ASTR 330
Homework #6 Solutions
1.) Using the radial velocity simulator, with the Star Properties, star mass set to 1 Msun , and the
System Orientation set to inclination of 90 degrees, find the period for a planet with the following semimajor axes (AU):
0.05 : 4.08 days
0.10 : 11.5 days
0.50 : 129 days
1.00 : 365 days
Does the eccentricity or planet mass change these values? Why or why not?
[2 pts each for 0.05-1.00, and 2 pts for written answer below]
The eccentricity does not change the period of the planet's orbit. Eccentricity tells us how elongated the orbit
is, but the semi-major axis doesn't change and it determines the period of the orbit.
2.) Using the Radial Velocity Simulator, set the star mass to 1 Msun, and the planet mass to 1 Mjup,
with an orbit semi-major axis of 1 AU. Now, find the amplitude of the radial velocity curve for the
following inclinations (dragging your mouse on the plot give radial velocity measurements, make sure
you are measuring the amplitude from the minimum to maximum).
90 deg. - 57 m/s
60 deg. - 50 m/s
30 deg. - 30 m/s
10 deg. - 10 m/s
5 deg. - 5 m/s
If our observations are limited to 10 m/s noise which of these planets might we be able to discover? The
first way to answer is that anything with an amplitude smaller than 10 m/s we couldn't find (which are
these?), but in the simulator you can turn off the theoretical curve and turn on the simulated
measurements increasing the number of points to see if you think you could find the pattern at each
setting.
[8 pts for above values, 2 pts for explanation below]
If we detect anything with a radial velocity greater than 10 m/s then we should discover everything with an
inclination greater than 10 degrees. Without the theoretical curve as guidance it is pretty difficult to pick out
the pattern at 30 degrees inclination without many points (~100). So 30 degrees inclination would be
difficult, but would with enough data would be possible.
3.) Using the radial velocity simulator, set the star mass to 1 Msun, and the planet mass to 1 Mjup,
with an orbit semi-major axis of 1 AU. Will an eccentric orbit be easier to discover? Vary the
eccentricity for this setting, and decide based on the amplitude of the curve, and also whether finding
a curve in noisy data becomes harder (turn off the theoretical curve again, and try to spot the curve in
noisy data).
[7 pts for enlarged amplitude explanation, 3 pts for explanation of finding curve in noisy data]
The theoretical curve for 0 eccentricity is a simple sine wave with an amplitude around 57 m/s, with
eccentricity of 0.8 the amplitude of the non-sinusoidal curve is ~95 m/s . So the overall amplitude is higher
for the eccentric orbits. This in principle should make detection easier, just based on the increased
amplitudes. However, since the eccentric orbits have pretty flat radial velocity curves, and one sharp peak, it
could be possible to mis-interpret data, or miss the peak entirely, and not find the true amplitude due to the
sharpness of the peak.
4.) Very recently a planet in the habitable zone of a Red Dwarf was found. The period was determined
to be 12.9 days, and the planet is only 5 times more massive than the Earth (Jupiter weighs 317 times
more than Earth). The planet is assumed to orbit at 0.073 AU, but that measurement assumes a mass
for the Red Dwarf star.
(a) What mass was used for that star in the scientist's calculations?
[[4 pts]
With a mass of 0.0157 MJ and a semimajor axis of 0.073 AU, the star must have a mass of ~0.31 Solar
Masses.
(b) For this observation what was the amplitude of the radial velocity curve (assuming zero
eccentricity and 90 deg. inclination)?
[4 pts]
The amplitude of the radial velocity curve is around 6 m/s for this scenario.
(c) What does the amplitude of this curve tell us about the noise level of the measurements and the
number made (make sure to experiment with the observations without the theoretical curve for this
one on the applet)?
[3 pts]
The noise on each observation must have been quite low, certainly below 5 m/s and probably closer to 1-2
m/s.
5. For the same planet and star properties as used in question #4 (with a star mass of 0.5, as it doesn't
go any lower), use the Planet Transit Applet for this question. The diameter estimated for this new
planet is 1.5 Earth radii, and Earth is 11 times smaller than Jupiter,
(a) Solve for the correct planet radii in terms of Jovian Radii, assuming zero eccentricity.
[2 pts]
The planets radii = 1.5*(1/11) = 0.136 Jupiter Radii.
(b) What range of possible inclinations will yield a transit down to 0.9995 (for example 84.1-95.9
degrees)?
[3 pts]
For a planet mass of 0.0157 Jupiter Masses, and a radii of 0.136 Jupiter Radii, around a star of 0.5 Solar
Masses, transits of depth 0.9995 are visible between 88.2-91.8 degrees.
(c)What does this tell us about the chances of observing a transit for this planet? Is there anything
about the successful Radial Velocity observation which suggest that the planet is more likely to have an
inclination closer to 90 than 0/180 (go back to that applet and experiment with the last setting,
changing the inclination)?
[5 pts]
This suggests that we need to be very lucky to observe this planet in transit, as we would generally expect
inclinations to be randomly sorted between 0-180 degrees, and only 3.6 degrees of that is viewable. 3.6/180
is not a big number, essentially a 2% chance. However, we discovered in question 2 that inclinations near 90
are more easily detected, so it is more likely that this system has an inclination between 30-150, but the odds
are still small. It would be nice if we could get the accurate mass for the star of 0.31, instead of 0.5, so that
would also increase the likelihood of transits.