Astronomy 210
Spring 2011
Homework Set #1: Solutions
1.
SOHO and the Angular Speed of the Sun
(a) As the Earth orbits the sun the apparent position of the sun against the starry background
traces a closed path. Since the sun completes one orbit through the celestial sphere in Δ
t
=
P
= 365
.
24 days, and one orbit consists of Δ
θ
= 360
◦
, the average angular speed is
ω
=
Δ
θ
Δ
t
=
360
◦
365
.
24 day
= 0
.
98
◦
/
day
≈
1
◦
/
day
(1)
(b) Compare images of the same star at diFerent times. Measure the length (on the screen or
paper printout) Δ
s
that the star moves in “image units” of cm (or, egad! inches). We want
to ±nd the angular displacement Δ
θ
this corresponds to.
To do this, we note that the image gives the Sun’s disk diameter, which we know corresponds
to an angular diameter
θ
⊙
= 0
.
5
◦
. Measure the length
D
⊙
of the Sun’s diameter in “image
units.” Now, all lengths in image units are
proportional to
the corresponding angular sizes.
So we must have
Δ
θ
θ
⊙
=
Δ
s
D
⊙
(2)
Thus we can solve for
Δ
θ
=
Δ
s
D
⊙
θ
⊙
(3)
²rom the timestamp on the images, you can also ±nd the elapsed time Δ
t
between the two
images.
Thus you can ±nd the star’s angular speed with respect to the Sun, which is the same as the
Sun’s angular speed with respect to the celestial sphere:
ω
⊙
=
Δ
θ
Δ
t
(4)
If all has gone well, your should indeed ±nd a speed very close to 1
◦
/day, or about 2 solar
diameters a day.
2. (a) We see that the Earth is more than half illuminated, but not entirely, so the phase is gibbous.
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 Spring '08
 Fields
 Astronomy, celestial pole, Equator, Celestial coordinate system, Circumpolar and Invisible Stars

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