Download The Sun and Planets Homework Solution 1.

The Sun and Planets
Homework Solution 1.
Exercise 1.
Object
Sun
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Spring Semester 2017
Prof Dr Ravit Helled
Scale of Solar System
Table 1: Scaled solar system model with a 1 cm Earth
Scaled Size (cm) Scaled Distance (cm)
Object
109
0
exercise ball
0.3
45
small pea
0.95
85
marble
1
118
marble
0.53
178
pea
10.96
611
big orange
9.13
1119
apple
3.98
2331
tennis ball
3.86
3532
tennis ball
(a) If the diameter of the Earth in your model is set to 1 cm, your model would have a
scale of 1-to-1.274 × 109 . In this case you would have to divide all other distances
and diameters by 1.274 × 109 to end up with a consistent model. You could have
used any other scaling, the only requirement is that your scaling is consistent. In
Table 1 we provide a consistent scaling for a model with the Earth’s diameter set
to 1 cm.
(b) Proxima Centauri (a.k.a., Alpha Centauri C) is a red dwarf star located approximately 4.25 light-years away from our Solar System. In centimeters this distance is
roughly 4 × 1018 cm. The star itself is only about 12% as massive as the Sun, with
a correspondingly small diameter of only 0.282 R . In centimeters this diameter is
roughly 2 × 1010 cm.
In our scaling above, we scaled the diameter of the Earth down to 1 cm. The Earth
has a diameter equal to 1.274×109 cm, so in order to scale the distance and diameter
of Proxima Centauri to our model, we must divide by this value:
1
· 2 × 1010 cm ' 15.7 cm
1.274 × 109
The scaled down version of the star measures just under 16 cm, which could be
represented by a handball (18.5–19.1 cm). But how far away from the center should
we place it?
1
· 4 × 1018 cm ' 3.14 × 109 cm
1.274 × 109
So, while Neptune is located about 3.5 meters away from the center of our model,
your Proxima Centauri handball must be placed 31’400 km away! The Earth only
1
has a circumference of 40’075 km at the equator, so you’d have to walk a little more
than three-quarters of the way around the Earth before you could complete your
model. Walking continuously without any breaks at an average walking speed of 4
km/h you would arrive in under 11 months.
Exercise 2.
Orbital Speed and Momentum
(a) The equation for orbital speed was given. As long as you kept your units in order,
you should arrived at or very near the values in Table 2.
(b) The orbital angular momentum of a planet is a function of its mass, distance, and
speed. The formula for the orbital angular momentum is:
L = mrv,
where m is the mass of the planet, r is the orbital radius (in this case we use the
semi-major axis), and v is the orbital speed that you calculated in part (a). If you
keep your units in order, your answers should look similar to the values in Table 2.
The total orbital angular momentum of the planets is about 3×1043 J·sec.
In the case of the Sun, we are interested in the rotational angular momentum. The
rotational angular momentum of a body is given by:
L = kM R2 ω,
where k is a constant between 0 and 1 that depends on the object, M is the mass
of the object, R is the radius of the object, and ω is the angular frequency. The
values of k for various objects can be found in any number of sources or derived
from scratch. In the case of the Sun, we assume that it is a solid sphere with k = 52 ,
M = 2 × 1030 kg, R = 6.96 × 108 m.
The angular frequency is related to the rotation period by ω = 2π
, where P is the
P
Sun’s rotation period. We gave you the rotation period in days, but this needs to
be converted to seconds: 24.5 days = 2116800 s. The resulting value will have units
of kg·m2 ·s−1 , but angular momentum is usually reported as J·sec:
L=
2
2π
2
· 2 × 1030 kg · 6.96 × 108 m
= 1.15 × 1042 J·sec,
5
2116800 s
If we compare the Sun’s angular momentum to the angular momentum of the planets, we find an interesting result. The Sun contains 99% of the Solar System’s mass,
but only about 3% of its angular momentum. If you look closer at the angular momentum of the planets, you’ll notice that roughly 60% of the Solar System’s angular
momentum is in the orbit of Jupiter.
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Table 2: Orbital speed and angular momentum of the planets
Object
v (m/sec)
L (J·sec)
Mercury
47’400
1.10×1039
Venus
35’000
1.85×1040
Earth
29’800
2.70×1040
Mars
24’100
3.85×1039
Jupiter
13’000
2.02×1043
Saturn
9’640
8.23×1042
Uranus
6’800
1.41×1042
Neptune
5’430
2.54×1042
Exercise 3.
Matryoshka Solar System
(a) The radius of the Sun is R = 0.00465 AU (taking the Sun’s radius in AU makes
the problem easy since the semi-major axes of the planets are usually reported in
AU).
a (R ) =
a
a (AU)
=
R
0.00465 AU
Your answers should match those in Table 3.
(b) The Galilean moons are the four largest moons of Jupiter—Io, Europa, Ganymede,
and Callisto. Note that in this problem we want the answers in units of Jupiter
radii, not solar radii as in part (a).
a (RJ ) =
a
RJ
The semi-major axes of the Galilean moons are usually reported in kilometers, so
you probably wouldn’t use astronomical units as in part (a). Your answers should
match those in Table 4.
(c) Your answer to part (c) will depend on your definition of “similar”. We graded your
answer based on whether or not you provided a good reason for your answer.
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Table 3: Semi-major axes of the planets in units of solar radii
Object
Semi-major axis (R )
Mercury
83.3
Venus
155.3
Earth
215.3
Mars
328.0
Jupiter
1120.0
Saturn
2061.0
Uranus
4132.4
Neptune
6470.0
Table 4: Semi-major axes of the Galilean moons in units of Jupiter radii
Object
Semi-major axis (RJ )
Io
5.9
Europa
6.4
Ganymede
15.0
Callisto
26.3
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