Astronomy 1
Math Extras (for pure enjoyment)
Mr. Kiledjian
1) Proof that 1 parsec = 3.26 light years and 1 light year is 9.5 x 1012 km and 1 light
year is about 63,240 AUs.
1 Light year is the distance that light travels in 1 year. It is used as a unit of distance to
measure the distance between stars and galaxies. The speed of light = c = 2.99792458 x
108 m/s. To calculate the distance of 1 light year, we use D = vt:
D = (2.99792458 x 108)(1 year)(365.25 days)(24 hrs)(3600 sec) = 9.46 x 1015 m!!!
The AU stands for astronomical unit and it is the average distance between the sun and
the earth. 1 AU = 1.495978706 x 1011 meters. Therefore, 1 Light year is equivalent to
1 Ly = (9.46 x 1015)/(1.495978706 x 1011) = 63,236.19 AUs.
How long does light take to reach Earth? To answer this, divide 1 AU by the speed of
light, so t = 1.495978706 x 1011/2.99792458 x 108 = 499.004783503 sec = 8.3167 min.
So, we can say that the Sun is 8 Light-Minutes away from Earth!!!
1 AU
 = 1 arcsec
1 Parsec
1 Parsec is the distance to a star whose
Parallax angle is 1 arcsecond, which is
1/60 of an arcminute, which is 1/60 of
a degree.
Tanbecause the angle is so small)
1 AU/1 Parsec = 1 arcsec Therefore, 1 Parsec =
1AU
1 LightYear
1.495978706 x 1011 m


 3.262 LightYears!!!!!
1arc sec  1
 rads  9.46 x1015 m


deg rees 
180 deg rees 
 3600
2) Calculating what Portion of the Sky is Seen by an Observer and what Portion is
North Circumpolar or South Circumpolar and How long it is seen:
If someone lives on the equator, they see 100% of the sky throughout the year, but no
portion of it is circumpolar for them. That means there are no stars or constellations that
they can always see at any time of the year. On the other hand, if someone lives on the
North Pole, they see the upper 50% of the sky, but 100% of what they see is North
Circumpolar for them; they can see it at any night during the year.
Therefore, assuming that the 88 constellations in the sky are distributed evenly, the person
at the equator will see all 88 during the year, 44 of them at any given night, and none of
them are seen always. However, the person at the north pole see the upper 44
constellations and sees them every night. How about if a person lives at a latitude of ???
Let’s assume for now, that  is positive, but the same arguments apply if it is negative.
Then, the person will see any star or constellation above ( - 90), because that forms
their horizon line, all the way up to 90. However, any star between (90 - ) and 90 is
always seen by them. Therefore, the formula becomes --------------->
% SeenDuringtheYear  100% 
90    90  100%  180   
180
% AlwaysSeen  %Circumpola r  100% 
180
90  90     100%   
180
180
Example: Los Angeles’ latitude is  = 340. Therefore, %Seen During the Year =
100% x (180 – 34)/180 = 81.11%. Multiplying this by 88 constellations, we get 71.4
constellations. In actual the number is 73. We see 73 of the total 88 constellations!!
%Circumpolar for LA = 100% x 34/180 = 18.89%. Multiplying this by 88, we get 16.62
constellations. In actual the number is 5. We see 5 constellations at any given night.
The math didn’t work that well here. That is because the distribution of constellations is
not even on the celestial sphere. They are more heavily concentrated at the center!!
The stars that are between ( - 90) and (90 - ) are visible for certain portions of the year
based on a continuum from 0 – 12 months. Let’s say such a star’s DEC angle is 
  (  90)  12months      90  12moths
Then, it would be visible for
(90   )  (  90)
(180  2 )
Ex: For LA, ( - 90) = (34 – 90) = -560 and (90 - ) = (90 – 34) = 560. So, anything
above 56 degrees is seen for 12 months, and anything below -56 degrees is never seen.
How about if a star’s DEC angle is  = 150. Then, applying the above formula, it would
be seen for 12 x (15 – 34 + 90)/(180 – 2(34)) = 7.61 months of the year. If  = -150, then
it would be seen for 12 x (-15 – 34 + 90)/(180 – 2(34)) = 4.39 months of the year!!!
3) Determining which months a star or constellation is visible:
In the previous problem, we found a way of calculating how many months a star is
visible for a person at any latitude. Now, we need a way to determine which months
these are. For that, we will need the RA angle of the star or constellation, and we will use
the fact that the Sun crosses the Vernal Equinox (RA = 0 hrs) on or about March 21.
Let’s assume that the star  = 150 in the previous problem has an RA = 19 hrs and the
star  = -150 has an RA = 4 hrs. When the sun is exactly at the opposite end of the
celestial sphere, the star is at the middle of its visibility period. For the 1st star above, the
sun has to reach RA = 19 – 12 = 7 hrs and for the 2nd star, the sun has to reach RA =
4 + 12 = 16 hrs. Now, let’s calculate how long it takes the sun to reach that RA angle.
7hrs
 365.25days  106.53days Now, add this to
24hrs
March 21st. March 21st is the 80th day of the year, so 80 + 106.53 = 186.53 which
corresponds to July 5 or 6. This is the middle of the period of its visibility. In the
previous problem, we calculated that it would be visible for LA people for a total of 7.61
months. Divide this time by 2 and subtract it from July 5 and add it to July 5. We get ---7.61months 365.25days
-------> 186.53days 

 70.72days which is March 11 or 12
2
12months
7.61months 365.25days
and 186.53days 

 302.34days which is October 29 or 30.
2
12months
So, the 1st star will be visible at nighttime from around March 11 – Oct. 30 for LA people!!
Time to reach RA = 7 hrs --->
16hrs
 365.25days  243.5days
24hrs
Adding this to March 21st, we get ----> 80 + 243.5 = 323.5 days which is Nov. 19 or 20.
This is the middle of its visibility period. In the previous problem, we calculated that it
would be visible for LA people for a total of 4.39 months. Divide this by 2 and add and
subtract it from November 19 as we did for the 1st star.
4.39months 365.25days
323.5days 

 256.69days which is September 13 or 14.
2
12months
For the 2nd star, the Time to reach RA = 16 hrs ---->
4.39months 365.25days

 390.31days which corresponds to January 25.
2
12months
star will be visible at nighttime from around Sept. 13 – Jan. 25 for LA people!!
323.5days 
So, the 2nd
4) How to Calculate the Circumference of the Earth:
You have to use an experiment similar to that done by Eratosthenes. Take a pole of
known height (let’s say 10 ft) and measure its shadow in LA and another city (let’s say
San Diego). Let’s say that its shadow comes out to be 6 ft in LA and 5.6 ft in SD. (it
should be shorter in SD since San Diego is closer to the equator). From this you can
calculate the angle of the shadow from the vertical pole using Trigonometry.
TanLA = 6/10 -----> LA = Tan-1.6 = 30.9638o
TanSD = 5.6/10 -----> SD = Tan-1.56 = 29.2488o
LA - SD = 1.715o. Therefore, LA and SD subtend an angle of 1.715 degrees from the
center of the Earth. Then, apply a simple ratio assuming the distance from LA to SD is
120 miles and using that fact that the angle around a complete circle is 360o.
C Earth DbetweenSD/ LA
(120miles )(360)

     C Earth 
 25,189.50miles
o
o
1.715
360
1.715
The real circumference of the Earth = 25,046.35 miles.
5) Proof of Kepler’s 3rd Law: T2  d3
F = GMsunmplanet/d2betweenSun/Planet = mplanetv2/dbetweenSun/Planet (Centripetal force equation)
G = Universal Gravitational constant = 6.67259 x 10-11, Msun = 1.98892 x 1030 kg
From the above equation, mplanet cancels out and v2 = GMsun/dbetweenSun/Planet
Also, the v = (distance traveled around the circle)/(1 period) = (2dbetweenSun/Planet)/T,
2
4 2 d betweenSun
GM sun
/ Planet

Therefore, v 
---------------------->
2
d betweenSun/ Planet
T
2
4 2 d 3
4 2 d 3

 2.97473293306  10 19 d 3
GM sun (6.67259 x10 11 )(1.98892 x10 30 )
So, we have shown that the period of a planet around the sun is proportional to its
distance from the sun cubed, which is what Kepler proved in his 3rd Law of Planetary
Motion. The “d” in the above equation is in meters and the “T” is in seconds. Now,
using this equation, let’s calculate how long it takes for the Earth to go around the sun
and come back to the same point with respect to the stars, called a sidereal year.
Therefore, T 2 
T 2  2.97473293306 x10 19 1.495978706 x1011 meters 3  9.95919456193  1014 
T  31558191.5862 sec  365.256847063days
The actual value is 1 sidereal year = 365.256363004 days because the path of the Earth
around the sun is somewhat elliptical and our equation assumes that the path is perfectly
circular. Also, the value of G we used is an approximate value and it is hard to determine
its exact value precisely.
6) Determining the difference between a Sidereal Day/Solar Day & Sidereal
Year/Tropical year and also the nature of the Gregorian Calendar!
One Sidereal Day is defined to be the amount of time required for the Earth to rotate
around once and face the same star and is equivalent to 23 hrs, 56 min and 4 seconds
approximately. It has to rotate a little bit more to face the sun since the Earth has moved
around the sun by some angle; therefore, one Solar Day is defined as the time to go
around once and face the sun and is equal to 24 hours by definition. Therefore, the stars
will be a little ahead of the sun (west of it) every day. [From this point on, any time we
refer to a day, it will be the Solar Day] The sun takes 365.256363004 days (solar days) to
go around the sun once. This is known as Sidereal Year. In the meantime, the Earth’s
axis precesses due to the gravitational pull of the Sun and Moon on its equatorial bulge,
and the gravity of the planets. The period of precession is about 25,772 years. Because
of this the Tropical Year which is the time needed from Summer Solstice to the next
Summer Solstice and Winter Solstice to the next Winter Solstice, etc, is shorter than the
Sidereal Year by a certain amount. This is equal to
1
365.256363004 days
 .0000388018004035 years 
 .014172604493days
25,772 years
1 year
 20.40855 min
So, whatever calendar one adopts should account for the Tropical year and not the
Sidereal or otherwise, the seasons will start earlier by about 20 minutes each year. The
value of a Mean Tropical Year = 365.256363004 - .014172604493 = 365.2421904 days.
The tropical year actually changes over the centuries but the Mean Tropical year for 2000
was 365.2421896698 days. Let’s see now how the Gregorian Calendar accounts for this.
If we assumed that a year has 365.25 days exactly, then in 4 years, we would have 3
regular years of 365 and 1 leap year of 366 days amounting to 1461 days. However,
365.2421896698 x 4 = 1460.9687586792 days, so during one 4-year cycle, we would
have over-added 1461 - 1460.9687586792 days = .0312413208 days. Since, there are 25
4-year cycles in a century, that would total to .0312413208 x 25 = .78103302 days. After
4 centuries, we would have over-added .78103302 x 4 days which is 3.12413208 days.
So, what the Gregorian Calendar does is to skip a leap year at the end of every century
divisible by 100, except every 4th century which is also divisible by 400. This would
mean that after 400 years, we would have over-added by only 3.12413208 – 3 =
.12413208. This is not a perfect fix since we have an additional .12 days over-added but
it is much better than over-adding 3.12 days per 400 years, which amounts to adding
nearly 10 days in 1200 years. So, from 325 (Council of Nicea) to 1582, a span of 1257
years, there had been an extra 1257 x 3.12413208/400 = 9.817585 days over added so
Vernal Equinox was beginning on March 11 instead of March 21, so 10 days were
skipped over from Thursday, October 4, 1582 to Friday, October 15, 1582. [After the
Gregorian correction, we will end up over-adding about 1 day every 8 cycles of 400-year
span which amounts to 3200 years. So, every 3200 years, we can skip an extra leap year.
Now, let’s prove that the sidereal day is shorter than the solar day by about 4 minutes.
A
Sun

A
Earth
We can write an equation of point A as a function of time,

R  d ES cos( rev t )  RE cos( rott   ), d ES sin( rev t )  RE sin( rott   )
Here, dES = 1 AU and RE = radius of Earth and rev = rotational velocity of the revolution
of the Earth around the Sun = 2/365.256363004 rads/day, and rot = rotational velocity
of the rotation of the Earth around its own axis which is an unknown for now, because we
can see that point A has to rotate more than 2 radians in 1 day to face the sun. In order
to calculate this, we use the equation,
d sin( rev t )  RE sin( rott   )
Tan  ES
Let’s apply this equation to t = 1 day!
d ES cos( rev t )  RE cos( rott   )
We can calculate  = revt = (2/365.256363004 rads/day)(1day) = .017202124162 rads,
so put this into the above equation and solve for the unknown variable rot!
  6378000 sin( (1)   )
1.495978706  1011 sin  2 (1)

rot
365
.
256363004


Tan(.017202124162 ) 
  6378000 cos( (1)   )
1.495978706  1011 cos 2 (1)
rot
365.256363004 

2573274229.06  6378000 sin( rot (1)   )
 .01720382114 
149575737163  6378000 cos( rot (1)   )

The answer to this comes out rot = 6.3003874351346 rads/day which makes sense since
it is greater than 2 rads/day. Now, the last step is to calculate how many days it takes
the Earth to rotate 2 radians at this rate, t = 2 radians/6.3003874351346 rads/day =
.997269671408 days = 23.9344721138 hours = 23 hours, 56 min, 4.0996 seconds!!


7) Predicting the Number of Hours of Sunlight at any Day during the Year:
The number of hours of sunlight depends on two factors which are: the latitude of the city
and the Dec angle of the sun. Let’s first assume that the date is around June 21st when the
axis of the Earth is pointed directly towards the sun and its DEC angle is 23.5 degrees on
the celestial sphere. The picture looks like the following:
Earth’s North Axis
23.5 degrees
Sun

d
Rcos
23.5
R

The portion of the day when the city is
lit is (Rcos + d), and tan23.5 = d/Rsin.
Therefore, d = R(tan23.5)sin
What really matters is the ratio of the portion that is lit to the total length = 2Rcos.
Therefore, we get -------------->
R cos   R sin  tan 23.5
# Hours of Sunlight onJune21 
 24hours  (1  tan  tan 23.5)12 hrs The
2 R cos 
latitude of LA = 34.5, so it should get, (1 + tan34.5tan23.5)12 = 15.59 hours on June 21
The latitude of the equator = 0, so it should get, (1 + tan0tan23.5)12 = 12 hrs on any date.
The latitude of the Arctic circle = 66.5, so it should get, (1 + tan66.5tan23.5)12 = 24 hrs
on June 21. Actually, anywhere above the Arctic circle will get 24 hrs on that day.
The latitude of the Antarctic circle = -66.5, so it should get, (1 + tan(-66.5)tan23.5)12 = 0
hrs. Actually, anywhere below the Antarctic circle will get 0 hrs on that day.
Now, we need a general equation that is valid for any day. To derive that, we need a
general expression for the Sun’s position on the celestial sphere on any day. On June 21,
the sun is at 23.5 degrees because the Earth is most tipped towards the sun. In problem 3,
we saw that the sun crosses the celestial equator on or about March 21 which is the 80th
day of the year, and we know that it makes a sine-wave curve in 1 year reaching its
maximum on June 21. Its DEC angle can be expressed by the following equation,
360(t  80)
180(t  80)
This angle is also the
  23.5 sin( (t  80))  23.5 sin
 23.5 sin
365.25
182.625
same as the angle of the tip of the Earth’s axis towards the sun. Therefore, in the above
equation, we can replace the 23.5 degrees by this angle, and we’ll have a general
expression for the amount of sunlight at any location during the year!!


 180(t  80)  
# Hours of Sunlight OnAnyDay  1  tan  tan 23.5 sin
 12 hrs
 182.625  


Let’s say, the day is December 1st, which is the 334th day of the year and you live at a
latitude of -25 degrees. Then,


 180(334  80)  
# Hours of Sunlight  1  tan(25) tan 23.5 sin
 12 hrs  14.28hrs
 182.625  


8) Predicting the Length of the Shadow of any Object at any Time of the Year:
The shadow of any object depends on the altitude of the sun which changes throughout
the course of the day depending on the month and day of the year. The picture is:
Sun
tanSun = H/L
Object
(H)
Sun
LShadow = H/tanSun
Shadow (L)
If the sun is directly over an object, then Sun = 90o and Lshadow = H/tan90 = 0. If the sun
is just barely rising or setting on the horizon, then Sun = 0o and Lshadow = H/tan0 =
Infinitely large!!. Now, we need to come up with a general formula to predict the Sun.
Sun
r



r
Sun

r = the distance between the city and the sun
the angle that the sun makes with the
horizontal.
= the altitude angle of the sun when it is

midway in the sky.
 = the azimuth angle of the sun measured

from the South direction
Sun = altitude angle of the sun at any time
City
/180 = t/T where t = number of hours elapsed since sunrise, and T = total number of
hours of sunlight for that city for that day as given by the equation
from the previous problem. Therefore,  = 180t/T


 180(t   80)  
’
T  1  tan  tan 23.5 sin
 12 hrs where t is the time since the beginning
 182.625  


of the year. For example, if the city is LA and the date is June 21, we calculated in the
previous problem that T = 15.59 hrs. Therefore, if sunrise occurs at 6:10 AM and the
time is 10:00 AM, then t = 3 hrs & 50 minutes = 3.833 hrs. Therefore, we get:
 =180(3.833)/15.59 = 44.260.
 = 90 + Sun - City where  is the dec angle of the sun and the latitude angle of the sun.
We can again use the Dec angle formula we had for the sun in the previous problem.
180(t   80)
where t’ is the time since the beginning of the year.
 sun  23.5 sin
182.625
For example, let’s say the city is LA again and it is June 21. Then, city = 34.5 and
sun = 23.5 since the sun gets highest in the sky on June 21. Then, we get:
 = 90 + 23.5 – 34.5 = 790. (This is the highest that the sun gets in the sky for LA).
On the other hand, if the city’s latitude had been 0 degrees, and the date had been March
21 or Sept 21, then  = 90 + 0 – 0 = 900 since the sun crosses the ecliptic on that date!!
Therefore, the sun gets exactly overhead by noontime for the equator on that day. Now,
let’s put all these together to come up with a general equation for the sun’s altitude.
From the above picture, we get the following relationships:
rsin = R = radius of arc made by the sun: Rsin = height of sun above the horizon line
rsin = Rsin = rsin sinherefore, we get:

 180t 
 Sun  sin 1 (sin  sin  )  sin 1 sin 90   sun -  city sin

 T 

As an example, let’s continue the example we have begun in LA at June 21 on 10:00
AM. Let’s say, we want to predict the shadow length for a 10 foot pole. We get:

 180(3.833) 
0
  43.24 then, we use the shadow
 15.59 

equation, Lshadow = H/tanSun = 10/tan43.24 = 10.63 feet!!!
 Sun  sin 1 sin 90  23.5 - 34.5sin
9) Proof that Moon’s Sidereal Period is 27.32 days:
We will use the same equation derived in #7 but this time putting in the mass of the Earth
instead of the mass of the Sun, since the moon goes around the Earth due to Earth’s
gravity, and then we’ll put in the average distance between the Earth and the Moon.
MEarth = 5.974 x 1024 kg, dAvgbetweenEarth/Moon = 384,400 km = 3.844 x 108 meters
4 2 (3.844 x10 8 ) 3
4 2 d 3

 5.62535863253x1012 sec 2       
GM Earth (6.67259 x10 11 )(5.974 x10 24 )
1day
1hr
T  2371783.8503 sec onds 

 27.45days !!!!
3600 sec onds 24hrs
T2 
10) Proof that Moon’s Synodic Period is 29.53 days:
When the moon has gone around once around the Earth, the Earth has also revolved around
the sun. Therefore, the phases of the moon don’t repeat every 27.32 days. It takes the
moon a couple extra days to catch up. Let’s calculate exactly how many extra days.
We’ll use the equation: FinalEarth = InitialEarth + Eartht, FinalMoon = Moont
InitialEarth = (27.32/365.25)2 = .46997 radians (this is how much the Earth is ahead by)
Earth = 2/365.25 = .01720 radians/day (this is the angular velocity of the Earth)
Moon = 2/27.32 = .2299848 radians/day (this is the angular velocity of the Moon)
Now, set FinalEarth = FinalMoon to find out how many extra days it takes the moon to catch up
.46997 + .01720t = .2299848t ---------> t = 2.21 days. Add this to 27.32 to obtain the
Synodic Period -----------------> Tsynodic = 27.32 + 2.21 = 29.53 days!!!!!!!!!!
11) Proof that Moon’s Angular Diameter is similar to Sun’s Angular Diameter:
The angular diameter of an object is roughly equal to its diameter divided by its distance
from us. We will use the following data.
When the Earth is close to the sun, Perihelion, Dperihelion = 1.470 x 108 km
When the Earth is far from the sun, Aphelion, Daphelion = 1.520 x 108 km
When the Moon is close to the Earth, Perigee, Dperigee = 363,300 km
When the Moon is far from the Earth, Apogee, Dapogee = 405,500 km
Diameter of Sun = 1.392 x 106 km, Diameter of Moon = 3476 km
Therefore, the maximum angular diameter of the Sun, ----------------------->
maxSun = Dsun/Dperihelion = (1.392 x 106)/(1.470 x 108) = .009469 rads = 32.552 arcminutes
Here, I used the fact that  rads = 180o and each degree is equal to 60 arcminutes of angle!
The minimum angular diameter of the Sun, -------------------->
minSun = Dsun/Daphelion = (1.392 x 106)/(1.520 x 108) = .009158 rads = 31.483 arcminutes
The maximum angular diameter of the Moon, ------------------------>
maxMoon = Dmoon/Dperigee = (3476)/(363,300) = .009568 rads = 32.892 arcminutes
The minimum angular diameter of the Moon, ------------------------>
minMoon = Dmoon/Dapogee = (3476)/(405,500) = .00857 rads = 29.461 arcminutes
Notice that, when the moon is at perigee, its angular diameter is even bigger than the
maximum angular diameter of the sun, and therefore, it can totally cover up the sun.
However, when the moon is at apogee, its angular diameter is even smaller than the
minimum angular diameter of the sun, and therefore, it can only partially cover the sun!!
The largest that the thickness of an Annular Solar Eclipse can be is (maxSun - minMoon)/2 =
(32.552 - 29.461)/2 = 1.5455 arcminutes.
12) Calculating the Maximum Size & Length of a Total Solar Eclipse
A solar eclipse happens when the Moon is crossing the Ecliptical plane and comes between
the Earth and the Sun (this is called New Moon phase.) The moon casts two kinds of
shadows on the Earth, the dark umbral shadow, and the lighter-colored penumbral shadow.
Those people who are inside the umbral shadow will be seeing either a Total Solar eclipse
or an Annular Solar eclipse; those people inside the penumbral shadow only see a Partial
Solar eclipse. In this problem, we want to calculate the maximum length of a Total Solar
eclipse at any one location. First, let’s calculate the diameter of the largest umbral shadow:
This occurs when the moon is close to us (perigee), and the sun is far from us (aphelion)!!
Sun
Moon
dfrom center of Sun to end of Shadow
D
Earth
.X
DUmbralLarg est
d from surfaceof Earth to end of Shadow

DMoon
d fromceneter of Moonto end of Shadow

DSun
d fromceneter of Sunto end of Shadow
  
3476
1.392  10 6
3476
1.392  10 6

 

d from surfaceof Earth to end of Shadow X  d Perigee X  d Aphelion
X  363,300 X  1.52  10 8
Solving for X in the above equation gives us: ----------------------->
3476X + 5.28352 x 1011 = 1.392 x 106X + 5.057136 x 1011 ----> 2.26384 x 1010 = 1388524X
and X = 16303.93 km. That means, that if the Earth was not in the way, the shadow of the
Moon would end 16,303.93 km beyond the center of the Earth!!! Then, we apply the ratio:
DUmbralLarg est
DUmbralLarg est
DMoon
3476

  

 .009156912575
X  REarth
X  d Perigee
16303.93  6378 16303.93  363300
DUmbralLarg est

DUmbralLarg est  (22681.92)(.009156912575)  207.7 km
Now, we are ready to calculate the longest time that someone can observe a total solar
eclipse. That time is given by the Longest Umbral Diameter calculated above divided by the
Velocity that the Moon’s shadow travels across the Earth’s surface. The moon revolves
around the Earth counterclockwise as viewed from the North Celestial Pole, and the Earth
rotates counterclockwise also. Therefore, the shadow travels from West to East at the surface
of the Earth and its velocity = VMoon’s Revolution – VEarth’s Rotation since the rotations are in the
same direction. To find the velocity of the moon, we’ll use the equation developed earlier to
prove Moon’s sidereal period. Therefore, we get: VMoon’s Shadow = Vmoon – VEarth’s Surface
VMoon's Shadow 
GM Earth
2REarth


d Perigee
TEarth's Rotation
(6.67259 x10 11 )(5.974 x10 24 )
1km 3600 sec 2 (6378km)




363,300,000meters
1000m
1hr
24hrs
3770.94 – 1669.76 = 2101.18 km/hr = 1313.24 mph. Therefore, the longest solar eclipse,
DUmbral L arg est
207.7 km
t LongestSolarEclipse 

 .098849 hrs  5.93 min
VMoon's Shadow 2101.18 km / hr
In actuality, they can get up to 7.5 minutes, but they usually are around 2 – 3 minutes.
13) Calculating the Maximum Length of a Total Lunar Eclipse
A lunar eclipse happens when the Moon is crossing the Ecliptical plane and the Earth
comes between the moon and the Sun (this is called Full Moon phase.) The Earth casts two
kinds of shadows on the moon, the dark umbral shadow, and the lighter-colored penumbral
shadow. Those people who are experiencing nighttime while the moon is going through
this shadow will see a Lunar Eclipse. In this problem, we want to calculate the maximum
length of a Total Lunar eclipse. First, let’s calculate the diameter of the largest umbral
shadow cast by the Earth. This occurs when the sun is close to us (perihelion).
dfrom center of Earth to end of Shadow
Sun
Earth
Moon
Start of Total Lunar Eclipse
DiameterShadowat the locationof the Moon
d from ceneterof Moonto end of Shadow

12756
d from centerof Earth to end of Shadow
12756
d from centerof Earth to end of Shadow



DEarth
d from centerof Earth to end of Shadow

DSun
d from ceneterof Sun to end of Shadow
1.392  10 6
d from centerof Earth to end of Shadow  d Perihelion
1.392  10 6
d from centerof Earth to end of Shadow  1.47  10 8
    
    
12756d + 1.875132 x 1012 = 1.392 x 106d -----> d = 1,359,536.09 km!!! This is the farthest
distance away from Earth that the shadow would end. Then, we argue that the moon will
take the longest to cross this shadow when it is closest to Earth (perigee), because it would
have the bigger portion of the shadow to cross. Applying the ratio again, we get:
DiameterShadowat the locationof the Moon
DiameterShadowat the locationof the Moon
12756


  
d fromceneter of Moonto end of Shadow
d fromcenter of Earth to end of Shadow  d Perigee 1359536.09
DiameterShadowat the locationof the Moon
12756
   DiameterShadow  9347.3 km
1359536.09  363300
1359536.09
For a total eclipse to take place, the front edge of the moon has to enter this shadow, and
the eclipse lasts until the back edge of the moon leaves this shadow. Therefore, the moon
has to travel a total distance DTotal Eclipse = Dshadow + Dmoon and the eclipse lasts ---->
D
 DMoon 9347.3  3476
t LongestLunarEclipse  Shadow

 3.40 hours
VMoon
3770.94
In order for Totality to occur, the Moon has to be completely engulfed by the shadow of the
Earth; therefore, the back edge of the moon has to enter the Umbral shadow, and totality
lasts until the front edge of the moon leaves the shadow. Therefore, the moon has to travel
a total distance of DTotality = Dshadow - Dmoon and the eclipse lasts -------------->
t LongestTotality 

DShadow  DMoon 9347.3  3476

 1.56 hours  93.4 min
VMoon
3770.94
14) Calculating the Maximum Length of a Transit of the sun by Mercury & Venus:
A Transit of the sun by Mercury and Venus occurs when either planet comes between the
Earth and the Sun. This event is analogous to a Solar eclipse when the moon comes
between the sun and the earth. We want to calculate the maximum length of this event,
which will occur when either planet has to traverse the full diameter of the sun, and when
the planets are at their farthest from the sun and moving at their slowest speed, and when
  
the Earth is close to the sun so that the angular diameter of the sun is the largest. We can
use a similar method that we used to calculate the max. length of a solar and lunar eclipse.
Earth
Sun
.
dfrom center of Sun to Surface of Earth
Mercury
Dis tan cethat Mercury has to cross
d fromceneter of Mercuryto surfaceof Earth


DiameterSun
d fromcenter of Sunto surfaceof Earth
Dis tan cethat Mercury has to cross
d Perihelion Earth  d AphelionMercury  R Earth
Dis tan cethat Mercury has to cross
1.471  10 8  6.98  10 7  6378


  
1.392  10 6
 
d Perihelion Earth  R Earth
1.392  10 6
     D  731,457.42 km
1.471  10 8  6378
A transit of the sun by Mercury starts when its front side enters in front of the sun until its
back side leaves the sun. Therefore, it has to travel the above distance + its diameter.
D  DMercury 731,457.42  4878
736,335.42km
t LongestTransitMercury 



VMercury
GM Sun
(6.67259 x10 11 )(1.989 x10 30 )
m/s
d AphelionMercury
6.98  10 7
736,335.42km 1000m 1 min


 8.9 min
m
1km
60 sec
1,378,913.11
s
So, the longest transit of the sun by Mercury lasts about 9 minutes. We can do a similar
calculation for Venus. We get the following calculations:
Dis tan cethat Venus has to cross
d fromceneter of Venus to surfaceof Earth


DiameterSun
d fromcenter of Sunto surfaceof Earth
Dis tan cethat Venus has to cross
d Perihelion Earth  d AphelionVenus  R Earth
Dis tan cethat Venus has to cross
1.471  10 8  1.089  10 8  6378


  
1.392  10 6
 
d Perihelion Earth  R Earth
1.392  10 6
     D  361,440.02 km
1.471  10 8  6378
So, Venus only has to cross about half the distance of Mercury!! Now, we get:
t LongestTransitVenus 
D  DVenus 361,440.02  12102


VVenus
GM Sun
d AphelionVenus
373,542.02km
(6.67259 x10 11 )(1.989 x10 30 )
m/s
1.089  10 8

373,542.02km 1000m 1 min


 5.64 min
m
1km
60 sec
1,103,953.22
s
The transit of Venus lasts shorter even though it moves slower. This is because it has less
distance to cross since it is closer to us and farther from the sun!!!
15) Calculating how many Transits of the sun occur by Mercury & Venus per
Century:
Here, we want to calculate the average number of times that Mercury or Venus come
between the sun and the earth. We will assume that their orbits are flat with the Earth’s but
in reality, Mercury’s orbit is tilted at 7 degrees and Venus’ is tilted at 3.39. That will cause
them to sometimes skim the top or bottom edge of the sun or completely miss the sun
altogether. Therefore, the calculation we perform will give the maximum amount of
transits in a century. We can use a similar argument as the problem where we showed that
the Moon’s synodic orbit is 2 days more than its sidereal. We start out assuming that the
Earth, Mercury, and the Sun are lined up and then calculate how long it will take for them
to get lined up again; in other words, we want the difference between their angles to be
multiples of 2!! We will also perform a similar calculation for Venus. We get:
FinalEarth = Eartht,
FinalMercury = Mercuryt,
FinalVenus = Venust
Earth = 2/365.25 = .0172024 radians/day (this is the angular velocity of Earth)
Mercury = 2/87.97 = .07142418 radians/day (this is the angular velocity of Mercury)
Venus = 2/224.7 = .02796255 radians/day (this is the angular velocity of Venus)
Now, set FinalMercury = FinalEarth + 2------> .07142418t = .0172024t + 2 -------->
t = 115.88 days. Dividing 365.25 by this number, we get n = 3.15 and multiplying this by
100 gives us 315 times. According to this, Mercury should transit the sun 315 times per
century. However, as mentioned before, the actual number should be less than this because
of the tilt of Mercury’s orbit which is high. Let’s perform a similar calculation for Venus.
FinalVenus = FinalEarth + 2------> .02796255t = .0172024t + 2-------> t = 583.93 days.
Dividing 365.25 by this number, we get n = .6255 and multiplying this by 100 gives us
62.6 transits of Venus per century. This number is also not accurate but it’s more accurate
than Mercury’s answer since the orbit of Venus is not tilted as much as Mercury’s orbit!
16) How to Prove that the Maximum Elongation Angle of Mercury is 280 and of
Venus is 470
Mercury and Venus are the only 2 planets that are always seen close to the sun. According
to the Ptolemaic model of the solar system, the reason for this was that there was an
imaginary line between the sun and the earth, and this line went through the center of the
epicycles of Mercury and Venus, insuring that Mercury and Venus were always close to the
sun. Copernicus’ heliocentric model replaced this ad hoc argument with a much more
pleasant and logical explanation based on geometry. According to Copernicus, the reason
for Mercury and Venus being close to the sun is that they are Inferior Planets to Earth and
their orbits are smaller than Earth. This forces them to be close to the sun. The picture is:
Venus
Mercury
Earth


Sun
Maximum angle of Elongation of Venus,  = Maximum angle of Elongation of Mercury
The maximum angle of elongation for both planets occurs when the Earth/planet line
makes a 90 degree angle with the planet/Sun line. Therefore, sin = dVenus/Sun/dEarthSun
and sin = dMercury/Sun/dEarthSun. The maximum value for both angles occurs when the Earth
is at its closest to the sun and the planets are their farthest distance from the sun. We get:
d AphelionMercury 6.98  10 7 km
sin  Max 

      Max  28.330
8
d PerihelionEarth 1.471  10 km
sin  Max 
d AphelionVenus
d PerihelionEarth
1.089  108 km

      Max  47.76 0
8
1.471  10 km
17) Proof that Moon’s Tidal Force is Twice as Strong as Sun’s Tidal force on Earth
The tidal force of the moon and sun is due to the difference of the gravitational force
between their centers and the front and back sides of the earth. We will use the previous
data given and also the radius of the Earth, Rearth = 6378 km = 6.378 x 106 meters and
mass of Moon = 7.35 x 1022 kg
FMoon 
GM moonM earth
d Moon/ Earth  rEarth 
2

GM moonM earth
d Moon/ Earth  rEarth 
2
 (6.67259 x10 11 )(5.974 x10 24 )(7.35 x10 22 )


1
1

  1.31668 x1019 Newtons

8
6 2
8
6 2 
(3.844 x10  6.378 x10 ) 
 (3.844 x10  6.378 x10 )
GM SunM earth
GM SunM earth
FSun 

 (6.67259 x10 11 )(5.974 x10 24 )(1.989 x10 30 )
2
2
d Sun/ Earth  rEarth  d Sun/ Earth  rEarth 

1
1


11
6 2
11
(1.4959787 x10  6.378 x10 6 ) 2
 (1.4959787 x10  6.378 x10 )

  6.04174956575 x1018 Newtons

Now, take the ratio of these numbers and their absolute values:
FMoon
1.31668 x1019

 2.179 Therefore, Moon’s tidal force is at least twice
FSun
6.04174956575 x1018
as strong.
18) Calculating the Range of Frequencies of the Water Hole:
The water hole is a range of frequencies used in the SETI project. It lies between the
absorption line of Hydrogen (H) and Hydrogen Peroxide (OH). The absorption line of
hydrogen has a wavelength, H = 21 cm = .21 meters. The absorption line of hydrogen
peroxide has a wavelength, 18 cm = .18 meters. Plugging these into the equation c
= f for EM waves, we can solve for the corresponding frequencies of these
wavelengths. We will use the value, c = speed of light = 2.99792458 x 108 meters/sec.
fH = c/H = (2.99792458 x 108 m/s)/(.21 m) = 1.428 x 109 Hz = 1,428 MHz (this would
be station 1428 on the FM dial, which usually doesn’t exist for most radios)
fOH = c/ = (2.99792458 x 108 m/s)/(.18 m) = 1.666 x 109 Hz = 1,666 MHz (this
would be station 1666 on the FM dial, which also usually doesn’t exist for most radios)
Therefore, this frequency range is free from human “noise”. If you ever get a radio with
these stations and hear conversation, then you are listening to Aliens!!!!!!!!!!!
19) How to Calculate the Velocity of a Star or Galaxy using the Doppler Shift:
When a star is receding from us (moving away), the frequency of the waves its sends out
to us seem to decrease and thereby, the wavelengths increase toward the Red end of the
spectrum. We call this a Red Shift. The opposite happens when a star is moving towards
us and its wavelength Blue shifts. To find this amount of shift, we have to use the
Relativistic Doppler Equation for light. Here is the equation for a receding star.
f observer 
1  v / c 
f source If the star is approaching us, just change the top sign to +
1 v / c
and the bottom sign to -. Here, fobserver is the frequency observed by Earth people, and
fsource is the frequency emitted by the star or galaxy. We can rewrite this equation in
terms of the emitted and observed wavelengths, using the equation c = f. We get ---->
1  v / c  c
1  v / c 
c
f observer 


     observer 
source
observer
1  v / c source
1 v / c
Now, we can use the regular Balmer lines of hydrogen, red = 656.3 nm, Green = 486.1
nm, and Violet1 = 434.0 and Violet2 = 410.2 nm. Let’s say you are observing the spectral
lines of a star and you notice that its hydrogen Balmer lines are all shifted toward the red
end. For example, you observe, RedObserver = 656.85 nm. However, you know that a
stationary star would give off 656.3 nm. That means this star is receding from us.
1  v / c 
2
 1  v / c  
1 v / c
 656.85 


656.85 
656.3   

 1.00168
   


1 v / c
1 v / c
 656.3 
 1 v / c 
1  v / c  1.00168  1.00168v / c    2.00168v / c  .00168    v  .00084c
That means that the star is moving away from us at .084% of the speed of light, which is
251,788.5 m/s = 566,524.12 mph. That is a fast speed!!!!!!!!!
2
The difference of the two wavelengths divided by the source wavelength is known as the
“Red Shift” factor z, where
1  v / c 
1  v / c 1  v / c 
1  v / c   1


1 
1 
z = source =
source
1 v / c
(1  v / c)1  v / c 
(1  v 2 / c 2 )
1
The term
is known as the relativistic factor When v/c is a small number
(1  v 2 / c 2 )
 (gamma) approaches 1, so z approaches 1 + v/c – 1 which equals v/c. [Ultimately,
when v =0, therefore, z = 0 and there is no red shift] . You can also write the above
equation in terms of v/c, and  so 
z 1 
1   
  (1 
 2 1
)     2 1

(1   2 )
So, for objects that are receding away from us at non-relativistic speeds, z = v/c =
source is a good approximation of the wavelength shift experienced by the star.
Applying this to the previous problem, we get, (656.85 – 656.3)/656.3 = v/c ------> v =
.000838c = .838% of the speed of light which is similar to the answer we got previously.
For objects in the universe with very large red shifts such as z = 8, their recessional
velocities are definitely relativistic.
1  v / c  -------> 81  1  v / c 2
z = v/c) – 1 -----> 9 =
(1  v 2 / c 2 )
(1  v 2 / c 2 )
81  81v 2 / c 2  1  2v / c  v 2 / c 2      82v 2 / c 2  2v / c  80  0 which is like solving
the quadratic equation 41x2 + x – 40 = 0 ------> (41x - 40)(x + 1)= 0, x = 40/41 = .9756,
so objects with this much red shift are going 97.56% the speed of light. They represent
the first stars and galaxies that were born after the recombination event of the Big Bang.
20) How to Calculate the Rotational Velocity of a star using the Doppler Shift:
When a star spins as well as moving away or towards us, its spectral lines widen due to
the fact that the side of the star spinning toward us experiences a blue shift and the side of
the star spinning away from us experiences a red shift. Let’s take the star in the above
example. Suppose that you noticed its Red Balmer Line is centered at 656.85 but it has a
width of .72 nm. That means the side spinning away from us has a wavelength ---->
SideSpinningAway = 656.85 + (.72/2) = 657.21 nm, and the side spinning towards us ------->
SideSpinningTowards = 656.85 – (.72/2) = 656.49 nm. Now, we can apply the same technique
as in the previous problem to find the velocity of recession of both sides. We get:
VSideSpinningAway = .00139c & VSideSpinningTowards = .000289c. These give us the total
velocity of each side with respect to Earth. By subtracting these by the velocity of the
star itself, we get the velocity of each side with respect to the star.
VSideSpinningAwaywithrespecttoStar = .00139c - .00084c = .000055c = 165,000 m/s = 165 km/s
VSideSpinningTowardswithrespecttoStar = .000289c - .00084c = -.000055c = -165,000 m/s = -165 km/s
This star is spinning at 165 km/s at its perimeter, which is an extremely fast spin rate. It
probably is a Pulsar. Let’s assume we know its radius to be .1% of the sun’s radius.
Rstar = .001Rsun = .001(6.96 x 105 km) = 696 km. Then, its Rotational Velocity of spin,
Spin = VEitherSide/RStar = 165/696 = .237 rads/sec. Its period of Rotation, T = 2/ =
2/(.237) = 26.5 seconds. If this pulsar was radiating energy towards the Earth, we would
see a pulse of light every 26.5 seconds. Many such pulsars have been discovered!!!
21) Proof that Earth’s Escape Velocity is 11 km/sec:
To calculate the escape velocity from any object, we have to set the sum of the Kinetic
Energy (KE) and Potential Energy (PE) of the escaping rocket equal to their sum at
infinity. The argument we use is that the total energy (E) of the rocket should be a
constant. Therefore, Etot = KEat EARTH + PEat Earth = KEat infinity + PEat infinity
GM rocket M Earth
M
v2
We’ll use the following equations for KE & PE, KE  rocket , PE  
2
d
where d = distance between the rocket and the Earth’s center. When the rocket is at the
surface of the Earth, d = Rearth = 6.378 x 106 meters , and when the rocket is at infinity, d
=  . To find the escape velocity, we assume that the rocket barely escapes the surface
of the Earth and by the time it makes it out to infinity, its velocity is zero; therefore, --------->KEat infinity = 0. Putting this all together, we get:
M rocket v 2  GM rocket M Earth
 GM rocket M Earth

 KE atInfinity 
 0  0  0     
2
REarth

v 2 GM Earth

 0      v EscapeVelocity 
2
REarth
2GM Earth

REarth
2(6.67259  10 11 )(5.974  10 24 )
6.378  10 6
= 11,180.28 meters/sec = 11.18 km/sec =25,155 mph. That’s a fast speed!!!
(Notice that the Mrocket in the above equation cancelled out, so this answer applies for any
size rocket or object). However, this applies only to objects which are initially stationary
such as in the air or on the North or South Pole. If a rocket is on the equator of the Earth, it
already has some tangential velocity to the Earth = 2R/T =
2x106)/ 23.9344721138 hours x (1 hr/3600sec) = 465.09 m/s, so part of the needed
11,180.28 m/sec is provided by this tangential velocity. If a natural satellite NEAR the
surface of the Earth wants to escape the Earth’s gravity, then it has even more Kinetic
Energy due to the gravity of the earth, so now we have to use the total energy of the
satellite to compute the escape velocity,
M rocket v 2  GM rocket M Earth
 GM rocket M Earth

 KE atInfinity 
 0  0  0     
2
2 REarth

GM Earth
v 2 GM Earth
(6.67259  10 11 )(5.974  10 24 )

 0      v EscapeVelocity 

2
2 REarth
REarth
6.378  10 6
This yields 7905.65 meters/sec = 7.91 km/sec. The reason for this is that its kinetic
energy mv2/2 can be computed due to the fact that GMm/R2 = mv2/R and therefore,
mv2/2 = GMm/2R, so its total energy PE + KE = -GMm/R + GMm/2R = -GMm/2R!
22) How to Calculate the Diameter of Earth’s Liquid Core:
When an earthquake happens anywhere in the world, it releases two kinds of waves, p
and s waves. The p waves are longitudinal primary waves, travel faster, and they can
travel through solids and liquids. The s waves are transverse secondary waves, travel
slower, and then can ONLY travel through solids. Therefore, when an earthquake
happens, it sends out waves in all directions and even through the Earth.
Epicenter
1
Both S& P waves
7
2
Liquid Outer Core
Both S & P waves
3
6
Both S and P waves
4
Only P waves
5
Only P waves
Cities that are at the end of waves 1, 2, 3, 6, or 7 receive both s and p waves. However,
cities between waves 3 to 6, such as 4 and 5 receive no S waves. All we have to do is
find out the location of the two cities at the end of rays 3 and 6. Let’s says the two cities
are 7500 miles apart. Using this info, we can find the radius of the Liquid core:
RLiquidCore
 .Center of earth

Rearth
Rearth
S = 8500 miles
We will use the equation, S = Rearth, where Reartth = 6378 km = 3986.25 miles.
8500 = 3986.25 ---->  = 2.1323 radians = 122.17 degrees. That means  in the above
picture = 180 -  = 180 – 122.17 = 57.830. Then, using Trig we get:
RLiquidCore = REarthcos = 6378cos57.83 = 3395.86 km.
The actual radius of the Liquid core of the Earth is about 3400 km !!!!!!!!!!!!!!!!!!
23) Calculating the Predicted Average Temperature of the Earth:
We can follow up on the previous calculations to predict what should be the average
temperature of the Earth. In the last problem, we saw that half the surface of the Earth is
lit at any time and receives 3.55357 x 1017 watts of power. Let’s assume that this power
gets distributed evenly around the Earth. Then the Earth reradiates some of that power
according to the same equation, P = AeT4, where e = emissivity of the Earth. [The
emissivity of an object is a measure of how much energy it emits and is between 0 and 1.
The Emissivity = (1 – Albedo) of an object, where albedo is a measure of how much
energy it absorbs in relation to how much it receives. A “black body” has an albedo of 0
and therefore, absorbs all of the energy falling on it. The sun acts as a “black body”
when it radiates energy and therefore, its albedo = 0 and e = 1 as we saw in the previous
problem. The Earth’s albedo has been measured to be about .37, and therefore, its
emissivity = 1 - .37 = .73!!] The Earth reaches an equilibrium temperature when the
power it receives equals the power that it reradiates. Therefore, we get ---------->
2
3.55357  1017  (5.6696  10 8 )(4REarth
)(.73)T 4  (5.6696  10 8 )4 (6.378  10 6 ) 2 (.73)T 4
     16796186996.2  T 4        TAvgEarth  360 Kelvin
From these calculations, we predict that the average temperature of Earth should be 360
Kelvin. Using the equation, K = C + 273, this is equivalent to 870 Celsius = 188.60 F.
The actual Mean Temperature of the Earth is about 287 K = 140 Celsius = 570 F!!! The
reason that our calculation predicted a higher temperature is that we assumed all of the
power received from the sun shines at 90 degrees (perpendicular to Earth’s surface) and
gets spread around the Earth. In reality, the sun’s beam shines at different angles
depending on the latitude of the Earth and the season; at some locations, it comes in at a
glancing angle because the Earth is spherical and it is also tilted at 23.5 degrees. We can
calculate this average angle of incidence from the above formula:
3.55357  1017 sin   (5.6696  10 8 )4 (6.378  10 6 ) 2 (.73)(287) 4
     sin   .40394         AvgIncidentonEarth  23.82 deg rees
24) How to Calculate the Eccentricity of Earth’s Orbit and the Oblateness of the
Earth:
The oblateness of a planet is a measure of how spherical it is. A planet which is perfectly
spherical has an oblateness of 0, but a planet that is very oval has a higher oblateness
closer to 1. The oblateness of the Earth is .0034 because the equatorial diameter of the
Earth is larger than its polar diameter. This is also true for most other planets. The
Requatorial  R polar Requatorial  R polar

equation for oblateness is -----------> Oblateness 
Requatorial  R polar
Raverage
2
The equatorial radius of the Earth = 6378.1 km and its polar radius = 6356.8 km. Putting
these into the equation for oblateness, we get -------->
6378.1  6356.8
Oblateness 
 .00335 (Saturn has the highest oblateness of .1)
6378.1  6356.8
2
The eccentricity is similar to oblateness and it describes how oval the orbit of the planet
is, not the planet itself. If the orbit is perfectly circular, the eccentricity = 0 and if the
orbit is very oval, the eccentricity is closer to 1. The eccentricity depends on the planet’s
closest approach to the sun (dPerihelion) and its farthest approach distance from the sun
(dAphelion). For the Earth, its closest approach is around January 3 of each year and its
farthest approach is around July 4 (independence from the sun!!!). The equation is:
d Aphelion  d Perihelion
For the Earth, dPerihelion = 1.471 x 1011 meters and
Eccentrici ty 
d Aphelion  d Perihelion
dAphelion = 1.521 x 1011 meters. Therefore, e = (1.521 – 1.471)/(1.521 + 1.471) = .0167
The most circular orbit belongs to Venus whose eccentricity is .007 and the most
elliptical orbit is Pluto’s with an eccentricity of .248.
25) How to Calculate the Diameter of the Moon:
To calculate the diameter of the moon, take a coin of and hold it in your hand. Go
outside when the moon is a Full Moon. Adjust the distance of the coin until you are able
to completely cover the moon. For example, let’s say you use a quarter whose diameter
is 1.56 cm. Let’s say you can completely cover the moon when the quarter is 122.3 cm
away from you. We will use the fact that dAvgbetweenEarth/Moon = 384,400 km!!
Then the diameter of the moon is --------->
Dquarter
Dmoon
 1.56 

     Dmoon  (384,400)
  4903.22 km
dis tan ceto moon dis tan ceto quarter
 122.3 
The actual diameter of the moon = 2Rmoon = 2(1738 km) = 3476 km!!!
26) How to Calculate the Diameter of the Sun:
To calculate the size of the sun, get a piece of cardboard and poke a small hole in it.
Then, tilt it towards the sun facing the sun at a 90 degree angle. On the other side of the
cardboard, place a piece of paper about a meter away also tilted perpendicular towards
the sun. You should see an image of the sun on the paper. Measure the diameter of that
image and the distance between the paper and the cardboard. Let’s say, the diameter of
the image comes out 1.23 cm = .0123 meters and the Distance between paper and
cardboard = 1.10 meters. We will use the fact that the average distance between the
Earth and the Sun = 1 AU = 1.495978706 x 1011 m. Then use the equation:
Dsun
DbetweenEarth / Sun

Diameter of Im age
 .0123 
     Dsun  (1.495978706  1011 )

DbetweenCarboard& Paper
 1.1 
1.673  10 9 meters  1.673  10 6 km
The actual diameter of the sun = 2Radiussun = 2(6.96x105 km) = 1.392 x 106 km
The reason that the above equation works, is that the Angular diameter of the image is
equal to the Angular diameter of the sun if the hole in the cardboard is relatively small!!
27) How to Calculate the Density of the Earth, Moon, and Sun:
The density of an object is the ratio of its mass to its volume. We will assume that the
earth, moon, and sun are perfectly spherical bodies even though they are not quite
spheres. The volume of a sphere is 4/3R3. Therefore, we get
M Earth
M
5.974  10 24 kg
kg 1000 g  1m 
3
DEarth 


 5496.97 3 

  5.50 g / cm
4 3
4
V
1
kg
100
cm
m


REarth
 (6.378  10 6 meters ) 3
3
3
The Earth is the densest planet in the solar system, because it has a massive iron core.
3
M Moon
M
7.35  10 22 kg
kg 1000 g  1m 
3
DMoon 


 3342.33 3 

  3.34 g / cm
4 3
4
V
1kg
m
 100cm 
RMoon
 (1.738  10 6 meters ) 3
3
3
The Moon’s density resembles the density of the Earth’s crust, since it is primarily rock!!
3
M Sun
M
1.989  10 30 kg
kg 1000 g  1m 
3



 1408.38 3 

  1.41g / cm
4 3
4
V
1
kg
100
cm
m


RSun
 (6.96  10 8 meters ) 3
3
3
3
DSun
The density of the sun is less than the moon’s and earth’s even though it is much more
massive than either one. This is because the sun occupies a greater volume; it is
composed mainly of hydrogen and helium which are the lightest elements. However, it
still is denser than water which has a density = 1 g/cm3. There is one planet which has a
density less than water and will float on water. Which planet is that????
28) Predicting the Equation governing the Temperature versus Spectral Class of
Stars & using Wein’s Law to estimate the Temperature of stars:
In this problem, we want to predict an equation that governs the Temperature of stars
versus their spectral types. Stars are classified by the kind of spectra they give as O, B,
A, F, G, K, M. It can be easily memorized by the pneumonic, “Oh, be a fine girl/guy and
kiss me!” They are further subclassified into 10 parts like O0, O1, O2, O3, … O9, B0, B1,
… B9, … M0, M1, M9. O0 stars are the hottest stars and are about 50,000 Kelvin. M9
stars are about 2200 Kelvin and are the coolest stars. Our Sun is a G2 star and its surface
temperature is 5800 Kelvin. Cooler than the M stars are the Brown Dwarf L and T stars
which aren’t really stars. L brown dwarfs range from 1300-2200 Kelvin and T brown
dwarfs range from 900 – 1300 Kelvin. Let x = 1 stand for the coldest Brown dwarf (T9)
which is 900 Kelvin. Then x = 2 corresponds to T8, x = 10 corresponds to T0, x =11
corresponds to L9, x = 20 corresponds to L0, etc. In this scale, x = 48 would correspond
with to our sun (G2) and would have a temperature of 5800 Kelvin, and x = 90 would
correspond with an O0 star and have a temperature of 50,000 Kelvin. Let’s try to fit this
data into an exponential function of the form ------> T = Aebx + C where A, b, & C are
some constants which we have to solve for. We can now write down 3 equations:
900 = Aeb(1) + C,
5800 = Aeb(48) + C, 50,000 = Aeb(90) + C. By combining the first
and second equation, we get:
900  C
5800  C
 A
     900e 48b  Ce 48b  5800eb  Ceb     
b
e
e 48b
900e 48b  5800eb
900e 48b  5800eb  C (e 48b  eb )      C 
e 48b  eb
We can also combine the 1st and 3rd equation and get another equation for C:
900  C
50,000  C
 A
     900e 90b  Ce 90b  50,000e b  Ce b     
b
e
e 90b
900e 90b  50,000e b
900e 90b  50,000e b  C (e 90b  e b )      C 
e 90b  e b
Then, we can set the 2 equations for C equal to each other and solve for b:
900e 48b  5800e b 900e 90b  50,000e b

     900e138b  900e 49b  5800e 91b  5800e 2b 
e 48b  e b
e 90b  e b
900e138b  50,000e 49b  900e 91b  50,000e 2b      0  4900e 91b  49,100e 49b  44,200e 2b
     0  49e 89b  491e 47b  442
Let z = e47b and the equation reduces to ---> 0 = 49z(89/47) – 491z + 442 which could be
solved on a Ti calculator using the Solver button. We get z = 12.09. Setting this equal
to e47b, we can now solve for b ----------> b = ln(12.09)/47 = .053!!! Now, we can solve
900e 90(.053)  50,000e (.053)
for C ------> C 
 457 . Then, we can finally solve for A.
e 90(.053)  e (.053)
A = (900 – 457)/e.053 = 420.13. Therefore, the general equation for T versus x becomes:
T = 420.13e.053x + 457 Kelvin. From this formula, we can predict the temperature for
any spectral type. Let’s say we want to know the temperature of the coolest start M9,
which corresponds to x = 21 in our scale. Then, T = 420.13e(.053)(21) + 457 = 1735.66,
which is cooler than what we mentioned above 2200 but close to it. The equation is not
meant to be an exact formula but an approximation to the behavior of stars.
We can combine the above equation with Wien’s law which relates the temperature of a
body with the maximum wavelength of radiation that it emits. The formula becomes:
3  10 6
3  10 6
According to Wien’s law, cool stars radiate
Max (nm) 

T
420.13e .053x  457
mostly in the high wavelength region like Red, Infrared, and Radio waves, whereas hot
stars radiate mostly in the low wavelength region like Violet, Ultraviolet, X-ray. The
visible spectrum Red, Orange, Yellow, Green, Blue, Indigo, Violet can be remembered
by the name Roy G. Biv. Red has a wavelength of about 700 nm = 700 x 10-9 m =
7 x 10-7 m and Violet has a wavelength of about 400 nm = 4 x 10-7 m. Let’s do an
example using the star Spica which is a B1 star, which corresponds to x = 79. We get:
T = 420.12e(.053)(79) + 457 = 28,111.41 Kelvin. Therefore, it will radiate Most Strongly
Max = (3 x 106)/28,111.41 = 106.72 nm which is a shorter wavelength than violet and lies
in the Ultraviolet range. However, the star will look violet to the human eye because we
can’t see ultraviolet waves!!
29) Calculating the Luminosity of the Sun & the Solar Constant and the Luminosity
of other stars using the Stefan-Boltzmann Law:
The power emitted by an object is given by the formula P = AeT4 where T is its
temperature in Kelvins, A = surface area of the object in meter2 = 4R2 for a spherical
object,  is the radiation constant = 5.6696 x 10-8, and e = emissivity of the object which
is between 0 and 1. We will use Rsun = 6.96 x 105 km = 6.96 x 108 meters and Tsun =
5,800 Kelvin. The emissivity of the sun is about 1 because it is considered a “black
body” object which radiates at all wavelengths. Therefore, the Luminosity becomes,
Lsun = Psun = (5.6696 x 10-8)4(6.96 x 108)2(1)(5800)4 = 3.91 x 1026 watts
This is the actual luminosity of the sun!!! It radiates 3.91 x 1026 Joules per second.
The solar constant is the amount of watts per meter2 falling on the Earth. Let’s assume
that the sun radiates uniformly in all direction in a spherical fashion. Then, part of this
energy falls on the Earth and illuminates half the surface of the Earth at any one time.
Let’s calculate the ratio of half the surface of the Earth to the Surface area of the sphere
extending from the sun to the earth. Then, multiplying this by Lsun, we get,
Amount of Power Falling on Earth ------------------------>
2
 4REarth



(6.378  10 6 ) 2 (3.91  10 26 )
2
L

 3.55357  1017 watts

 sun
2
11 2
2(1.495978706  10 )
 4d betweenEarth / Sun 


The solar constant is this number divided by Half the Area of the Surface of the Earth:
C = (3.55357 x 1017/(4R2Earth)/2 = (7.10714 x 1017)/(4(6.378 x 106)2) = 1390.3 watts/m2.
The actual Solar Constant is 1370 w/m2. Imagine if we could harness some of that power!!
Now, we go on to calculate the luminosity of other stars. We use the Stefan-Boltzmann
law which states that the luminosity of any object is proportional to its temperature to the
power 4 and to its radius to the power 2, Lstar  T 4 R 2 This is because the surface area of a
spherical object is 4R2 and each square area of its surface radiates proportional to its
temperature to the power 4. We can use this equation to calculate the relative luminosity
4
2
T   R 
of a star compared to the sun. We get: Lstar  Lsun  star   star 
 Tsun   Rsun 
As an example, if a star is 3 times hotter than the sun (Tstar = 3 x 5800 = 17400 Kelvin), and
8 times its size, then Lstar = Lsun(3)4(8)2 = 5184Lsun. It’s 5 thousand times brighter!!
The hottest stars are about 50,000 Kelvin and so they are 50,000/5800 = 8.62 times as hot
as the sun, since the sun’s surface temperature is 5800 Kelvin. The biggest stars are
about 1000 times bigger than the sun and are called Bright Supergiants 1a. If a star was
the hottest that it could be and the biggest, then its Lstar = (8.62)4(1000)2 = 5.52 x 109Lsun.
However, there is no star that is that bright; that’s because the really hot stars are not
1000 times bigger than the sun. In the next problem, we will show that the brightest stars
are about 883,883.48 times the luminosity of the sun. Let’s assume that a 50,000 kelvin
star has this absolute luminosity; what would its size be?? We solve it like this ---------->
883,883.48 = (8.62)4R2 -------------> R = 12.65. So, really hot and bright Main Sequence
stars are only about 12 to 13 times the size of the sun. As these star die, they move off
the main sequence and their temperature eventually cools to about 4000 kelvin which is
about .7 the temperature of the sun. What would their size be then? We get -------------->
883,883.48 = (.7)4R2 ------> R = 1918. These are the supergiant stars. (In the above
calculations, we assumed that their luminosity stays constant, but that is not necessarily
true!!) We can do a similar calculation to the above to calculate the size of the Red
Dwarfs, which are the cool and dim Main Sequence stars. They tend to be about 2500 –
3000 Kelvin which is ½ the temperature of the sun, and their luminosity is about 2 x 10-4.
What would their size be? 2 x 10-4 = (.5)4R2 -----------> R = .0565 They are about 5%
the size of the sun which is still about 5 times bigger than the White Dwarfs which are
dead remnants of low or middle-mass stars. White Dwarf tend to be about the size of the
Earth which is about 1% the size of the sun!!!!!!!!
30) Calculating the Lifespan of our Sun and other Stars:
The lifespan of a star depends mainly on how massive it is. The more massive it is, the
more molecules and atoms it has, and one would think that it should live longer.
However, massive stars also produce a lot of energy, fusing millions and billions of tons
of hydrogen every seconds and they run out of “fuel” quickly. The formula for the mass
dependence of Luminosity is known as the Mass-Luminosity relationship
Lstar = Lsun(Mstar/Msun)3.5. For example, if a star is 10 times heavier than the sun, its
luminosity, L = Lsun(10)3.5 = 3162.28Lsun. The heaviest stars are about 50 times the mass
of the sun and the lightest stars are about .08 times the mass of the sun; therefore,
LBrightest Star = 503.5LSun = 883,883.48 LSun & LDimmest Stars = (.08)3.5 LSun = 1.45 x 10-4 LSun
To calculate the lifespan of a star, we know that a star starts its death process when it runs
out of hydrogen. Most stars use the proton-proton chain as well as other processes to
create energy; our sun primarily utilizes the proton-proton chain. It fuses 4 hydrogen
atoms into 1 helium atom and produces 4.3 x 10-12 Joules of energy per each helium atom
produced. Our sun’s mass is about 1.989 x 1030 kg, 74% of which is Hydrogen by mass.
About 25 – 30% of the sun’s mass is concentrated at its core, and that is where energy
creation takes place. Earlier, we calculated that the luminosity of the sun was about
3.91 x 1026 watts. Therefore, we get:
M hyrdogenin Core .25  .74  1.989  10 30 kg 1000 grams 1moleH 6.02  10 23 H 4.3  10 12 Joules
t Lifespan Sun 





Lsun
1kg
1gram
1mole
4 Hatoms
26 Joules
3.91  10
sec
1hr
1day
1year
 6.090  1017 sec 


 19.3 billion years
3600 sec 24hrs 365.25days
This crude calculation doesn’t take into account many other complex factors, such as not
all of the present hydrogen is utilized before it starts its death process, and its luminosity
gets higher as time goes on while it is still in its Main Sequence Stage. The actual life
expectancy of our sun on the Main Sequence is about 10 billion years.
Now, to predict the lifespan of other stars, we just divide their relative mass compared to
the sun divided by their relative luminosity compared to the sun, and we multiply that
ratio by 10 billion years. We get:
t Lifespan Star
M
 star 
Lstar
M star
3.5
M 
M
 sun   sun 
Lsun  M star 
2.5
2.5
M 
 10 sun  billion yrs
 M star 
M 
Lsun  star 
 M sun 
Therefore, tlongest lifespan = 10(1/.08)2.5 = 10(12.5)2.5 = 5,524.3 billion yrs (this is longer than
the life of our known universe!!!!!!!!!). tshortest lifespan = 10(1/50)2.5 = 565.7 thousand yrs
(this is very short compared to our sun, but it is still longer than a human lifespan!!!)
31) Calculating the Hipparchus-Scale Absolute and Apparent Luminosity for Stars:
The Hipparchus scale is a logarithmic scale of measuring luminosities and it gets its name
from the Greek astronomer Hipparchus who was the 1st cataloguer of stars and their
brightness. Hipparchus gave the brightest stars a number 1 and the dimmest stars a
number 6; therefore, as the numbers went higher the stars got dimmer. Today, we still
keep the same tradition and we classify both the Absolute and Apparent luminosity of
stars on the Hipparchus scale. First let’s deal with the absolute luminosity of stars.
The sun gets a 4.83 on this scale. Brighter stars get numbers less than this and Dimmer
stars get numbers larger than this. The equation is --------------->
L 
L AbsoluteHipparchus  2.5 log 10  star   4.83
 Lsun 
If a star is 100 times brighter than the sun, then LAbsoluteHipparchus = -2.5log10100 + 4.83 =
-2.5(2) + 4.83 = -5 + 4.83 = -.17
If a star is 10,000 times brighter, then LAbsoluteHipparchus = -2.5log1010,000 + 4.83 =
-2.5(4) + 4.83 = -10 + 4.83 = -5.17
If a star is 1,000,000 times brighter, then LAbsoluteHipparchus = -2.5log101,000,000 + 4.83 =
-2.5(6) + 4.83 = -15 + 4.83 = -10.17 (This is the upper limit of the brightness of stars)
Notice from above that every 100 fold difference in the brightness of a star changes its
Hipparchus scale luminosity be 5!!! Therefore, if a star is 100 times dimmer, than its
Hipparchus Scale brightness should be 4.83 + 5 = 9.83 and 10,000 times dimmer should
be 9.83 + 5 = 14.83. These are the dimmest stars.
In our sky the brightest star is Deneb which is 3.2 x 105 times as bright as the sun.
Therefore, LAbsoluteHipparchusDeneb = -2.5log10(3.2 x 105) + 4.83 = -8.93!! The dimmest star
is GJ 1111 which is 2.7 x 10-4 times as bright as the sun. Therefore, LAbsoluteHipparchusGJ1111
= -2.5log10(2.7 x 10-4) + 4.83 = 13.75. Now, let’s go on to Apparent Luminosity.
The Apparent Luminosity of a star is how bright it appears to us. Therefore, it is a
measure of the intensity of light that reaches us. Since light goes out from a star in a
spherical fashion, its intensity drops as 1/r2 away from a star. We get:
L Apparent 
Lstar
2
4d Star
/ Earth
For the sun, we get:
3.91  10 26 watts
 1390.32 watts / m 2 This is the same as the
4 (1.495978706 x1011 ) 2 meter 2
Solar Constant. The sun is the brightest looking object in the sky for us. A star with the
same luminosity as the sun but twice as far away from us will look ¼ as bright and three
times as far away will look 1/9 as bright.
L Apparent 
We can make a similar Hipparchus scale reading for apparent luminosities. Since the sun
is the brightest appearing object, it gets the most negative number on this scale, -26.7!!!
Then, we can form an equation for the apparent luminosity of other stars based on their
distance from us and their absolute luminosity as compared to the sun. The equation is:
L star




2
 L ApparentStar 
4d Star / Earth 

  26.7  2.5 log 10
L ApparentHipparchus  2.5 log 10 

  26.7 
L

L
sun
ApparentSu
n




2
 4d Sun / Earth 
L d2



L star (1AU ) 2
  26.7
 2.5 log 10  star 2Sun / Earth   26.7  2.5 log 10 
2
2 
 L sun d Star / Earth 
 L sun d Star / Earth (63236.19 AU ) 
Here, we are using the fact that 1 Ly = 63236.19 AUs. Let’s see an example:
Let’s take Deneb again. It’s Luminosity = 3.2 x 105Lsun and its distance from us = 3230
light years. Using the above equation, we get the following for its apparent luminosity:
 (3.2  10 5 )(1AU ) 2

  26.7  1.09 This is a bright
L ApparentHipparchus  2.5 log 10 
2
2 
 (3230) (63236.19 AU ) 
looking star. Later, we will calculate the brightness of the brightest looking star.
For the other star, GJ1111, L = 2.7 x 10-4 and its distance from us = 11.82 light years.
 (2.7  10 4 )(1AU ) 2 
  26.7  11.59 This is a very dim
L ApparentHipparchus  2.5 log 10 
2
2 
 (11.82) (63236.19 AU ) 
looking star. As a matter of fact, it is not visible to the naked eye. Any star whose
apparent luminosity is greater than 6 on the Hipparchus scale is not visible to the naked
eye. Let’s calculate the apparent luminosity of Sirius which is the brightest looking star
in the sky. LSirius = 26.1Lsun and dSirius = 8.61 light years. We get:


(26.1)(1AU ) 2
  26.7  1.56 This is the most
L ApparentHipparchus  2.5 log 10 
2
2 
 (8.61) (63236.19 AU ) 
negative star on the Apparent Hipparchus scale. We can also calculate the apparent
luminosity of planets and the moon on this scale. Let’s use the Moon as an example!
The Luminosity of the moon is due to the reflection of sunlight from its surface. The
moon’s average albedo is .07 which means that it reflects 7% of the sun’s power reaching
it. The sun’s power reaching the moon is equal to the solar constant at the surface of the
moon times half the surface area of its surface. The moon looks the brightest during Full
Moon phase when it is the farthest from the sun. The solar constant is the same as the
apparent brightness of the sun as viewed from the full moon. Therefore, we get:
3.91  10 26 watts
 1383.21 watts / m 2
11
8 2
2
4 (1.495978706 x10  3.844  10 ) meter
LReflected from Moon = .07CMoon(4R2Moon)/2 = .07(1383.21 watts/m2)(2(1.738 x 106 m)2) =
1.838 x 1015 watts!!!!!!!!
L ApparentSun from Moon 
This acts as the absolute luminosity of the moon even though it is not a star. We now get:
 L d2

 (1.838  1015 )(1.495978706 x1011 ) 2 
  26.7
L ApparentHipparchusMoon  2.5 log 10  moon 2 Sun/ Earth   26.7  2.5 log 10 
(3.91 x 10 26 )(3.844  108 ) 2


 Lsund Moon/ Earth 
= -11.33!!! The Full Moon is the brightest looking object after the sun.
32) Combining Hipparchus Scale Absolute and Apparent Luminosity into a Single
Equation:
We can combine the 2 equations we obtained for Hipparchus scale luminosities from the
previous problem and obtain another interesting equation. From the 1st equation we get:
 ( LAbsolute Hipparchus 4.83) 


 2.5



 Lstar 
Lstar
  4.83   
LAbsoluteHipparchus  2.5 log 10 
 10 
Lsun
 Lsun 
Then, plug this expression into the Apparent Luminosity equation:


Lstar (1AU ) 2
  26.7 
L ApparentHipparchus  2.5 log 10 
2
2 
L
d
(
63236
.
19
AU
)
sun
Star
/
Earth


(
L

4
.
83
)




  AbsoluteHipparchus



 2.5
2

2


 2.5log 10 10
  log 10 d   log( 63236.19)   26.7  L AbsoluteHipparchus  4.83  5 log 10 d





 5 log( 63236.19)  31.53  L AbsoluteHipparchus  5 log 10 d  7.5252
Often times, LApparent Hipparchus is written as mv and LAbsolute Hipparchus is written as Mv. Then,
this equation becomes, mv = Mv + 5log(d) – 7.5252 where d is measured in light years.
It is more common to see this equation written where d is measured in parsecs. We’ll use
the conversion that we derived earlier, 1 parsec = 3.262 light years. Now, we get:
mv = Mv + 5log(3.262d) – 7.5252 = Mv + 5log3.262 + 5log(d) – 7.5252 ------->
mv = Mv + 5log(d) – 5 = Mv + 5log(d/10) where d is measured in parsecs.
According to this equation, if the distance to a star is 10 parsecs, then:
mv = Mv + 5log(10/10) = Mv Therefore, if all stars were 10 parsecs away from us, then
their apparent brightness would equal their absolute magnitudes.
We could also rewrite this equation in terms of the distance d. We get:
m  Mv
d
d
mv  M v  5 log     v
 log     d  10
5
 10 
 10 
  mv  M v  1 
  5  






This equation allows one to calculate the distance to a star in units of parsecs if they
know their Hipparchus scale apparent and absolute magnitudes. As an example, let’s
take the star Betelgeuse. Its mv = +.45 and its Mv = -5.14; therefore,
d  10
 .45( 5.14) 
1

5


 131.22 par sec s 
3.262 ly
 428 ly !!!
1 par sec
33) Calculating the Distance to a Star or Galaxy using Cepheid Variables:
When a star is in the process of dying, it goes through the instability strip and begins to
pulsate; these stars are called variable stars. Low mass stars become RR Lyrae variables,
medium mass stars become Type 2 Cepheid variables, and high mass stars become Type
1 Cepheid variables. The pulsation period of these variable stars is tied to their mass and
luminosity. The more luminous the star, the larger is the pulsation period and the slower
it pulsates. Therefore, if one observes the pulsation period of these stars, they can
determine their absolute luminosity. Then, they can use a photometer to determine their
apparent luminosity. Once both of these values are known, the distance to the variable
star can be determined, and therefore the distance to the galaxy that houses the star can be
determined. So, variable stars are a great Distance Indicator. Here is the graph that
governs the Luminosity versus Pulsation period relationship:
Luminosity
100,00010,000-
Type 1 (Classical) Cepheids
1000Type 2 Cepheids
100RR Lyrae
100
0
.5
1
3
10
30
100
Period (days)
We can predict a general equation that governs the Luminosity versus period relationship
for these 3 kinds of variable stars:
80  20
RR Lyrae variables: L  20 
(T  .3)  L  85.7T  5.7 where .3  T  1 day
1  .3
Type 2 Cepheids:
7000  200
L  200 
(T  1)  L  174.4T  25.6 where 1  T  50 days
40  1
Type 1: L  2000 
80000  2000
(T  3)  L  804.1T  412.4 where 3  T  200 days
100  3
Once we discover one of these variable stars and measure its pulsation period, then we can
calculate their absolute luminosity. Then, we change their absolute luminosity to the
Hipparchus Scale. We can also use a photometer to calculate their Apparent Luminosity on
the Hipparchus Scale. Then, we use the following formula to calculate the distance to the
variable star and the galaxy that hosts the star ---------------------->
d  10
 
 



mv  M v 


 1 
5




Let’s do an example. Suppose we discover a Type 1
Cepheid variable, and we measure its pulsation period to be 85 days. From this, we calculate
its absolute luminosity: L = 804.1(85) – 412.4 = 67,936.1. Then, we convert this to the
Hipparchus scale------------->
L 
 67,936.1Lsun 
  4.83  7.25  M v
L AbsoluteHipparchus  2.5 log 10  star   4.83  2.5 log 10 
Lsun
 Lsun 


Suppose the star is a very dim looking star and its apparent luminosity is measured to be
mv = 9.5. This means that it is not visible to the naked eye since mv is greater than +6.
Now, we get:
d  10
 




9.5 7.2 5 
1
5






=22,387.21 parsecs = 73,027.1 LY.
This star is perhaps at the other edge of our Milky Way galaxy or in a nearby galaxy in our
own Local Group!!!!!!!!!
34) Calculating the Schwarzchild Radius of a Black Hole:
The Schwarzchild radius is the radius of the event horizon of a black hole where the
escape velocity is greater than the speed of light and therefore, even light is trapped. The
formula for the Schwarzchild radius of a black hole is the following:
2GM Black Hole
RSchwarzchild 
The Mass of regular black holes tend to be between 3 – 15
c2
times the mass of the sun. However, supermassive black holes at the center of galaxies
could be million or billion times the mass of the sun. Let’s calculate the schwarzchild
radius of a black hole where Mblack hole = zMsun, where z is an integer greater than 3.
2(6.67259 x10 11 ) z (1.989 x10 30 )
RSchwarzchild 
 2,953.4 z meters  2.95 z km
(2.99792458  108 ) 2
Therefore, if z = 3, then Rs = 8.85 km. (This is the radius of the smallest black hole)
if z = 15, Rs = 44.3 km (This is the radius of the heaviest and largest regular black hole)
The Milky way is believed to house a black hole at its center that is 4 million solar
masses. What would its schwarzchild radius be? z = 4 x 106
Rs = 2.95(4 x 106) km = 1.18 x 107 km x
1 AU________
= .079 AU’s. That is a
huge monstrous black hole!!
1.495978706 x 108 km
The Virgo galaxy houses a 3 billion solar mass black hole. z = 3 x 109. It’s size is ---->
Rs = 2.95(3 x 109) km = 8.85 x 109 km x
1 AU________
= 59.2 AU’s.
8
1.495978706 x 10 km
That is a tremendously large black hole. It would eat up our whole solar system!!!!!!
35) Calculating the Mass of a Binary Star system:
Binary stars are two stars that are relatively close to each other and are gravitationally
bound to each other. They revolve around one another’s center of mass. We can use the
Modified form of Kepler’s 3rd law, M1 + M2 = d3/T2 where M1 and M2 are the masses of
the stars in terms of solar masses, d is the length of the average separation between the
two stars in units of AU’s, and T is the period of revolution of the stars about their
common center in years. Let’s say that we have determined the distance of their average
separation to be 3 AU and the period to be 9 months which is .75 years. Then,
M1 + M2 = 33/.752 = 48 solar masses. This is a heavy binary star system!!!
Furthermore, if we do further investigation, we can calculate the radius of orbit of each
star separately and determine their individual masses. In the above example, let’s say
Star A has an average radius of orbit = .5 AU’s and Star B has an average radius of orbit
= 2.5 AU’s. Their radii have to add up to 3 AU since their average separation equals 3
AU. We know that the heavier star has to move with a smaller orbit and the lighter star
with a bigger orbit, and the ratio of their masses equals the reciprocal ratio of their
individual radii. Therefore we get:
M star A 2.5

 5    M star A  M star B  48    5M star B  M star B  48    M star B  8
M star B
.5
Therefore, the lighter of the two, Mstar B = 8 solar masses and the heavier of the two,
Mstar A = 40 solar masses.
36) Calculating the Diameter of a Star using Eclipsing Binaries:
Eclipsing binaries are two stars that revolve around each other in such a way that their
plane of orbit points towards the Earth. That way they successively eclipse each other;
this event is also called Occultation. By measuring how long they each eclipse each other
and how much the total luminosity of the pair drops, we can calculate their relative
brightness and size!!
37) Calculating the Hubble Time from the Hubble Constant:
The Hubble Time tH is the reciprocal of the Hubble Constant H, and it gives the age of
the Universe if the Universe had no matter or energy in it. In other words, if it was
expanding at a constant rate. In such a universe, the velocity of expansion would be
linearly proportional to the distance away from any point, v = Hd, where H = the slope of
the velocity versus distance graph; this slope is known as the Hubble constant, and it is
one of the most important constants in Astronomy. Presently, its value is around
71 km/sec/Mpc. This means that if a galaxy is 1 million (Mega) parsecs away from us, it
moves away from us at a velocity of 71 km/sec. If we reciprocate this constant, we get
the Hubble time. The actual age of our universe is related to the Hubble time but less
than it. For an accelerating universe, Age of Universe > 2/3tH!!!
1
1
1
1sec
3.262 Lightyears 9.46 x1015 meters
6
tH 


 Mpc 
 10 par sec s 

km / s
km
H
71km
1 par sec
1Lightyear
71
71
Mpc
sec
1km
1hr
1day
1year




 1.377  1010 years  13.77 billion years!!!
1000meters 3600 sec 24hrs 365.25days
Therefore, the age of our universe > 2/3(13.77) = 9.18 billion years!!!.
