Homework 2 Solutions– ASTR1110H – Fall 2009

 

1.     KeplerÕs first law states that planets orbit the Sun in elliptical orbits with the Sun at one focus.  This implies that the distance of a planet from the Sun varies as it orbits the Sun.  The second law states that Òequal areas are swept out in equal timesÓ.  This means that a planetÕs speed varies in its orbit, going faster as it nears the Sun.  This law is a consequence of the conservation of angular momentum.  Finally, the third law states that the period squared is equal to the semimajor axis (the average distance of the planet from the Sun) cubed.  This law implies that the further a planet is from the Sun, the slower it goes around the Sun.

 

2.     KeplerÕs second law is just a manifestation of the Law of Conservation of Angular Momentum.  Angular momentum can be expressed as the product of the mass distribution of an object and its distance from the spin axis.  As one quantity increases, the other decreases in a way that keeps the product the same.  Think of the figure skater and how she speeds up and slows down by redistributing her mass from the spin axis.  Planets do the same, speeding up when their mass is near the Sun, and slow when they are far away.

 

3.     Astronomically speaking, the problem with KeplerÕs laws is that they deal only with the 5 known planets at the time and their motion about the Sun. They are not general enough to apply to the orbits of other bodies (the Moon going around the Earth,  for instance).

 

4.     Philosophically speaking, the problem with KeplerÕs laws is that, while they explain, what the planets are doing, they say nothing about WHY the planets obey these strange rules.

 

5.     Mercury and Venus are always near the Sun as viewed from the Earth.  In the heliocentric model this is a consequence of these two planets being interior to the EarthÕs orbit.   The outer planets exhibit retrograde motion.  In a heliocentric model this is produced by the Earth moving faster in an inner orbit and catching up and passing the outer planet.

 

 

6.     Quickest at perihelion; slowest at aphelion.

7.     KeplerÕs Third Law:   p2 = a3

But you must be careful in using this formula.  In KeplerÕs formulation, which is OK to use for anything going around the Sun – but only the Sun – the period must, repeat: MUST, be entered in units of years.  There is no other time unit allowed.  In a similar fashion, the semimajor axis has units of AUs, only; not meters, not inches, not miles, not furlongs.  So, in this problem, a=46 AUs.   Now plug and chug:

 

p2 = (46)3

 

p = 312 years

 

 

8.     Here, you are given the period in units of weeks, but you must convert that to years for KeplerÕs Third Law to work.  Thus 2 weeks = 2/52 years, = 0.0385 years.

Now, just solve (0.0385)2 = a3

a = 0.11 AU

 

If you have trouble solving these equationsÉsee me.

 

 

9.     Here, you are given an aphelion distance, but not the semimajor axis.  You must solve for the semimajor axis.  Look around for more info.  You are told the eccentricity is 0.1.  Recall

Aphelion distance = semimajor axis (1 + eccentricity)

50 = a(1 + 0.1)

a = 45.45 AU

Now, solve for p

p2 = a3 = (45.45)3

p = 306 years

10.  This is a variation on the above problem (how clever of me to assign it!).  Here, you are given the aphelion and the perihelionÉhow can you use them to get the semimajor axis?  Recall that 2 times the semimajor axis is equal to the perihelion + the aphelion.

a= 52.5 AU  and so, with that, you can solve for p.  p = 380 years

You can use either the perihelion or the aphelion distance, coupled with the semimajor axis, to get the eccentricity.  IÕll use the aphelion distance:

 

Aphelion = semimajor axis (1 + e)

65  = 52.5(1 + e)

e = 0.24