ASTR 3220 - Homework 3 Solutions
Due Thursday Feb. 5
Total of 10 points.
Show your work if a calculation is required. Give units for your answer
if applicable. Do your own work.
1. Magnitudes (... more practice) (2 points) -
Consider the following 5 stars:
(A) m = +1 M = +2
(B) m = +8 M = +6
(C) m = +4 M = -1
(D) m = +10 M = +6
(E) m = -1 M = -1
(a) Which star(s) is/are intrinsically brighter than the sun?
Ans: A, C, E since M less than M(sun) = +4.8
(b) Which star is the most distant?
Ans: C , since m - M = 5 is the largest (m-M = 5log d - 5).
(c)* What is the distance (pc) to star (C)?
Ans: 100 pc , since m - M = 5 = 5log d - 5, so log d = 2, so d = 10^2 = 100 pc.
(d)* What is the ratio of the APPARENT brightness of star(E)
to star (A)?
Ans: The ratio is (2.512)^2 = 6.3, since their apparent mags. differ by +1 - (-1) = 2 and
star (E) has a lower magnitude, so it is brighter.
(e) Which star(s) is/are visible to the naked eye?
Ans: A, C, E since they all have m < 6 (m = 6 is the naked eye limit).
2.* Kepler's Laws (3 points) -
(a) Two neutron stars each of mass 2 solar masses are
orbiting each other.
The orbital semi-major axis is 2 AU. What is the orbital period (years)?
Ans: By Kepler's 3rd law, P = sqrt(2) = 1.414 years. Kepler's 3rd law gives
m1 + m2 = a^3/P^2, so 2 + 2 = (2)^3/P^2, so P^2 = 8/4 = 2, so P = sqrt(2) = 1.414 yr
(b) If two stars are initially orbiting each other with a period P1 and the orbital
semi-major axis is then doubled, what is the ratio of the new period P2 to
the original period P1 (that is, P2/P1 = ?).
Ans: By Kepler's 3rd law, P2/P1 = sqrt(8) = 2.83. By Kepler's 3rd law, (P2)^2/(P1)^2 = (a2)^3/(a1)^3,
since the mass term m1 + m2 cancels out.
But a2 = 2 * a1 (given), so (a2)^3 = (2 * a1)^3 = 8 * (a1^3), so (a2)^3/(a1)^3 = 8 .
Thus (P2)^2/(P1)^2 = 8, so P2/P1 = sqrt(8) = 2.83.
(c) If a low mass star of mass 0.1 solar masses is orbiting a high mass star of
10 solar masses in a nearly circular orbit with eccentricity e = 0.1, what
is the ratio of the orbital speed of the low mass star at periastron to
its orbital speed as apastron? (Note: At periastron the low mass star is
closest to the primary, and at apastron it is farthest.)
Ans: Angular momentum J is constant throughout the orbit, which is just an equivalent
statement of Kepler's 2nd law.
Since e = 0.1, the orbit is nearly circular (e = 0 would be a circle),
so to a very good approximation,
J = m*v*r, where m is the mass of the low mass star, v is the orbital speed
of the low mass star when it is a distance r from the high mass star (where the
high mass star is almost exactly at the focus). If r(min) is the distance from
the high-mass star to low-mass star at periastron, and r(max) is the distance
at apastron, then equating J at these two points gives:
m* v * r(min) = m * V * r(max), where v is the speed at periastron and V
is the speed at apastron. Since m appears on both sides of the equation, it cancels
out and we have v/V = r(max)/r(min) . But as given in class, for an ellipse one
has r(max) = a(1 + e) and r(min) = a(1 - e), so v/V = a(1 +e) / a(1 - e), so
v/V = (1 + e)/ (1 - e) = (1 + 0.1)/ (1 - 0.1) = 1.1/0.9 = 1.22. Thus, the low
mass star is moving 1.22 times faster at periastron.
3. Kirchhoff's Laws (2 points) -
For the situations described in (a), (b), and (c) below,
would you see continuous emission, absorption lines, or emission
lines? Note: There may be more than one possibility; list all possibilities.
(a) the hot (approx. 1 million K) low-density gas of a supernova remnant (assume no
intervening material between you and the remnant).
Ans: emission lines (hot low-density gas). Very weak continuous emission might also be
present, depending on the density of the gas.
(b) a compact, high-density white dwarf of temperature 50,000 K (assume no
intervening material between you and the white dwarf).
Ans: continuum (solid material)
(c) a hot star (30,000 K) viewed through an intervening cloud of cold
hydrogen gas (50 K)
Ans: continuum plus absorption lines.
(d) Is it possible to see absorption lines without also seeing
continuous emission?
No. In order to have an absorption line,
light at a particular wavelength has to be removed from light that is already present.
If there is no light present to begin with, then you cant have
an absorption line. In order to see an absorption line, you must have continuous
emission on either side of the line to recognize that absorption has occurred.
4. Space Velocity (1 point) -
Can the radial velocity of a star ever be larger than its space velocity?
(Explain why, or why not, in two or three short sentences.)
Ans: This is not possible. By the Pythagorean theorem,
(Vspace)^2 = (Vradial)^2 + (Vtransverse)^2. All terms in this
equation are non-negative, since they are squares. The above can
be rewritten as (Vspace)^2 - (Vradial)^2 = (Vtransverse)^2.
If Vradial were larger than Vspace, then the left side of this
equation would be negative, which is impossible, since it must
equal the right side (Vtransverse)^2 , which is non-negative.
5. Light and Radiation (2 points) -
(a) Why does the apparent brightness (or energy flux) of light from a star
decrease as the inverse square of the distance? (Explain in
two or three short sentences, adding a figure if you wish.)
Ans: Because the light energy gets spread out over a larger and
larger surface area as the distance increases. It's easiest to visualize
the light getting spread out over the surface of a sphere with the star at
the center of the sphere. In that case, the area over which the light gets
spread out at distance d is just the area of a sphere of radius d, which is
4*pi * (d^2).
(b) Although electromagnetic radiation at very short wavelengths can
exist (for example, gamma rays of wavelength 10^(-14) meters),
the wavelength can never go to zero. Explain why, in two or
three short sentences.
Ans: This is impossible since it would require infinite energy.
The energy of a photon is E = h*f = h*c/(wavelength), and if
wavelength = 0 then E is infinite. Infinite energies are
phsically unrealistic. Also, if the wavelength were 0, you really
would not have a wave any more.
* Problems marked with an asterisk require a numerical calculation.
Show all steps in your calculation.