Jonathan M Marr, Ronald L. Snell, Stanley E. Kurtz

Jonathan M Marr, Ronald L. Snell, & Stanley E. Kurtz Solutions Manual for Fundamentals of Radio

Astronomy: Observational

Methods Chapter 1-6 (no Solutions for chapter 7) 1 / 4

C H A P T E R1

Introductory Material

• The cosmic radio signals were too weak to be detected above the noise of the radio

receivers. Successful detection occurred, later, by amplifying the signal input to the receiver.

2. a. By Equation 1.1,=c== (3:0010

8 m s 1

)=(22:210

9 Hz) = 0:0136 m = 1:36 cm

b. By Equation 1.2,E=h= (6:62610

34

J s)(22:210

9 Hz)

= 1:4710

23 J

• The energy of each photon is given by Equation 1.2. Equating N radio photons with
• UV photon we haveNradiohradio=hc=UVand so the number of radio photons

needed is Nradio=

3:0010

8 m s 1

(2:0010

9 Hz)(15010 9 m) = 10 6 radio photons

• continuous, emission-line, absorption-line
• elemental composition (also recessional velocity and temperature)
• RA2>RA1and RA increases to the east.

Therefore, source 2 is further east.

• LST = RA of transiting meridian = 4h 0m 0s
• Using Equation 1.4,

HA=LSTRA= 4h 00min 00s5h 30min 00s =1h 30min =1:5h

• Both have Dec = 0 and so are on the equator.

Therefore, by Equation 1.3, = (RA)15 = 1:5h15 = 22:5

• Imagine a sphere of radiusRand consider an arc along the equator. Let the arc

contain 1/24th of the circumference, and so the arclength equals 2R=24. The arc length also corresponds to 1 hour in RA (since there are 24 hours of RA around the full circle). The angle at the center of the sphere subtended by this arc is 1/24th of 360

and so equals 15

. At the equator, then, 1 h of RA = 15

of angle at the center of the sphere.

1 2 / 4

2Solutions Manual for Fundamentals of Radio Astronomy: Observational Methods

Now consider an arc running parallel to the equator but at a higher latitude. Let represent the latitude of this arc. By considering the right triangle with vertices at the sphere's center, a point on the arc, and the right angle on the sphere's rotational axis (in the same horizontal plane as the arc), we see that the radius of the circle in the plane parallel to the equator isRcos. Let this arc also be 1/24th of the way around, so that in terms of RA this arclength also corresponds to 1 hour of RA. In relation to the radius of the sphere, though, this arclength equals 2Rcos=24. Since the angle at the center of the sphere is proportional to arclength divided by radius (in radians, the angle equals the arclength divided by the radius, when arclength and radius and in the same units) and since this angle is subtended by a smaller arclength, while the radius is kept constant, this angle is smaller than the angle subtended by the arc in the equatorial plane. Since the arclength at latitudeis a factor cossmaller, the angle is reduced by factor cosand therefore 1 h of RA now corresponds to 15

cos.

b. = cos:Hence, = 50

00 =cos(60

) = 100

00 =1 m 40 s

• To an observer at the North pole, the celestial equator is on the horizon and so its

altitude 0

• The star is 30

north of the equator, which is at zero altitude and so this star's altitude is 30

.

• altitude of NCP = latitude of observer. So,(NCP)= 42

• altitude of equator = 90

lat = 90

42

= 48

• altitude of star at transit = altitude of Equator + Dec of star

= 48

• 30

= 78

• 0

, by denition

• 270

• A bit east of southeast and almost half-way up from the horizon.

10.sphere=

4 (2) 2 = 3 6= 4 Equation 1.6 assumes thatis small. The correct approach is to use Equation 1.7 with A equal to the surface area of the sphere. Then sphere= 4R 2 R 2 = 4

• a.

Sun=

4

0:533

radians 180

2

= 6:8010

5 sr b.Sun= ( 1 2

1:39210

6 km) 2

(1:49610

8 km) 2

= 6:8010

5 sr 3 / 4

Introductory Material3

• =majminfor a rectangle

= (0:1

0:05

)

radians 180

2

= 1:5210

6 sr

• The limit on the size scale of the surface deviations of a reector are proportional to

the wavelength of the radiation. The long wavelengths of radio waves mean that the demand of a radio telescope's reector in terms of how smooth it must be is much less stringent than that of a visible-wavelength telescope.

• To get good reections, one only needs the surface irregularities to be smaller than

1/20th of the wavelength, i.e. z < =20. At the longer radio wavelengths, such as 21 cm, the dish can be a mesh with holes large enough to see through (such as 1 cm), whereas the mirror of a visible-wavelength telescope cannot have irregularities (or holes) larger than tens of nanometers.

• An equatorial mount is the same thing as alt-az mount but with the whole structure

tilted so that the azimuth axis points at the north celestial pole instead of the zenith.

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