ASTRO 1 HW3 Solution

5.34

(a) TPele = (320 + 273) K = 593 KUsing Wien’s Law,

λmax =0.0029 K m

T=

0.0029

593= 4.89× 10−6 m = 4.89 µm

[1 µm = 10−6 m] The wavelength is in the infrared.

(b) TIo = (−150 + 273) K = 123 KUsing Stefan-Boltzmann law, we can find the ratio of energy per second per square meter surface, i.e.energy flux,

FPele

FIo=σT 4

Pele

σT 4Io

=5934

1234= 540

For each square meter of surface, Pele radiates 540 times more energy per second than that of Io.

5.37

(a) To find the wavelength of P∆, i.e. the 4th wavelength in the Paschen series, use the Bohr formula,

1

λ= R

(1

N2− 1

n2

)where R = 1.097× 107 m−1

Paschen series: set N = 34th wavelength of Paschen series: set n = 3 + 4 = 7

⇒ 1

λ= (1.097× 107)

(1

32− 1

72

)λ = 1.01× 10−6 m = 1.01 µm

(b) The P∆ line is the transition between the 3rd and 7th energy level.

(c) With the wavelength of 1.01× 10−6 m = 1.01 µm, the line is in the infrared.

ASTRO 1 – HW3 Solution 2

5.43

λrest = 486.133 nm, λobs = 486.112 nmWavelength shift ∆λ = λobs − λrest = (486.112− 486.133) nm = −0.021 nmThe negative value means the light observed is shifted towards shorter wavelength, i.e. blueshifted.

Using the Doppler shift equation,v

c=

∆λ

λrest

⇒ v =∆λ

λrestc =

(−0.021 nm

486.133 nm

)(3× 108 m s−1) = −1.30× 104 m s−1 = −13.0 km s−1

The negative sign means the star is coming towards us, at a speed of 13.0 km s−1.

5.44

For the red light, λrest = 700 nmTo be observed as green light, λobs = 500 nmWavelength shift ∆λ = λobs − λrest = (500− 700) nm = −200 nm

Using the Doppler shift equation,v

c=

∆λ

λrest

⇒ v =∆λ

λrestc =

(−200 nm

700 nm

)(3× 108 m s−1) = −8.57× 107 m s−1 = −8.57× 104 km s−1

This means you are going at a speed of 8.57 × 104 km s−1 towards the light if your claim is true. Youdefinitely deserve a ticket.

6.34

(a) We calculate the diffraction limited angular resolution by

θ = 2.5× 105 λ

D

With D = 40 m and λ = 550 nm = 550× 10−9 m,

θ = (2.5× 105)× 550× 10−9

40= 3.44× 10−3 arcsec

(b) Let θ40 and θ10 be the diffraction limited angular resolution of the 40-m and 10-m telescope respectively,

θ10

θ40=D40

D10=

40 m

10 m= 4

The diffraction limited angular resolution θ of the Keck 10-m telescope is 4 times larger than that of the40-m telescope, i.e., the angular resolution at the diffraction limit of the Keck 10-m telescope is poorer.Note that the larger the θ, the poorer the angular resolution.

Without adaptive optics, however, observations on the ground cannot reach the diffraction limited an-gular resolution due to turbulence in the air (i.e., atmospheric distortion). The true angular resolutionis therefore poorer than the diffraction limited angular resolution.(See Universe section 6-3 on Limits to Angular Resolution.)

ASTRO 1 – HW3 Solution 3

6.36We use the small angle formula,

D =αd

206265

To observe the Jupiter moon with the Hubble Space Telescope, using d = 6.28 × 108 km and the angularresolution of α = 0.1 arcsec,

D =0.1× (6.28× 108)

206265= 304 km

The smallest surface feature which we can resolve on Jupiter’s moon using the HST is therefore 304 km.

If we look at our Moon with an unaided eye (resolution = 1 arcmin), using the Earth-Moon distance ofd = 384400 km and angular reoslution of α = 1 arcmin = 60 arcsec,

D =60× 384400

206265= 112 km

The smallest feature on the Moon which can be resolved with unaided human eye is 112 km. This is is 2.7times smaller than observing Jupiter’s moon with the HST.

6.40

(a) Given the frequency ν = 557 GHz = 557× 109 Hz, and speed of light c = 3× 108 m s−1,

ν =c

λ⇒ λ =

c

ν=

3× 108

557× 109= 5.39× 10−4 m = 0.539 mm

This wavelength is at the infrared/microwave.

(b) The use of satellites is necessary since the atmosphere is opaque to this wavelength.(See Universe Fig. 6-25.)

(c) At λ = 5.39× 10−4 m, to achieve an angular resolution of θ = 4 arcmin = 4(60) arcsec = 240 arcsec, weneed a primary mirror with diameter D,

θ = 2.5× 105 λ

D⇒ D = 2.5× 105λ

θ= (2.5× 105)

5.39× 10−4

240= 0.561 m

6.41The astronomers are looking for absorption lines.The distant galaxy emits a continuous spectrum. The light from this distant galaxy then passes through theionized oxygen gas surrounding the Milky Way, for which the gas absorbs light at particular wavelengths.Absorption line spectrum is therefore expected.