orbits is interpreted. Binaries, aside from being interesting in their own right, provide
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the most direct way of measuring stellar masses (and a very good way of measuring
radii as well). This problem leads you through the puzzle, which is a classic of scientific
inference. Remember, we’re dealing with something that looks like a dot in the sky!
Brightness
Time (d) (as above).
(a) Refer to the accompanying diagram, which shows the light curve and radial velocity
curve of an eclipsing, double-lined (e. g., both stars are visible in the spectrum)
spectroscopic binary. Figure out from the velocity curves when each star is in front
and in back; then answer: Which star is brighter? Which star has the larger radius?
(b) Let’s look at how we might determine masses for these stars. Suppose you have two
stars, which we’ll call 1 and 2, orbiting each other. For simplicity assume their orbits
are circular. The only ‘fixed’ point in the system is the center of mass. Let’s say that
star 1 revolves at a distance of r1 from the center of mass, while star 2 revolves at a
distance of r2 from the center of mass. Note that the center of mass lies between the
two stars; while star 1 is on one side, star 2 is on the other.
Suppose the two stars have masses m1 and m2 . They are separated by a center-tocenter distance a; clearly a = r1 + r2 . Because we’re taking the orbit to be circular,
a is a constant. Because we’re measuring r1 and r2 from the center of mass, we have
m1 r 1 = m 2 r 2 .
Now:
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(i) What is the centripital force needed to keep star 1 in its orbit, in terms of m1 ,
r1 , and its velocity in orbit v1 ?
(ii) What is the mutual gravitational attraction between the two stars?
Note that v1 = 2πr1 /P , where P is the period of the orbit (the time for one revolution);
show that
a
,
r1 =
1 + m1 /m2
and hence prove that
a
3
2π
P
2
= G(m1 + m2 ).
This is the modern form of Kepler’s third law.
(c) Define K1 and K2 to be the velocity amplitudes of the two stars, as for stars A and
B in the diagram (if you look at the diagram, the K for star A is 200 km/s, the K for
star B is 100 km/s). Suppose the orbit of the two stars is inclined to the line of sight;
that is, suppose we view the plane of the orbit from some inclination angle i, where i
is the angle between the plane of the orbit and the plane of the sky. Thus i is zero for
a face-on orbit and 90 degrees for an edge-on orbit. Note that if i < 90 degrees the
observed velocity amplitude K will be an underestimate of the true velocity in orbit;
hence show that
2πr1 sin i = K1 P.
From this and the corresponding expression for star 2, show that
a=
(K1 + K2 )
P.
2π sin i
Finally, show that Kepler’s third law can now be rewritten in terms of observed
quantities:
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K1 + K 2
P = 2πG(m1 + m2 ).
sin i
(d) Argue from the data that for the system in the diagram, the inclination can’t be a lot
less than 90 deg. Find mA + mB in grams, and convert to solar units.
(e) Notice the relative sizes of KA and KB . Which star is more massive? (Be a little
careful here). Find mA /mB given KA and KB . From the known values of mA + mB
and mA /mB , find mA and mB separately.
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