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Rutgers University *
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Course
341
Subject
Physics
Date
Dec 6, 2023
Type
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6
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Physics 341: Problem Set #6
due November 1 by 2:00 PM in PDF format on Canvas [40 pts]
You are encouraged to work in groups on these problems, but each student must write up the
solutions individually. You must also list your collaborators on your solutions, and cite any
external sources you used (other than the course notes or textbook). Using online solutions
to these or similar problems is not allowed.
Show your work and explain your reasoning. The number of points for each problem is listed;
partial credit will be given. Be careful with units.
Questions asking about concepts require at most a few sentences to answer. Please make
your answers clear and
concise
.
Figure 1: Radial velocities of the star TrES-3, from Sozzetti et al. (2009, ApJ, 691, 1145).
The period is 1.31 days.
1. TrES-3, shown in Figure 3, is a transiting extrasolar planetary system, with an edge-on
circular orbit. At what orbital phase does the planet go in front of the star? [5 pts]
A. 0.25
B. 0.50
C. 0.75
D. 1.00
E. either 0.50 or 1.00; can’t tell which with just the radial velocity curve
Answer: B
1
2. Imagine that the Kepler space telescope has discovered an “Earth-like” planet outside
of our solar system that is twice the radius of our own Earth, but the same mass. The
newly discovered planet happens to have a moon the same size, mass, and distance as
our own moon. Which of the following is true? [5 pts]
A. The magnitude of the gravitational force on the new Earth from its moon is less
than the magnitude of the gravitational force in our own Earth-Moon system.
B. The tidal force on the new Earth from its moon is larger than the tidal force in our
own Earth-Moon system.
C. The tidal force on the new Earth from its moon is exactly the same as in our own
Earth-Moon system.
D. The tidal force on the new Earth from its moon is less than the tidal force in our
own Earth-Moon system.
Answer: B
3. Shown is the light-curve for a planet transiting in front of a star. The time difference
t
B
−
t
A
is 32 minutes and the time difference
t
C
−
t
A
is 227 minutes. [10 pts]
(a) The star has a radius of 1
.
5
R
⊙
. What is the radius of the planet in units of the
radius of Jupiter?
2
The fraction of the light blocked by the planet is
f
= 0
.
02
. This is related to the
radii of the star and the planet as
f
=
R
2
planet
R
2
star
R
planet
=
p
fR
star
=
√
0
.
02(1
.
5
R
⊙
)
=
(0
.
21
R
⊙
)
6
.
955
×
10
8
m
1
R
⊙
1
R
J
7
.
149
×
10
7
m
=
2
.
06
R
J
Alternatively, one can see that the radius of the planet and the star must be given
by:
2
R
star
=
(
v
star
+
v
planet
)(
t
C
−
t
A
)
2
R
planet
=
(
v
star
+
v
planet
)(
t
B
−
t
A
)
.
Therefore, we can solve for the planet’s radius
R
planet
R
star
=
t
B
−
t
A
t
C
−
t
A
R
planet
=
t
B
−
t
A
t
C
−
t
A
R
star
=
32
227
(1
.
5
R
⊙
)
=
(0
.
141)(1
.
5
R
⊙
)
6
.
955
×
10
8
m
1
R
⊙
1
R
J
7
.
149
×
10
7
m
=
2
.
06
R
J
Finally, one could solve directly for
v
planet
+
v
star
in this step, and then use the
previous equations to solve for
R
planet
, solving for (b) as in intermediate step.
(b) The star is measured to have a maximum radial velocity of 0
.
5 km/s, how fast is
the planet moving?
We know that the time difference
t
B
−
t
A
is related to the velocities of the planet
and the star by
2
R
planet
= (
v
star
+
v
planet
)(
t
B
−
t
A
)
.
Therefore,
v
planet
=
2
R
planet
t
B
−
t
A
−
v
star
3
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