Download HW8 - UCSB Physics

21.34) Recall that the equation for time dilation is given by
∆t
∆t0 = q
1−
Setting ∆t0 = 10 years and ∆t = 8 years we have
r
1−
(1)
v2
c2
v2
8
=
2
c
10
(2)
Squaring both sides gives
1−
v2
= .64
c2
v2
= .36
c2
Taking the square root of both sides shows that v = .6c
21.36)Let us start with the Lorentz length contraction formula which is given as
r
v2
L = L0 1 − 2
c
Setting L = .6m and L0 = 1m, we have
r
v2
1 − 2 = .6
c
Solving this in the same manor as above gives v = .8c
10.47)To calculate the Schwarzschild radius of each given black hole we begin with the equation
rs =
2GM
c2
(3)
(4)
(5)
(6)
(7)
Using this let us first calculate the three requested figures. The given mass of the earth, the sun, and the
black hole of NGC 4261 are 5.97 ×1024 kg, 1.99 ×1030 kg (1 M ), and 1.2 ×109 M , respectively. Direct
substitution into the equation above gives the values (a),(b),(c) for the earth, the sun, and the black hole to
be
(a)The Schwarzschild radius is given by
rs = 8.87 × 10−3 m
Now, to calculate the corresponding density we imagine a spherical black hole with volume V =
Then the density is just ρ = M
V . Direct substitution into this equation yields the value
ρ = 2.04 × 1030 kg/m3
(8)
4
3
3 πrs .
(9)
(b)repeating for the sun gives a Schwarzschild radius
rs = 2970m
(10)
ρ = 1.82 × 1019 kg/m3
(11)
with a corresponding density of
(c)Lastly, for the super-galaxy we find a Schwarzschild radius
rs = 3.56 × 109 km
1
(12)
with a corresponding density of
ρ = 12.6kg/m3
(13)
21.50) Let us begin with the proof. In the previous solution it was stated that the volume of a black
hole is given by V = 43 πrs3 . To find out the mass dependency of this equation we need to recall that the
Schwarzschild radius is defined as rs = 2GM
c2 . Substituting into our equation for V we find
32 πG3 3
4 2GM 3
M
π( 2 ) =
3
c
3 c6
Then, recalling that density is given by ρ = M/V we have
V =
ρ=
M
=
V
M
32
3
πG3
c6
M3
=
1 3c6
M 2 32πG3
(14)
(15)
Which is what we set out to show. Now to answer the second part of the problem we must set ρ = 1000kg/m3
and solve for the given mass.
1000
kg
1 3c6
=
m3
M 2 32πG3
(16)
Thus, we have
r
M=
1
3c6
kg = 2.7 × 1038 kg
1000 32πG3
2
(17)
ASTRO 1 – HW8 Solution
1
ASTRO 1 HW8 Solution
22.30
(a) Volume of a cylinder = πr2 h, where r is the radius, h is the height of cylinder. Given the Galaxy disk
has diameter of 50 kpc, and thickness of 600 pc, we have r = 50/2 = 25 kpc, and h = 600 pc.
V = πr2 h
⇒
Vdisk = π(25 × 103 )(600) = 1.18 × 1012 pc3
(b) Volume of a sphere = 4πr3 /3, where r is the radius of the sphere. With r = 300 pc,
V =
4 3
πr
3
⇒
Vsphere =
4
π(300)3 = 1.13 × 108 pc3
3
(c) The region 300 pc around the Sun is the spherical region we considered in part (b).
Probability of the supernovae to occur 300 pc around the Sun =
Vsphere
1.13 × 108
= 9.58 × 10−5
=
Vdisk
1.18 × 1012
If there are about 3 supernovae per century in our Galaxy, there is 1 supernova per 33.3 years on
average (100/3 = 33.3 years). For 1 supernova to occur in the sphere of 300 pc in radius, there will be
1/9.58 × 10−5 = 1.04 × 104 supernovae in the whole Galaxy.
Time interval for 1 supernova to occur in the sphere = (1.04 × 104 )(33.3) = 3.46 × 105 years
22.31
(a) For the RR Lyrae star, we are given L = 100L and b = 1.47 × 10−18 b . Let d be the Sun-Earth
distance, i.e. d = 1 AU and use b = L/(4πd2 ),
s
s
2
b
L
1
L d
b
=
⇒ d = d
= (1 AU) (100)
= 8.25 × 109 AU
b
L
d
L
b
1.47 × 10−18
(b) To also express the distance in parsec, note that 1 pc = 3.26 ly, and 1 ly = 63240 AU,
1 ly
1 pc
9
9
d = 8.25 × 10 AU = (8.25 × 10 AU)
= 40000 pc
63240 AU
3.26 ly
22.37
The orbital velocity is v = 400 km s−1 = 400 × 103 m s−1 .
The radius of the orbit is r = 20000 pc = 20000(3.09 × 1013 )(103 ) m = 6.18 × 1020 m.
(a) To find the orbital period P ,
P =
2πr
v
⇒
P =
2π(6.18 × 1020 )
= 9.71 × 1015 s = 3.08 × 108 years
400 × 103
(b) To find the mass of the galaxy, we use the equation in Box 22-2 in Universe,
M=
rv 2
G
⇒
M=
(6.18 × 1020 )(400 × 103 )2
= 1.48 × 1042 kg = 7.45 × 1011 M
6.67 × 10−11
ASTRO 1 – HW8 Solution
2
22.45
(a) Sagittarius A* is a supermassive black hole with M• = 4.1 × 106 M , and therefore comparatively, the
mass of the stars S0-2 and S0-19 is negligible.
M1 + M2 =
a3
P2
⇒
M• =
a3
P2
⇒
a = (M• P 2 )1/3
For S0-2, with P = 14.5 years, a = ((4.1 × 106 )(14.52 ))1/3 = 952 AU .
For S0-19, with P = 37.3 years, a = ((4.1 × 106 )(37.32 ))1/3 = 1790 AU .
(b) To calculate the angular size of the semi-major axis as seen from Earth, use the small-angle formula,
D=
αd
206265
⇒
α=
206265D
d
which we will set D = a = semi-major axis. Since Sagittarius A* is at the Galactic center, and the
distance between the Milky Way and the Galactic center, i.e. Sagittarius A*, is around 8 kpc, we set
d = 8 kpc = 8(103 )(3.26)(63240) = 1.65 × 109 AU.
For S0-2, with a = 952 AU,
α=
206265(952)
= 0.119 arcsec
1.65 × 109
α=
206265(1790)
= 0.224 arcsec
1.65 × 109
For S0-19, with a = 1790 AU,
High-resolution of infrared images are required due to the presence of dust, and the angular sizes of the
two stars are very small.