Masses, Distances and Times in the Universe
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Masses, Distances and Times in the Universe
Lu =
∫
∞
Rm
n
L ( Rm )
4π R 2 dR = 4π Rm2 L ( Rm )n[ R]∞Rm → ∞
( R / Rm )
51
(3.1)
But, before making reasonable estimates of the mass for a typical star
(like our Sun), a typical galaxy (like our Milky Way) and the observable
universe, let us remind ourselves what is the mass of the Earth and the
mass of the Sun, using simple newtonian arguments. The mass of the
Earth can be obtained using the quantity of the gravitational acceleration
on its surface g ≈ 980 cm/sec2, the quantity of Newton’s universal
gravitational constant, G ≈ 6.67 × l0-8 cm3 g-1 sec-2, and the Earth’s radius,
RT ≈ 6378 km, which is fairly well known since the time of Erathostenes.
The values of g and G can be obtained without much difficulty from
experiments performed on Earth’s surface. We have
F = mg = G
mMT
RT2
g
, i.e. MT = RT2 = 5.97 × 1027 g
G
(3.2)
On the other hand the mass of the Sun can be obtained from the fact
that, to keep the Earth in its elliptic but very nearly circular orbit around
the Sun, it is necessary that the effective centrifugal force be exactly
compensated by the gravitational attraction,
MT (2π RST / TT )2
M M
(2π / TT )2 3
= G T 2 S , i.e. M S =
RST
RST
G
RST
(3.3)
33
= 1.99 × 10 g
Where RTS is the Earth-Sun distance, and use has been made of the
fact that νT = 2πRST / TT, i.e., the Earth’s average velocity, obtained
dividing the length of the orbit, 2πRST, by the period of time necessary to
complete one revolution, TT = 365 × 24 × 60 × 60 sec.
In what follows we will show how very simple physical arguments3
can give us excellent estimates of the mass of stars and galaxies. We will
use the dimensionless quantities
α ≡ e2 / ℏc ≃ 1/137
(3.4)
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The Intelligible Universe
and
α g ≡ Gm 2p / ℏc ≃ 6.3 × 10 −39
(3.5)
which measure, respectively, the strength of the electrostatic interaction,
and the much smaller gravitational interaction, in terms of physical
constants (see table in Sec. 2.2). We will also make use of the virial’s
theorem,4 which states that for a system in equilibrium in a central field,
Ekin =
1
Epot
2
(3.6)
where Ekin = kinetic energy, Epot = potential energy. A star is made up of a
plasma of electron and ions (basically protons in the hydrogen burning
phase of the life of a star, which is the longest one) and we can equate
the kinetic energy of particles in the surface of this plasma which we
assume to have spherical shape to a representative electrostatic ionization
energy, so that
( Ekin )S ≈ k BT ≈
e2
α rC
(3.7)
The gravitational potential energy of a proton with mass mp, on the
other hand, is given by
m M
(3.8)
( Epot )S ≈ G Rp S
S
Then, making use of the virial theorem, Eq. (3.2), we can write,
dividing by ħc both sides of the equation,
2
1 e2 1 Gm p N S
≃
α ℏc 2 ℏcc RS /rC
(3.9)
where NS is the number of protons within the sphere of radius Rs, and,
therefore
NS ≃
4π 3 4π 3
RS
rC , i.e. ( RS /rC ) ≃ N S1 / 3
3
3
(3.10)
Substituting (RS/rS) = NS1/3 in Eq. (3.9), and taking into account the
definitions in Eq. (3.4) and (3.5), we get
Masses, Distances and Times in the Universe
1
NS ≃ 2
αg
53
3/2
≃ 5.65 × 1057
(3.11)
which is the number of protons (baryons) in the plasma sphere which
makes up the star. The star’s mass, hence, will be given by
M S = N S m p ≃ 9.4 × 1033 g
(3.12)
which is in fairly good agreement with the solar mass evaluated above,
Eq. (3.3). Note that the expression for the mass of a typical star as given
by Eqs. (3.11) and (3.12) involves only general physical constants, like
the combination αg = Gmp2/ ℏ c and nip, a truly remarkable result. Figure 3.1
depicts the equilibrium configuration of a star.
The mass of a protogalaxy, i.e. a compact galactic mass presumably
in the very first moment after formation, can be estimated in a similar
way. In this case gravitational pressure can be assumed to be
compensated by radiation pressure from the heated up plasma, and one can
conceive a near equilibrium situation when the gravitational free fall
time, tg, for particles in the surface of the quasi-spherical matter
distribution, is comparable5 to the characteristic dissipation time, td, of
radiant energy coming away from the hot plasma. If tg « td, the
protogalactic plasma would collapse towards the center of mass before
reaching quasi-equilibrium conditions. If, on the other hand, tg » td,
radiation pressure would push the protogalactic matter away from the
center.
The gravitational free fall time can be given in terms of the
protogallactic mass MG, (which must be the same as the gallactic mass),
and the protogallactic radius RPG, (which may be much smaller than the
future galactic radius), taking into account the newtonian expression for
uniformly accelerated motion,
2
RPG ≃ 12 (GMG / RPG
)tG2
(3.13)
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The Intelligible Universe
Fig. 3.1. Equilibrium configuration of a star leading to an estimate of its mass.
M S ≈ (2 / α g )3 / 2 m p ≈ M sun
from which tg can be eliminated as
3
t g ≃ (2 RPG
/ GMG )1/ 2
(3.14)
The characteristic dissipation time can be estimated dividing the total
radiation energy, Er, enclosed by the protogallactic sphere, by the rate at
which radiation energy is given up at its surface, i.e.,
Er
1 RPG
td ≃
≃
(3.15)
2
4
c 4π RPG (σ T ) 3 c
where as given by Stefan–Boltzmann’s law, the energy density per unit
volume in terms of the absolute temperature T, is equal to (σT4), and,
therefore
4π 3
(3.16)
Er =
RPG (σ T 4 )
3
Equating tg and td, and taking into account that MG = NG mp, where
NG in the number of protons in the whole protogalaxy, one gets
3
2
2 RPG
18 RPG m p c
1 RPG
≃
,
i
.
e
.
N
≃
G
GN G m p 9 c 2
Gm2p ℏ
ℏc
Now we can make use of the fact that
(3.17)
Masses, Distances and Times in the Universe
m p c m p me c m p 1
ℏ = m ℏ = m αr
e
e
c
55
(3.18)
and taking into account that
(RPG / rc ) ≈ N G1 / 3
(3.19)
a relationship fully analogous to that in Eq. (3.10), we can finally get
3/2
N G ≃ (18m p /me )
1
αα g
3/ 2
≃ 3.3 × 1067
(3.20)
the number of protons (baryons) in the whole protogalaxy. The galaxy
mass is, therefore
MG = N G m p ≃ 5.5 × 10 43 g
(3.21)
This is of the order of 1010 typical star masses, since, as shown in
Eq. (3.12), Ms ≈ 9.4 × 1033 g. Again, the expression of MG as given by
Eqs. (3.20) and (3.21) involves only numerical factors of order unity and
combinations of general physical constants, such as (mp/mc) ≈ 1836, the
mass ratio of proton and electron, α = e2/ ℏ c ≈ 1/137 (dimensionless),
αg = Gmp2/ ℏ c ≈ 6.3 × 10-39 (dimensionless), and mp, the proton mass. It
should be mentioned that Eq. (3.21) agrees fairly well with estimates of
the total mass of our own galaxy, the Milky Way or Via Lactea, which
contains about 1011stars, and therefore has a mass
M VL = 1011 M SUN ≈ 2 × 1044 g
(3.22)
The mass of the observable universe, MU cannot be calculated, at least
for the time being, in a way similar to those used to estimate Ms and MG,
because, as we know, the whole universe is expanding at fantastic speed,
and we are in no way near an equilibrium situation. However one can
make a rough estimate of its total mass if one takes seriously current
observational estimates of its present average density, ρ0, its present
radius, R0 ≈ τ0/C, and its spatial curvature k < 0, corresponding to an open
universe with less mass than that necessary for a hypothetical future
phase of contraction after the present expansion.
Using, for instance,
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The Intelligible Universe
ρ0 ≃ 2 × 10 −31 g/cm 3
(3.23)
R0 ≃ 1.6 × 10 28 cm (τ 0 ≃ 1.75 × 1010 years)
(3.24)
one can substitute ρ 0, and R0 in the expression of MU to get
MU ≃ ρ0V0 ≃ ρ0
4π 3
R0 ≃ 3.8 × 1055
3
(3.25)
which is of the order of 1012 galactic masses, equivalent roughly to the
actually observable number of galaxies. It may be noted that, strictly
speaking, the expression used for the volume
V0 ≃
4π 3
R0
3
(3.26)
corresponds to the euclidean volume (flat space |k| = 0), somewhat
different from the more general relationship7
V0 ≃
∫
4π r 2 dr 3
R0
0 (1+ |k | r 2 )1 / 2
1
(3.27)
which would be the correct general relativistic expression for a noneuclidean open space. However, the definite integral within square
brackets, which is elementary and can be found in tables, becomes (4π/3)
if |k |2 = 0, and does not differ by more than 15% from this value for |k | =
+1, or |k | = −1, corresponding, respectively, to a closed and an open
universe. Errors in the estimates of ρ 0 and R0 are larger than 15%. The
results thus obtained for number of baryons and masses of succesively
larger objects in the cosmos are summarized in Table 3.1.
Table 3.1
Object or System
Star
Galaxy
Observable Universe
Calculated number of
barions
Ns ≈ 5.6 × 1057
NG ≈ 3.3 × 1067
Mass (gr)
Ms ≈ 9.4 × 1033
MG ≈ 5.5 × 1043
MU ≈ 3.8 × 1055
Number of units in
larger object
(MG/Ms) ≈ 1010
(MU/MG) ≈ 1012
(MU/MU) ≈ 1
Masses, Distances and Times in the Universe
57
3.2. Distances
As mentioned in the historical introduction, the oldest estimates of
astronomical distances go back to the time of Erathostenes and
Aristarchus of Samos. While Erathostenes, using a very simple method,
made a very good estimate of the Earth’s radius, within of 5% the correct
value, Aristarchus went from the Earth to the moon and from there to the
sun, in two steps, so to speak. The latter’s arguments were correct, but
some of the observational data he used lacked the necessary accuracy,
and, consequently, his numerical estimates for sizes and distances to
moon and sun were seriously flawed. We can reproduce the basic
reasoning of Erathostenes and Aristarchus and use modern observational
data to obtain successively the Earth’s diameter, the distance to the moon
and the distance to the sun.
Fig. 3.2. Erathostenes method to measure the Earth’s circumference. The distance from
S (Syene, Southern Egypt) to A (Alexandria) was SA ≈ 5000 stades ≈ 800 km, and
the angle between the vertical and the direction of sunrays at A, on the particular day
and time at which no shadow was present at S, could be measured as a = 7.2°. Therefore
2πRT = (α/360) AS ≈ 40,000 km.
It may be noted that the greek geometer had the good luck of relaying
on the distance between two cities, Syene, directly under the ecliptic in
Southern Egypt, and Alexandria, in the Mediterranean cost, connected by
the Nile river, the only great river running South to North almost along a
meridian line in the Northern Hemisphere. Thanks to this happy
coincidence he was able to measure with very good accuracy the distance
between both cities, 5000 stades. The angle made by sunrays with the
The Intelligible Universe
58
vertical at Alexandria (in the same year’s day and time in which no
shadow could be seen at noon at Syene) was 7.2º, i.e. one fiftieth of a full
circle, and consequently, the Earth’s circumference was estimated as
250,000 stades or 40,000 km.
Figure 3.3 shows another happy coincidence, perspicaciously noted
by Aristarchus, which paved the way for an estimate of the size of the
diameter of the moon’s orbit. As he noted, it takes, in a lunar eclipse,
almost the same time for the moon to begin to reappear after full
Fig. 3.3. Lunar eclipse. As noted by Aristarchus the path of the moon within the cone of
full shadow produced by the Earth is just about two lunar diameters. This fact pointed the
way to an estimate of the Earth-moon distance in terms of the lunar diameter.
occultation as it took her to disappear after making contact with the
shadow cone produced by the Earth. This implies that the length of the
path travelled during the eclipse was about two lunar diameters (more
exactly n = 1.8 times the lunar diameter). On the other hand the maximum
time for a lunar eclipse is tEL = 1 hour, 42 min. i.e. 102 min, relatively
small in comparison with the time needed for the noon to complete a full
orbit around the Earth, which is TL = 27.3 days. Therefore, assuming that
the orbital velocity is constant, we have
t EL
nDL
=
=, i.e.
TL 2π DTL
dTL 27.3 × 24 × 60 1.8
=
≃ 110.45
102
DL
2π
(3.28)
Masses, Distances and Times in the Universe
59
where DL is the lunar diameter and dTL the Earth–moon distance. On the
other hand the apparent angular diameter of the moon as seen from the
Earth’s surface is directly measurable, and is given by
θL =
dTL
DL
= 9.205 × 10 −3 rad
1
− DT
2
i.e.
dTL
1 1 DT
=
+
DL θ L 2 DL
(3.29)
where DT is the Earth’s diameter. Combining Eqs. (3.28) and (3.29) one
gets
DT
= 0.276
DL
(3.30)
which means that the moon’s linear diameter in fairly large, between
one forth and one third of the Earth’s linear diameter. Also, introducing
Eq. (3.30) into Eq. (3.28), one gets
dTL
= 30.46
DL
(3.31)
which, taking into account that DT = (40,000/π) km leads to dTL = 384 ×
106 m as the Earth–moon distance.
Finally, to estimate the Earth-sun distance in terms of the Earth-moon
distance, Aristarchus considered the triangle made up by Earth, moon
and sun when exactly half moon is viewed from the Earth. The angle a
subtended from the center of the Earth to the centers of moon and sun is
very approximately the same, although slightly smaller, than the
corresponding angle subtended from the point nearest the moon on the
earths surface (see Fig. 3.4). While it is only eight arc minutes (8’)
smaller than a right angle, due to the very large distance from sun to
Earth, this difference was not beyond the observational capabilities even
at Aristarchus time but he did measure it with a large error.
The Intelligible Universe
60
Fig. 3.4. Triangle made up by Earth, Moon and Sun when exactly half moon is viewed
from the Earth, which allows an estimate of the Earth–sun distance, dTS, in terms of the
Earth–moon distance, dTL, and the angle α.
From Fig. 3.4 it can be seen that
cos α =
dTL
dTS
(3.32)
and, taking into account that the observable value of a is 89 degrees and
52 minutes, very close to a right angle, one gets
dTS =
dTL
= 149 × 109 m
cos α
(3.33)
a very large distance, indeed.
Also, taking into account that the angular dimension of the sun is
close to that of the moon, it may be expected that the linear diameter of
the sun be about (dTS/dTL) times the angular diameter of the moon. Using
previous results and the fact that
θs =
DS
D
≃ S ≃ 12.88 × 10 −3 rad
1
dTS − 2 DT dTS
(3.34)
one can readily check that
DS
= 109.1
DT
(3.35)
i.e. that the size of the sun is more than one hundred times that of the
Earth, or RS = RSUN = 6.96 × 105 km.
Masses, Distances and Times in the Universe
61
These were the first steps of man into its cosmic neighbourhood but,
as we will see immediately, these large distances were puny in
comparison with the far vaster distances between stars and between
galaxies.
The radius of a typical star can be calculated from Eq. (3.10) as
Rs = N S1 / 3rc ≃ 8.9 × 105 Km
(3.36)
where Ns ≈ 5.65 × 1057 and rc ≡ (1/α) ( ℏ /mec) = 0.5 × 10-8 cm have been
used.
This is very close to RSUN as given by Eq. (3.36). The size of a typical
star remains fairly constant during most of its long lifetime. On the other
hand the same argument applied to a protogalaxy would give, using now
NG ≈ 3.3 × 1067, something like
RG = N G1/ 3rc ≃ 1.6 × 10 9 Km
(3.37)
which is far smaller than the radius of an actual galaxy of comparable
mass. Fortunately we can use observational data on the rotation speeds of
outermost stars in an actual spiral galaxy (Andromeda) to get a realistic
value for the present radius of a typical spiral galaxy, RG.
Notwithstanding the dark matter (or missing mass) question, it can be
argued that the centrifugal acceleration of the outermost star is
compensated by the acceleration of gravity towards the galactic center,
and, therefore
M S vS2
M M
= G S2 G , i.e.
RG
R G
RG =
GMG
vS2
= 9.1 × 10 21 cm = 9.1 × 1016 Km (3.38)
where G = 6.67 × l0-8 cm-1 g-1 sec2, MG = 5.5 × l043 g and vs = 200 ×
l05 cm/sec have been used. This is equivalent to
RG c
365 × 24 × 60
2
= 9.17 × 10 3 light year
(3.39)