Calculations 1

If you still have the papers I put in your mail box, I’d like to add those onto here. 1.1

Show that the Sun produces 10000 times less energy per unit mass than an average human giving out about 1W kg-1 L☼ 3.86∗10 26 W 1026 W W = ≈ 30 =10−4 30 M ☼ 1.99∗10 kg 10 kg kg To compare the sun to a human, divide the energy of the human by the energy of the sun. W 1 kg =10,000 −4 W 10 kg

1.2 The red supergiant star Betelgeuse in the constellation Orion has Teff≈3500K and a diameter of .045”. Assuming that it is 140pc from us, show that its radius R≈700R☼, and that its luminosity L≈105L☼.

π −7 =2.18∗10 rad Diameter of Betelgeuse =.045”→ 648,000 Using equation [1.2]

Distance Diameter of object (km) α ( ¿ rad )= object (km) ¿ ¿ d∗α= D=2.18∗10−7∗4.32∗1015=9.4∗106 km

Divide the diameter by 2 for the radius, so R=4.7*108km=675R☼ Use equation [1.3] L=4 π R2 σ SB T 4 13 2

−8

−2

−4

4

35

8

L=4 π∗( 4.7∗10 ) m∗5.67∗10 W m K ∗( 3500 K ) =2.36∗10 W =6.1∗10 L☼

1.15(A) - Part b includes the use of a table that I did not include.

If a galaxy has an absolute magnitude M, use the equations [1.1] and [1.27] to show that its apparent magnitude m is related to the redshift z=Vr/c of equation [1.19] by m=M + 5 log10z +C, where C is a constant same for all objects. F=

[1.1]

L 2 4πd

−1 V r d=h [1.27] 100

λobs Vr [1.19] 1+ z= λ e =1+ c

∴ z=

Vr c

Take the distance modulus m−M =5 log10

−1

d=h

( 10dpc ) → 5[ log

10

( d )−log 10 ( 10 pc ) ]=5 [ log 10 ( d )−1 ]

z C

m−M =5 [log 10 ( h−1 ) + log 10 ( z )−log 10 ( c )−1] −1 If C=5[log 10 ( h )−log 10 ( c )−1] , C is a constant for all objects.

∴ m=M + log 10 ( z ) +C

1.10”

2.7

1.62”

Assuming that the supernovae is a circle (e=0), to find the angle of inclination i, take sin ( i )=1.10 } over {1.62 →i=sin−1 ( 1.10 } over {1.62 ) =43̊

A BA i

Point A will be observable before point B due to the fact that is it closer to us. Since the object, as a whole, is so far away, the non-radial differences in points are negligible. Velocity=

Distance Time

Distance to point A is d=Radius−Radius∗sin ⁡( i) Velocity is c, the speed of light. So −¿=

R−R∗sin ⁡( i) R (1−sin ( i ) ) = c c t¿

That would elude to +¿=

R(1+ sin ( i ) ) c t¿

To find the radius, add the two times together. +¿=

R−R∗sin ( i ) R+ R∗sin ( i ) 2 R+ R∗sin (i ) −R∗sin ( i ) + = c c c −¿+ t ¿ t¿

+¿ −¿ +t ¿ t¿ ¿ c¿ 2R +¿= → R=¿ c −¿ +t ¿ t¿ The distance can be found from using the determined radius and the measured radius, that is half of the largest measurement of the observed ring, so 1.62” is the diameter, .82” is the radius. radius ( arcseconds )=

rad ius ( m ) distance ( m )

tan ⁡¿ .82)} =4.38* {10} ^ {17} m=14.2pc tan ⁡¿ radius ( m ) 6.19∗1015 d= = ¿ tan ( radius ( arcseconds ) )

Considering that mv=3, use the apparent magnitude of the sun as well as its distance, to find the luminosity in terms of L☼, equation [1.10] L1

m1−m2=−2.5 log 10

( )

F1 4 π d 12 L1 d 2 2 =−2.5 log 10 =−2.5 log 10 ( ) F2 L2 L2 d 1 2

( )

2

4 π d2

L=

d1 d2

2

(m1−m2) −2.5

∗10 2

L☼