Download Exercise 17 Gravitation______Grade:______.

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 = t to 2t, area swept out = A2 = 2A1,
from time t = t to 3t, area swept out = A3 = 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.
When the comet is at the aphelion (A), it moves with the lowest speed. (1A)
Its speed increases as it approaches the Sun.
(1A)
When it comes to the perihelion (C), it moves with the highest speed. (1A)
Then when it moves away from the Sun, its speed decreases. (1A)
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 periods of rotation. (Since T2  a3)
6. A complete cycle of phases appears in Figure f but not in Figure e. (Full Venus cannot be observed from the
Earth in the geocentric model because Venus is always between the Sun and the Earth while in the geocentric
model, the Sun is sometimes between the Venus and the Earth and thus full Venus is possible.)
Therefore, the geocentric model could not explain Galileo’s observation but the heliocentric model could.
7. (a) On a celestial sphere, Mars generally moves from west to east throughout a year. (1A)
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. (1A)
(b) (i)
(Include epicycle and deferent, and with the Earth at the centre.) (1A)
The motion of a planet is composed of two separate circular motions as shown in the figure above.
When the planet moves from B to C, retrograde motion occurs. (1A)
(1A)
(ii)
(Concentric circular paths with the Sun at the centre.)
(1A)
The Earth moves faster than Mars. Sometimes the Earth passes Mars.
(1A)
When this happens, Mars appears to move backwards in the sky as observed on the Earth, as shown in
the figure above. (1A)
8. (a)
(b)
MC
He found that Venus showed a complete cycle of phase.
(i) No
(ii) In the Ptolemaic model, the labels ‘deferent’ and ‘epicycle’ are interchanged.
Besides, the Sun moves around the Earth in an orbit outside the deferent of Venus.
1-5 A A B D B
6-10 C D C C B
11-13 B C D
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. Point D is the perihelion, so the planet must move fastest at D.
8. Statement 2, Coperinicus still retained epicycles with the deferent centered on the Sun.
Statement 3, Kepler discovered the elliptical orbit.
9. By Kelper’s third law,
TA 2
a A3
=
TB 2
aB3
=>
TA 2
TB 2
=
a A3
aB3
a
T
=> A =  A
TB
 aB
13. Statement 3 is explained by Kepler’s third law.
3

2 2
2
 =   =
1
1

3