LONG QUESTIONS: MARKING SCHEME 1. A moon is orbiting a planet such such that the orbit is perpendicular perpendicular to the surface of the planet where an observer is standing. After some necessary scaling, suppose the orbit satises the following equation 9
(
Let
x 2
+
√ 3 y 2
) ( 2
− 4 + 25
)
2
−√ 3 x y + =225 2
2
r be the radius of the moon. Assume that the period of rotation of the
planet is much larger than the orbital period of the moon. Determine where
tan
θ 2 ,
θ is the elevation angle when the moon looks largest to the observer.
Answer and Marking Scheme: N!"ice N!"ice "he s"andard s"andard #ersi!n #ersi!n !$ !$ "he "he !r%i"s !r%i"s he ellipse may be obtained from a standard standard ellipse ellipse
&
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¿
x − 4
¿ ¿ ¿2 ¿ ¿
by rotating the standard ellipse with
respect to the origin, counterclockwise, !by
π / 3 radians". hen
c =√ a −b = √ 16 =4 2
2
' Es"a%(ished re(a"i!n %e"ween c!!rdina"es %e$!re and a$"er "rans$!rma"i!n. ¿ ¿ #or any point ( x , y ) on the ellipse, let ( x , y )
' &
be its coordinates before the transformation. #rom the equation of the ellipse, we can easily see that ¿ x √ 3 y x = +
2
¿
y=
2
[] ¿
−√ 3 x y +
2
[ ][ ] 1 2
x ¿ = y −√ 3
!$n fact,
2
2
√ 3 2
1 2
x
[ ]= − x y
2
3 + √ y 2
√ 3 x y 2
+
. "
2
) O%"ain e*+ressi!n $!r ca(c,(a"ing dis"ance $r!m a +!in" !n an e((i+se "! "he $!ci . %onsider standard ellipse
) &
y
(¿¿ ¿)2 3
2
=1
( x ¿− 4 )2 5
2
%hoose any point
( x ¿ , y ¿ ) on the ellipse. Let d 1 and d 2
+¿
be the distances from any point
( x ¿ , y ¿ ) on the ellipse to the foci ( 0,0 )
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y
y
¿
¿ ¿ and 2 2 ¿ 2 ¿ 2 d 1=( x ) +¿ d 2=( x −8 ) +¿ ¿ ¿
d ( 8,0 ) &
and
¿
hus,
y
¿
¿ ¿
¿
2
2
d 1=( x ) +¿ - Iden"i$. and charac"eri/ed "he c!!rdina"es when "he +(ane" (!!ks (arges" . he following is the position of the moon when it looks largest. At that point, its second coordinate equals r . hus, the
( x , r )
coordinates be
( 16 ( x¿ ) + 72 x ¿+ 81) = x +r
0
.
2
2
S!(#e e1,a"i!n
2
25
' &
"! !%"ain #a(,e !$ x .
0
herefore ¿ ¿ 16 ( x ) + 72 x + 81 ) ( d= = x + r 2
2 1
2
25
obtain the value of 'ubstituting
25 d
2
2 1
2
x
2
=25 x + 25 r =16
2
2
herefore, we can
x 2
+
x .
to the equation, we have
2
(
¿ x r √ 3 x = + .
by solving the above equation for
¿ x r √ 3 x = +
2
where
r √ 3 2
) ( 2
+ 72
x 2
+
r √ 3 2
)+
81 =¿
or 3 / 13
2
+ 25 r 2= 4 x 2 + ( 8 r √ 3 + 36 ) x + ( 12 r 2 + 36 √ 3 r + 81 )
2
−( 8 r √ 3 + 36 ) x + ( 13 r 2−36 r √ 3− 81 )=0
25 x 21 x
hen use quadratic formula to obtain value of
x to get larger value of
x in term of
tan
θ
r . %hoose smaller
r 2 x . he smaller one is
=
4 r √ 3 + 18 15 √ −r + 4 r √ 3 + 9 x = − 21 21 2
2 3ind "he e*+ressi!n !$
tan
θ
0
2 .
(ence tan
θ
r 2 x
= =
21 r 4 r √ 3 + 18 −15 √ −r
2
+ 4 r √ 3 + 9
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). wo massive stars A and * with mass
m A and
m B are separated by a
distance d . *oth stars are orbiting each other with respect to their center of gravity whose orbits are circular. 'uppose the stars lie on the X +Y plane !see #igure )" and are moving under gravitational force.
#igure )
a. %alculate the speed of star A and its angular velocity. An observer lies on the Y + Z plane !see #igure )" see the stars from the large distance with angle θ relatively to the Z +ais. (e-she measure that the velocity component of A to the line of his-her sight has the form K cos ( ωt + ε ) , with K and ε are positive. b. press the value
3
K / ωG in term of
m A ,
mB , and
θ where G is
the universal gravitational constant. he observer can then identify that the star A has mass equal to where
30 M S
M S is the 'un/s mass. 0n the other hand, he-she observe that the 5 / 13
star * produces +rays, so it could be a neutron star or a black hole. hese situations depend on m B 2 1" $f mB < 2 M S , then * is a neutron star. )" $f
m B > 2 M S , then * is a black hole. c. A
measurement
has
been
done
by
the
observer
which
results
3
1 K = M ωG 250 S . $f the value of
calculate
the
probability
∫ sin x dx =−cos x + C
of
cos θ
*
to
has the same probaiity, then be
a
black
hole.
!(int2
3se
"
Answer and Marking Scheme: a he center of gravity of the stars is relatively to the star A given by 4 m A r A = d m A + mB
) &
and since the orbit of A is a circle, then
F AX =
G m A m B 2
d
2
m v = A A r A
'o, we get
v A =mB
√
G ( m A +mB ) d
he angular velocity of A is given by
ω=
v A r A
=
√
G ( m A + mB ) d
3
"
% 4
$n %artesian coordinate system, the velocity of A is
v A ( t )= v A ⃗
(−sin ( ωt + ε ) ^i+ cos ( ωt +ε ) ^ j )
) &
3nit vector of the observer is
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^ + sin θ j^ r^ P =cos θ k v A in the line of the observer sight is given by ⃗
so the component of
v A ∙ r^ P= v A sin θ cos ( ωt + ε ) ⃗
'ince the component of
v A in the line of the observer sight is ⃗
K cos ( ωt + ε ) , then K = v A sin θ #inally, we have 3
3 mB K 3 sin θ = 2 ωG ( m A + m B )
c 4
&
#rom the result in b., namely eq. !44", we get 2
2 1 32 64 K 3 ( m A + m B ) sin θ= < = 3 250 8 125 ωG mB 3
'ince
θ ∈ [ 0, π ] , then
sin θ < 0,8 . hus, the probability of * is a black sin θ < 0,8 for
hole is the same as the probability of −1
si 0,8 ! 53
"
or
127
!44"
"
θ ∈ [ 0, π ] . 'ince
, then
B: 127
"
∫ sin θ dθ 5 3"
P ( ¿ #"$% )=1 − 180
"
=1 −0,6 =0,4
∫ sin θ dθ "
0
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8 / 13
5. 'uppose a static spherical star consists of N neutral particles with radius R !see #igure 1".
Figure 1
with
0 &θ & π ,
0 & ϕ & 2 π , satisfying the following equation of states
P' = ( k
) *−) 0 ln ( ) * / ) 0 )
!1" where P and V are the pressure inside the star and the volume of the star, respectively, and k is *olt6mann constant. ) * and ) 0 are the temperature at the surface that
r = * and the temperature at the center
r = 0 , respectively. Assume
) * &) 0 .
a. 'implify the stellar equation of states !1" if
+ ) =) * −) 0 ! 0 !this is called ideal
star" !(int2 3se the approimation ln ( 1 + x ) ! x for small x " 'uppose the star undergoes a quasi+static process, in which it may slightly contract or epand, such that the above stellar equation of states !1" still holds.
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b. #ind the work of the star when it epands from
) *
process where
) 0
and
' 1
' 2 in isothermal
to
are constants.
he star satises rst law of thermodynamics 2 ,= - Mc +
!)"
where Q, M, and W are heat, mass of the star, and work respectively, while c is the light speed in the vacuum and - M / M 0ia$− M iitia$ . $n the following we assume
) 0 to be constant, while
) * /)
C v in term of M and at
c. #ind the heat capacity of the star at constant volume constant pressure
( 1 + x ) ! 1 + x Assuming that
C 1 epressed in
varies.
C v and T !(int2 3se the approimation
for small x "
C v is constant and the gas undergoes the isobar process so the
star produces the heat and radiates it outside to the space. d. #ind the heat produced by the isobar process if the initial temperature and the nal temperature are ) i dan ) 0 , respectively. e. 'uppose there is an observer far away from the star. 7elated to point d., estimate the distance of the observer if the observer has 8.19 error in measuring the e:ective temperature around the star. ;ow we take an eample that the star to be the 'un of the mass
*⨀ , its luminosity !radiation energy emitted per unit time" 'un distance, f.
M ⨀ , its radius
2⨀ , and the arth+
d⨀ .
$f the sunlight were monochromatic with frequency
5 3 10
14
(6, estimate the
number of photons radiated by the 'un per second.
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C v of the 'un assuming its surface temperature
g. %alculate the heat capacity
runs from <<88 = until >888 = in this period.
Answer and Marking Scheme: a Dening 4
+ ) =) * −) 0 and
&
+ ) ! 0 , we have
+ ) ln ( 1 ++ ) / ) 0 )
P' = ( k
using
!5"
ln ( 1 + + ) / ) 0 ) ! +) / ) 0 , we then obtain
P' = ( k ) 0
!?"
% $f the gas undergoes isothermal process where 4 then the work has ' 2
∫
. = P d' = ( k ' 1
) *−) 0 ln ( ) * / ) 0 )
ln
) * and the
) 0 are constant, form
0
( ) ' 2 ' 1
2 2 c 4 = M c he internal energy of the star is ! 4 ( ) )= M () ) c for ideal 4 star". hus, the constant volume heat capacity of the star has the form + C v = = + M c 2 + ) ' + ) '
' 0
( ) ( )
for small
+ ) . hen, using rst law of themodynamics, the constant
pressure heat capacity of the star is +, + M C 1 = = + ) P + )
( ) ( )
for small
+ ) . Dening
'
2
c + P
+ ' + ' =C v + P + ) + )
+ ) =) 2−) 1 , then
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(
P + ' = ( k
) 1−) 0 + + )
(
ln ( ) 1+ + ) ) / ) 0
)
−
) 1−) 0 ln ( ) 1 / ) 0 )
)
3sing the approimation
(
)
ln ( ) 1 + + ) ) / ) 0 ! ln
(
+ ) 1+ ) 1 ln ( ) 1 / ) 0 )
( )+
−1
)
! 1−
) 1 ) 0
+ ) ) 1
+ ) ) 1 ln ( ) 1 / ) 0)
then we have
P
where
(
( ) −) 0 ) / ) + ' ( k 1− = + ) ln ( ) / ) 0 ) ln ( ) / ) 0 )
) 1 /) . #inally, we obtain
(
( ) −) 0 ) / ) + + M + ' ( k 2 1− = c + P =C v + + ) P + ) ' + ) ln ( ) / ) 0 ) ln ( ) / ) 0 )
( ) ( )
C 1 =
d 4
'ince
) ) ' &
C v is constant, the heat produced by the star is given by # =C v ( ) 0 −) i ) + P + '
¿ C v ( ) i−) 0 ) + ( k
(
) 0 −) 0 ln ( ) 0 / ) 0 )
e he error 8,19 , namely 4
) 0
) %00 =
) i−) 0 ln ( ) i / ) 0 )
) 0
) 0 −) %00
then
−
=
1 1000
999 ) 0 . #rom 'tefan law of black body radiation 1000
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, #
( )
999 =5 ) %00 =5 2 1000 4 π r0
where
4
4 4
) 0
r 0 is the observer/s distance which is given by 1 /2
( )( )
# r 0= 4 π5 $4
1000 999
2
−2
) 0
5
nergy per second radiated by the 'un 2⨀ = (67 where N is the number of photon. hus 26 2⨀ 3.96 3 10 ( = = =1.195 3 1045 photons − 34 14 6 7 6.6261 3 10 3 5 3 10
g 4
nergy per second radiated by the 'un is proportional to mass defect of the 'un
2⨀ =+ M c
6
2
hus, 2 2⨀ 3.96 3 1026 +Mc 23 C v = = = 8 / K =7.92 3 10 8 / K + ) + ) 6000 −5500
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