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Physics Test 1 (Astronomy and Space Science) Suggested Answers
1. (a)
L
LSUN

R2 T 4
2
4
RSUN TSUN
(1M)
24004
LSUN
57774
L = 2.69 x 104 LSUN
L
 9502 
(1A)
(b) The absolute magnitude of a star is the apparent magnitude of the star when it were placed at 10 pc from
Earth. (1A+1A)
(c) (i) m – M = 5 log d/10
0.4 - (-6.0) = 5 log d/10
d = 190.5 pc
(1M)
(1A)
(ii) 1.6 – (-2.7) = 5 log d/10
d = 72.4 pc
(d)
apparent brightness of Bellatrix
 2.512(1.6  0.4 )  3.02
apparent brightness of Betelgeus e
(1M+1A)
(e)
1A for lower peak, 1A for peak shift to right
2.
(a)
Length of the semi-major axis of the orbit of Comet Halley:
(0.6  35.1)
=
2
= 17.85 A.U.
[1M]
By the Kepler’s third Law,
 4 2  3
a
T2 = 

 GM 
 4 2
(75.3  365  24  60  60)2 = 
 GM

 (17.85  1.5  1011)3


[1M]
M = 2.0  1030 kg
(b)
[1A]
By conservation of mechanical energy, (Kc – Kf) is the difference in gravitational energy of Halley
when it is closest to and furthest away from the Sun.
(Kc  K f )
GM
GM


=
 
11 
11
mH
0.6

1.5

10
35.11.5 10


[1M]
= 1.5  109 J
(c)
MC
[1A]
Yes.
[1A]
Kepler’s second law states that the line segment joining the centre of the Sun and the centre of Halley
should swept-up equal areas in any given period of time.
[1A]
It means that when Halley is furthest away from the Sun, its speed must be lower than the speed it has
when it is closest to the Sun. [1A]
Therefore, Kc – Kf > 0。
[1A]
1-5 B B C A D
6-10 A B C C B
11-15 D C C B C
16 C
Explanations to selected mc
1. The celestial sphere completes one revolution every day.
3. Let D = Diameter of sun
D = 1AU x 0.5 x /180
 = S/r = 7.8 x 1012 x (D/2) / 7.0 x 1013 = 4.86 x 10-4 rad
6.
By Kepler’s third law:
T2 = Ka3
for the same central star where K is a constant,
 Tx

 Ty





2
a
=  x
 ay





3
With this formula:
Option A is incorrect as planet D has the shortest orbital period.
Option B is correct as the orbital period of planet B is
 5 


 0.5 
D.Option C is incorrect as the orbital period of planet A is
3
= 31.6 times the orbital period of planet
 10 
 
5
3
= 2.83 times the orbital period of planet
B.
Option D is incorrect as the orbital period of planet C is
 15 
 
 10 
3
= 1.84 times the orbital period of planet A.
9. Explaining planetary motion with a single unified theory is the contribution of Newton’s law of gravitation.
10. Kepler’s second law states that the line segment joining the centre of the Sun and the centre of Halley should
swept-up equal areas in any given period of time. Angular speed from P to P’ is higher than that from Q to Q’.
11. Parallax p = 0.75/2 = 0.375”.
d = 1/p = (1 / 0.375) x 206265 = 5.50 x 105 AU
14. Luminosity of a blackbody is proportional to R2T4.
Statement 1, similar colors implies similar temperature, but the size (radius) may be differ a lot, so the two stars
may have different luminosities.
Statement 2, two stars of similar colors implies they have similar surface temperature, not core temperature.
Statement 3, two stars of similar surface temperature have similar intensity since I = T4.
15. Consider the doppler effect:
 vr

0
c
40 000

=
8
(3  10 )
(656.1 10  9 )
Δλ = 0.088 nm
Since the star is moving towards the Earth, the line should be blue-shifted.
16. To escape the gravitational attraction of the Sun, total mechanical energy of the object must be positive, i.e.
1 2
GMm
mv 
> 0, and rv2 > 2GM.
2
r
Only option C satisfies this condition.