Download Explanations to selected mc

Detailed solution of Exercise 2 (Astronomy through history)
1. (a) Semi-major axis = (0.1 + 0.4) / 2 = 0.25 AU
(b) T2 = a3 = 0.253 => T = 0.125 year
2. (a) Since the comet follows a very elongated path, like the following:
Sun
Thus the average distance of the comet from the Sun is the semi-major axis a which is given by
T2 = a3 => 1252 = a3 => a = 25 AU
(b) The farthest distance of the comet from the Sun is approximately 2a = 50 AU
3. (a) Since the radius vector sweeps out equal area in equal time interval, so
from time t = 0 to t, area swept out = A1,
from time t = 0 to 2t, area swept out = A2 = A1 + A1 =2A1,
from time t = 0 to 3t, area swept out = A3 = A1 + A1 + A1 = 3A1,
Therefore ratio A1: A2 : A3 = 1 : 2 : 3
(b) The student is not correct. The distance traveled is average speed × time. Although the time interval 3t is
the longest, the average speed during this time interval may be low if the planet comes far from the Sun, so the
distance traveled may not be the largest.
4. (a) 1.5 x 109 km = 1.5 x 1012 / 1.50 x 1011 = 10 AU
1.4 x 109 km = 1.4 x 1012 / 1.50 x 1012 = 9.33 AU
(b) Semimajor axis = (10 + 9.33) / 2 = 9.67 AU
(c) Orbital period T is given by
T2 = a3
T=
9.673  30.1 years
5. If the hypothesis is true, the earth and the Vulcan must have the same period of rotation. But according to
Kepler’s 3rd law, different planets in the same solar system have different orbits and so must have different
periods of rotation. (Since T2  a3)
6. (a) On a celestial sphere, Mars generally moves from west to east throughout a year.
However, there is a period when Mars’s eastward motion stops and moves westwards before reversing
direction again. This backward loop is called retrograde motion.
(b)
Mars moves uniformly around a small circle, called an epicycle. The centre of the epicycle moved uniformly around
the Earth on a larger circle called a deferent. When the planet moves from B to C, retrograde motion occurs.
(c) (i) Earth travels faster
(ii) Between c and d
MC
1-5 A A B D B
6-10 C C A B D
11-15 B B D C B
16-18 D C B
Explanations to selected mc
1. T2 = a3
2.832 = 23 (By trial and error)
2. Statement 2, geocentric model uses epicycles and deferent to explain retrograde motion of planets.
Statement 3, geocentric model assumes that the Earth is at the center of the circular orbits of the planets.
3. Option B, a planet moves uniformly around epicycles with the center of the epicycle moved uniformly around
the Earth on deferent. This can explain retrograde motion.
5. Galileo also discovered that the Milky Way is made up of numerous stars.
6. a1 : a2 : a3 = 1 : 2 : 3
a13 : a23 : a33 = 13 : 23 : 33
Since T2 = a3,
T12 : T22 : T32 = 13 : 23 : 33
T1 : T2 : T3 = 13/2 : 23/2 : 33/2 = 1 : 23/2 : 33/2
7.
Given,
aA
2
aB
2
3
 TA   a A 
      23  8
 TB   a B 
 TA 
  8  2 2
 TB 
Hence 
8.
The speed of planet is the highest if the distance to the star is shortest. This is
from Kepler’s 2nd law: radius vector sweeps out equal areas in equal times.
9.
Acceleration = gravitational force / mass. Gravitational force is the smallest
when the distance is greatest (at S).
11. Most of the right hand side of Venus is illuminated
12. T2 = a3
For planet A, 2632 = aA3
=>
aA = 41.05 AU
aB = aA / 2 = 41.05 /2 = 20.52 AU
So TB = (20.523)1/2 = 93 years
14. Statement 3 is explained by Kepler’s third law, not second law.
16. Copernican’s model assumed circular orbit.