Physics 134
Homework Set 4 Solutions
Due 2/6/12
1. We observe that oppositions of Saturn repeat every 378 days.
(a) Use this information to find the sidereal period of Saturn, in (Earth) years. You
may take a year to mean 365.25 days and neglect the distinction between sidereal
and solar years.
We use our expression relating synodic period (time between oppositions of a
superior planet) and sidereal period
S
S
=
−1 .
P6
P⊕
Solving for the unknown P6 we find
P⊕ S
S
= P⊕
.
S − P⊕
S − P⊕
P6 =
We seek Saturn’s sidereal period in years, i.e.
S
378
P6
= 29.6 .
=
=
P⊕
S − P⊕
378 − 365.25
(b) Use your answer to the previous question to find the radius of Saturns orbit,
which we will assume is circular, in AU (Astronomical Units).
We use Kepler’s third law for circular orbits
P 2 = KR3 .
Dividing this relation for Saturn by the relation for Earth the constants drop out
and we find the scaling relation
P6
P⊕
2
=
R6
R⊕
3
,
or upon taking the third root of both sides
R6
=
R⊕
P6
P⊕
2/3
= 29.64705882/3 = 9.58 .
This is the radius of Saturn’s orbits in units of the radius of Earth’s orbit - AU.
(c) At opposition, Saturn’s equator appears to subtend an angle of 19.4”. Use this
to find the equatorial radius of Saturn, in km. You will need to use the fact that
1 AU = 1.496 × 108 km.
We will use the small angle formula here. Saturn’s radius r6 and its angular
radius α6 as seen from Earth are related to the planet’s distance D6 at any time
by
r6
α6
=
.
D6
206265”
At opposition, Sun, Earth, and Saturn lie in a straight line so that the distance
between the planets is
D6 = R6 − R⊕ = 8.58 AU ,
and the angular radius is given as α6 = 19.4”/2 = 9.7”. We then have, converting
AU to km,
r6 =
9.7”
α6
D6 =
8.58 × 1.496 × 108 km = 60, 362 km .
206265”
206265”
2. One trajectory for sending a spacecraft from Earth to Mars is along an elliptical orbit
that has the Sun at one focus, the Earth (at launch) at its perihelion, and Mars (at
arrival) at its aphelion. The craft would be launched from Earth (well, from lowEarth orbit, but on Solar System scales this is the same thing), its rockets firing until
it (rapidly) reaches the speed appropriate to the requisite orbit. At this point the
engines turn off, and the spacecraft can ”coast” until it reaches the vicinity of Mars.
Then rockets again fire, moving the craft into an orbit about Mars or a landing on its
surface. This minimal energy trajectory, possibly modified to allow for “gravitational
slingshot” interactions, is in fact used by many space missions.
(a) Assuming Earth and Mars orbit in circular orbits, draw the two circular orbits
as well as the elliptical orbit described above. Be sure you locate the ellipse so
that its perihelion is along Earth’s orbit and its aphelion along Mars’s orbit.
(b) The radius of Earth’s orbit is 1 AU; that of Mars’s orbit 1.52 AU. What is the
semimajor axis of the ellipse you found in (a)?
From the figure, the major axis of our ellipse (recall the major axis of an elliptical
orbit goes through aphelion, perihelion, and both foci) has length 2.52 AU. The
semimajor axis is thus half of this or 1.26 AU.
(c) Neglecting the acceleration phases, what is the time for a one-way trip from Earth
to Mars along such a trajectory?
With engines off, the craft is in orbit around the Sun, and we can use Kepler’s
third law for solar orbits
2
Psc
= Ka3sc ,
or upon dividing by the same equation for Earth,
2 Psc
asc 3
,
=
1 yr
1 AU
we find that the period of this orbit is (2.52/2)3/2 = 1.41435 yr. This is the time it
would take the craft, left to its own devices, to return to the point whence it was
launched (though Earth would no longer be there). By symmetry, the one-way
trip from this initial point to aphelion would take half this time, or about 0.71
yr.
3. Comets orbit the Sun, as we mentioned, in highly elliptical orbits. Let’s see how
elliptical, using Kepler’s laws.
(a) Halley’s comet orbits with a sidereal period of 76 years. At its nearest approach
to the Sun (perihelion) it is 0.586 AU from the Sun. How far is the comet from the
Sun at aphelion? Halley’s period makes it, for most people, a once in a lifetime
event. I saw it in 1986. Your next chance will be in 2061.
As we saw above, if Dp is distance from Sun at perihelion and Da distance at
aphelion, then
Dp + Da = 2a ,
since line from perihelion to aphelion is the major axis (goes through both foci).
Kepler’s third law tells us that with period P measured in years and semimajor
axis a in AU we have
2
3
P =a =
Dp + Da
2
3
.
We are given Dp and P so we solve the above for Da
P 2/3 = (Dp + Da )/2 ,
or
Da = 2P 2/3 − Dp = 2 ∗ (76)2/3 − 0.586 = 35 AU .
(b) Comet Lovejoy, first discovered in December 2011, grazed the Sun at a perihelion
distance of only 0.00554 AU and – to the surprise of many – survived. Its aphelion
distance is estimated to be about 137 AU. When will it next come near the Sun?
Same equation, here we have Da and Dp so solve for P
P =
Dp + Da
2
3/2
=
137.00554
2
3/2
= 1603 yr .
We will not get to see this comet again in our lifetime.
(c) Comet Hyakutake was a glorious sight when I saw it in 1996 at its perhihelion
distance of 0.23 AU. Its period is about 70,000 years (sorry, I doubt you will
get to see this one). What is its aphelion distance from the Sun? This is a
great example of what astronomers mean when they say long-period comets are
important probes of the far reaches of the Solar System.
Here we are back to the situation in (a) and we find plugging in the numbers for
Hyakutake
Da = 2P 2/3 − Dp = 2 ∗ (70, 000)2/3 − 0.23 = 3400 AU .
Long-period comets indeed come from deep space. Recall that Neptune orbits at an
average distance from the Sun of about 30 AU to get a sense of perspective. The
nearest star, Proxima Centauri, lies 4.2 light-years, or 265,000 AU from the Sun.