Download Homework #1 - Department of Physics and Astronomy

Astronomy 204 - Homework #1 - Solutions
1.
-
Chapter #1, Problem #1 – Assume that the orbital plane of a superior planet is
inclined 10E to the ecliptic and that the planet crosses the ecliptic moving
northward at opposition. Make a diagram similar to Figure 1-1B, showing the
retrograde path of this superior planet.
The planet should initially and finally be moving eastwards
Due to the relatively large inclination, there shouldn’t be any loop.
The planet should cross the ecliptic at opposition – this is the midpoint of the
westward motion.
north
east
ecliptic
opposition
west
Path of planet
south
2.
Chapter #1, Problem #2 – Imagine that you are observing the Earth form
Jupiter. What would you observe the Earth’s synodic period to be. What
would it be from Venus? (Hint: See Appendix 3.)
There is really no need for a calculation for this problem. The synodic periods of Jupiter
and Venus are given in Appendix 3. It is important to note that the synodic period of the
Earth as seen from Jupiter is equal to the synodic period of Jupiter as seen from the earth.
Since we are talking about the sun-Earth-Jupiter alignment, it doesn’t matter which planet
you are viewing.
3.
Chapter #1, Problem #10 – An object is observed from the Earth to have a
synodic period of 1.5 years. What are the two possible values for the semimajor axis of the object’s orbit.
This object (likely an asteroid) could have either an inferior or exterior orbit.
Superior – Since the Earth has orbited the sun 1.5 times, the superior planet will have
only orbited 0.5 times. Thus, the orbital period of the planet is 3 years and a = P2/3
= (3 years)2/3 = 2.08 AU
Inferior
1 1 1
= −
S P E
1 1 1 1 1 5
= + = +
=
P S E 1 1.5 3
P = 0.6 yr
2
3
2
a = P = ( 0.6 yr ) 3 = 0.71AU
4.
F=
Chapter #1, Problem #12 – Compare the mutual gravitational force between
you and the following two objects: (a) another person with mass 100 kg
located 1 m from you and (b) mars at opposition.
Gm1m2
m 
= Gmme  22 
2
r
r 
 (100kg ) 
Fanother person = Gmme 
= 100Gmme
2 
 (1m ) 
 ( 0.64 ×1024 kg ) 
 = 113.8Gmme
FMars = Gmme 
 ( 0.75 x1011 m )2 


Thus, the two are very comparable.
Proponents of astrology contend that the position of the planets (particularly at the
moment of our birth) profoundly affects the direction of our lives and what type of
person we will be. Well exactly how do the planets affect us. The above calculation
illustrates that the gravitational affect that the obstetrician has on a newborn baby is
comparable to that of Mars even when Mars is closest.
5.
Chapter #1, Problem #13
(a) Venus has a maximum elongation of 47E. What is its distance from the sun in
astronomical units?
rVenus = (1 AU) x sin 47E = 0.73 AU
(b) Mars has a synodic period of 779.9 days and a sidereal period of 686.908 days.
On February 11, 1990, Mars had an elongation of 43E west. The elongation of
Mars 687 days later, on December 30, 1991, was 15E west. What if the distance
of mars from the Sun in astronomical units?
This is a complicated, multi-step problem. You should first draw a picture.
S
E1
M
15E
43E
E2
- First note that the p E1-S-E2 can be solved from the data related to time.
 360° 
720° − 
 ( 687 days ) = 43°
 365days 
- Next note that ª E1-S-E2 is an isosceles triangle. Apply the law of cosines to solve for
the E1-E2 distance of 0.73AU.
- Note that p M-E1-E2 = 111.5º and p M-E2-E1 = 53.5º
- Apply the law of sines to ª E1-M-E2 to get E1-M = 2.27 AU
- Apply law of cosines to ª E1-S-M to get S-M = 1.68
Note that this is much different than the value of 1.52 AU given for the semi-major axis
of Mars which led Kepler to conclude that Mars had an elliptical orbit.
6.
Create a plot in Matlab of a scaled version of the orbit of Pluto. The plot
should be appropriately labeled and have the proper semi-major axis and
eccentricity. Turn in hard copies of your m-file and a plot.
First experiment with the m-file shown below which graphs conic sections. Use the help
embedded in Matlab to determine the meaning of any commands whose function isn’t
apparent.
% m-file to graph conic sections
clear
e1 = 0.5;
e2 = 1.0;
e3 = 2.0;
a1 = 1.0;
p = 1.0;
a3 = 1.0;
npoints = 1000;
theta=-pi:2.0*pi/(npoints-1):pi;
for i = 1:npoints,
rp1(i) = a1*(1-e1^2)/(1+e1*cos(theta(i)));
rp2(i) = (2*p)/(1+cos(theta(i)));
end;
theta2=-2.0*pi/3.0:4.0*pi/(3.0*(npoints-1)):2.0*pi/3.0;
for i = 1:npoints,
rp3(i) = a3*(e3^2-1)/(1+e3*cos(theta2(i)));
end;
hold on
grid on
title('Conic Sections')
axis([-10 10 -10 10])
polar(theta,rp1,'r-')
polar(theta,rp2,'b-')
polar(theta2,rp3,'g-')
One possible solution based on the script above is:
% m-file to graph the orbit of pluto
clear
e1 = 0.249;
a1 = 39.44;
npoints = 1000;
theta=-pi:2.0*pi/(npoints-1):pi;
for i = 1:npoints,
rp1(i) = a1*(1-e1^2)/(1+e1*cos(theta(i)));
end;
hold on
grid on
title('Orbit of Pluto')
axis([-50 50 -50 50])
text(-1.05,0.0,'0', 'color','yellow') %Draw in sun
polar(theta,rp1,'r-')