Astro 301
3/28/10
1
EndofLecture Problems: Answers
Lecture 2.
The problems make use of the small angle formula:
D
=
"
d
206 265
.
In the first case, we are solving for the distance
d
, given the Moon’s angular size
α
=
0.5°. Converting to arcesconds,
α
= 0.5
×
3600 = 1800’’. We also need the linear diameter,
which is D = 3476 km. We then get d
≈
400 000 km.
In the second problem, we solve for
α
, given the distance
d
(=1 AU) and the diameter
D
= 1.4
×
10
6
km, of the Sun:
=
206 265
D d
.
D
and
d
must be in the same units, hence
=
206 265
#
1.4
#
10
6
km
( )
1.5
#
10
8
km
( )
, giving
α
≈
1900’’.This is about the same
size as the Moon, which is why we have eclipses.
Lecture 3
Due to geometrical dilution, the solar flux at radius
r
is
F
=
L
!
4
r
2
. Therefore, the flux
reaching Mercury will be
F
(Mercury) = (1 AU/0.39 AU)
2
F
(Earth)
≈
6.6
F
(Earth).
Similarly, for Neptune
F
(Neptune) = (1 AU/30 AU)
2
F
(Earth).
The energy of a “typical” photon will be
E
=
hc
/
λ
, where
h
= Planck’s constant,
c
= speed
of light and
λ
(wavelength)
≈
500 nm. Looking up the values of the constants (and
remembering to convert wavelength to meters),
E
≈
4
×
10
–19
J.
Since the Sun’s luminosity is
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 Spring '10
 DrAndrewRobinson
 Astronomy, Moons, Planet, Astronomical unit, Celestial mechanics

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