Astronomy Quiz: Exoplanet Detection Methods

20 questions · exam conditions

0:00

Exoplanet Detection MethodsQuestion 1 of 20

An exoplanet orbits a G-type star. Its transit light curve is perfectly symmetrical and U-shaped. The radial velocity curve for the star is perfectly sinusoidal. What can be inferred about the planet's orbit?

The orbit has a high eccentricity, and the observer's line of sight is aligned with the semi-major axis.

The orbit is circular, and its inclination is exactly 90 degrees relative to the plane of the sky.

The orbit has zero or very low eccentricity, and its inclination is close to 90 degrees.

The orbit is retrograde, and it has a moderate eccentricity that is averaged out by the observations.

← Back to quizzes

Astronomy Quiz

Astronomy Quiz: Exoplanet Detection Methods

Practice Exoplanet Detection Methods in Astronomy with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Exoplanet Detection Methods, giving you a quick way to practice the rules, question types, and explanations that matter most for Astronomy.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The orbit has a high eccentricity, and the observer's line of sight is aligned with the semi-major axis.
The orbit is circular, and its inclination is exactly 90 degrees relative to the plane of the sky.
The orbit has zero or very low eccentricity, and its inclination is close to 90 degrees. (correct answer)
The orbit is retrograde, and it has a moderate eccentricity that is averaged out by the observations.

Explanation: A perfectly sinusoidal radial velocity curve is characteristic of a circular orbit (zero eccentricity). A highly eccentric orbit would produce an asymmetric, non-sinusoidal velocity curve. The presence of a U-shaped transit indicates that the orbital inclination

i

is close to 90 degrees (edge-on), allowing the planet to pass in front of the star from our perspective. While an exact 90-degree inclination produces a central transit, a symmetrical transit and sinusoidal RV curve primarily point to a circular or near-circular orbit with a high inclination.

Question 2

Observations of a star reveal periodic variations in the timing of its planet's transits; sometimes the transits occur slightly early, and sometimes slightly late, relative to a perfect clock-like schedule. What is the most likely cause of these Transit Timing Variations (TTVs)?

The host star's rotation is causing a precession in the planet's orbital plane.
The gravitational influence of at least one other planet in the same system is perturbing the transiting planet's orbit. (correct answer)
The planet's orbit is highly eccentric, causing its orbital speed to vary significantly.
The presence of large, long-lived starspots on the host star is interfering with the measurement of the transit midpoint.

Explanation: Transit Timing Variations (TTVs) are caused by the gravitational tugs of other bodies, usually other planets, in the same star system. These tugs cause the transiting planet to speed up or slow down slightly in its orbit, leading to transits that are not perfectly periodic. This effect is a powerful tool for discovering and measuring the masses of non-transiting planets in a system. While eccentricity and starspots can affect the shape and measured depth of a transit, they do not cause the periodic early/late variations characteristic of TTVs.

Question 3

A planet is found via radial velocity around a star identical to the Sun. The RV curve has a period of one year and an amplitude indicating a Jupiter-mass planet. Why would the transit method likely fail to detect this planet, even with perfect and continuous observation?

A Jupiter-mass planet is not large enough to cause a detectable transit against a Sun-like star.
The one-year orbital period means the transit would last too long to be distinguished from stellar variability.
The planet's orbital inclination is statistically unlikely to be aligned with Earth's line of sight. (correct answer)
The radial velocity signal from a Jupiter-mass planet at that distance would be too small to be real.

Explanation: For a planet orbiting a Sun-like star at 1 AU (one-year period), the probability of its orbit being aligned for a transit is only about 0.5%. This means that for every 200 such planets that exist, only one will, on average, produce a transit observable from Earth. Therefore, even if the planet is confirmed by RV and is large enough to produce a detectable dip, the geometric alignment is statistically very improbable. A Jupiter-sized planet would cause a very easily detectable ~1% dip in brightness, so choice A is incorrect.

Question 4

The radial velocity method is used to discover a planet around a Sun-like star. The measurement indicates a minimum mass $m \sin i$ of 1 Earth mass. Why is it highly unlikely that the true mass of this planet is 100 Earth masses?

Planets with 100 Earth masses are physically impossible to form around Sun-like stars.
A 100 Earth-mass planet would produce a radial velocity signal too large to be interpreted as 1 Earth mass.
A 100 Earth-mass planet would have a radius so large that it would have been detected by a transit survey first.
For the true mass to be 100 times the minimum mass, the orbital inclination $i$ would have to be extremely small. (correct answer)

Explanation: When you encounter radial velocity exoplanet detection questions, focus on the relationship between the measured minimum mass

m \sin i

and the orbital inclination angle

i

. The radial velocity method detects the wobble of a star caused by an orbiting planet, but it only measures the component of motion along our line of sight. The key insight is understanding what

\sin i

represents geometrically. Since the true mass equals

(m \sin i) / \sin i

, if the minimum mass is 1 Earth mass but the true mass is 100 Earth masses, then

\sin i = 1/100 = 0.01

. This corresponds to an inclination angle of approximately 0.6 degrees, meaning the orbital plane would be nearly perfectly face-on to Earth. While theoretically possible, this represents an extremely unlikely geometric configuration. Option A is wrong because 100 Earth-mass planets (super-Earths or mini-Neptunes) can and do form around Sun-like stars. Option B misunderstands the measurement process—a more massive planet at a small inclination angle would produce the same radial velocity amplitude as a less massive planet at a larger inclination. Option C incorrectly assumes all large planets would transit; transit probability depends on orbital geometry and distance, not just planetary size. Option D correctly identifies that achieving a 100-fold difference between minimum and true mass requires an extremely small orbital inclination, which is statistically very improbable. Study tip: Remember that

\sin i

approaches zero as inclination approaches zero degrees (face-on orbits), making very large mass ratios geometrically unlikely in radial velocity surveys.

Question 5

An astronomer detects a periodic signal using the radial velocity technique, but no corresponding transit is observed. Which of the following is the LEAST plausible explanation for the absence of a transit?

The planet's orbital plane is not sufficiently aligned with the observer's line of sight.
The planet is too small to produce a detectable change in the star's brightness.
The planet has a very low density, making it a large but low-mass 'super-puff' planet. (correct answer)
The observation window was not long enough to cover the specific time of the predicted transit.

Explanation: A 'super-puff' planet has a very large radius for its mass. While its low mass might produce a smaller RV signal, its large radius would produce a very deep, easily detectable transit. Therefore, a low-density planet that is massive enough to be found with RV is an excellent candidate for transit detection, making this the least plausible reason for a non-detection. The other options are all common reasons for non-detection: orbital misalignment (the most common reason), a planet too small to block enough light, or simply bad timing/incomplete phase coverage.

Question 6

A planet is detected via the radial velocity method, yielding a sinusoidal velocity curve with a semi-amplitude of $K$ and a period of $P$ . If the inclination of the planet's orbit to the line of sight, $i$ , is later determined to be 45°, how does the planet's true mass $m$ relate to the initially calculated minimum mass, $m_{min}$ ?

$m = m_{min}$ because inclination does not affect the mass calculation from the velocity curve.
$m = m_{min} / \sin(45°)$ , meaning the true mass is greater than the minimum mass. (correct answer)
$m = m_{min} \times \sin(45°)$ , meaning the true mass is less than the minimum mass.
$m = m_{min} \times \cos(45°)$ , meaning the true mass is less than the minimum mass.

Explanation: The radial velocity method measures the component of the star's velocity along the line of sight, which is

v_{obs} = v_{true} \sin i

. This leads to a mass determination of

m \sin i

, which is the minimum mass

m_{min}

. Therefore,

m_{min} = m \sin i

. To find the true mass

m

, one must rearrange the formula to

m = m_{min} / \sin i

. Since

\sin(45°) \approx 0.707

, the true mass is

m = m_{min} / 0.707 \approx 1.414 m_{min}

, which is greater than the minimum mass.

Question 7

Two planets, Kepler-X and Kepler-Y, are discovered via the transit method. Kepler-X produces a transit depth of 1.0% and has an orbital period of 20 days. Kepler-Y produces a transit depth of 0.5% and has an orbital period of 200 days. Assuming both host stars are identical, which conclusion is most strongly supported?

Kepler-X is larger in radius than Kepler-Y, and it orbits closer to the host star. (correct answer)
Kepler-Y is larger in radius than Kepler-X, but it orbits farther from the host star.
Kepler-X has a higher mass than Kepler-Y, and it has a higher surface temperature.
Kepler-Y has a lower density than Kepler-X, and it orbits in a more inclined plane.

Explanation: Transit depth is proportional to

(R_p/R_s)^2

. Since the stars are identical, a larger transit depth implies a larger planetary radius. Kepler-X has a depth of 1.0%, while Kepler-Y has 0.5%, so Kepler-X is larger. According to Kepler's Third Law, for a given star, a shorter orbital period implies a smaller semi-major axis (closer orbit). Kepler-X's 20-day period is much shorter than Kepler-Y's 200-day period, so it orbits closer. The transit method alone does not provide information about mass or density.

Question 8

A candidate transit signal is found with a period of 30 days and a depth of 0.8%. Follow-up radial velocity measurements are taken, but they show no corresponding wobble at a 30-day period. However, they do show a small wobble with a 60-day period. What is the most likely identity of the candidate signal?

A planet in a 30-day orbit whose mass is too low to be detected by the current RV instrument.
A background eclipsing binary star system with an orbital period of 60 days. (correct answer)
A transiting planet in a 30-day orbit being perturbed by a second, non-transiting planet in a 60-day orbit.
A large, rotating starspot on the star's surface which has a rotation period of 30 days.

Explanation: This scenario is a classic false positive. A background eclipsing binary, located behind the target star, can create a transit-like signal. If its orbital period is 60 days, it will have two eclipses per orbit (primary and secondary). If the stars are different sizes, these eclipses may have different depths. It is possible that only the primary eclipse is deep enough to be detected, creating a signal with a period of 60 days. The observer might misinterpret alternating deep and shallow eclipses as a single type of event, folding the data incorrectly to find a 30-day period. The RV signal from the target star would be unaffected, but the background binary itself might be detectable if it's bright enough, hence the 60-day RV signal. This explains both the apparent transit period and the true RV period.

Question 9

Consider a planet of a fixed radius transiting two different main-sequence stars, Star A and Star B. Star A is more massive and larger than Star B. How will the transit depth and duration for the transit across Star A compare to that across Star B, assuming the planets orbit at the same physical distance?

The depth will be smaller and the duration will be shorter.
The depth will be smaller and the duration will be longer. (correct answer)
The depth will be larger and the duration will be shorter.
The depth will be larger and the duration will be longer.

Explanation: Transit depth is

\Delta F/F \approx (R_p/R_s)^2

. Since the planet's radius

R_p

is fixed and Star A is larger (larger

R_s

), the ratio

R_p/R_s

is smaller, resulting in a smaller (shallower) transit depth. For transit duration, Star A is more massive, so at the same orbital distance, the planet must orbit faster (from Kepler's laws,

v = \sqrt{GM/r}

). However, the path across the larger star's disk is also longer. The duration is roughly

2R_s/v

. Since

v

scales with

\sqrt{M_s}

and the path scales with

R_s

, the duration scales with

R_s / \sqrt{M_s}

. For main-sequence stars, radius scales roughly with mass (

R \propto M^{0.8}

). Therefore, duration scales roughly as

M^{0.8} / M^{0.5} = M^{0.3}

. Since Star A is more massive, the duration will be longer.

Question 10

Kepler-X is larger in radius than Kepler-Y, and it orbits closer to the host star. (correct answer)
Kepler-Y is larger in radius than Kepler-X, but it orbits farther from the host star.
Kepler-X has a higher mass than Kepler-Y, and it has a higher surface temperature.
Kepler-Y has a lower density than Kepler-X, and it orbits in a more inclined plane.

Explanation: Transit depth is proportional to

(R_p/R_s)^2

Question 11

An astronomer measures the radial velocity of two stars, Star A and Star B, which are known to be equally massive. Both stars host a planet with the same orbital period. The amplitude of the radial velocity curve for Star A is twice the amplitude for Star B. What can be inferred about the planets?

The planet orbiting Star A has twice the radius of the planet orbiting Star B.
The product $m_p \sin i$ for Planet A is twice the value for Planet B. (correct answer)
The orbital inclination of Planet A is twice the orbital inclination of Planet B.
The planet orbiting Star B is twice as far from its star as the planet orbiting Star A.

Explanation: The amplitude (K) of the radial velocity curve is proportional to

m_p \sin i

and inversely proportional to the square root of the star's mass and the orbital semi-major axis. Since the stellar masses and periods (which implies the semi-major axes) are the same, the amplitude K is directly proportional to the quantity

m_p \sin i

. If the amplitude for Star A is twice that for Star B, then the minimum mass (

m_p \sin i

) of Planet A must be twice that of Planet B. We cannot separate the mass from the inclination with this information alone.

Question 12

An astronomer observes a star using the transit method and finds a periodic dip in brightness. If the star were actually an unresolved binary system of two identical stars, and the planet transits only one of them, how would the measured planetary radius compare to the planet's true radius?

The measured radius would be approximately correct because the transit depth depends only on the size of the star being transited.
The measured radius would be overestimated by a factor of approximately $\sqrt{2}$ because the total luminosity is doubled.
The measured radius would be underestimated by a factor of approximately $\sqrt{2}$ because the total light from the system dilutes the transit depth. (correct answer)
The measured radius would be underestimated by a factor of approximately 2 because the total area of the stellar disks is doubled.

Explanation: The transit depth (

\Delta F/F

) is proportional to the square of the ratio of the planet's radius to the star's radius,

(\frac{R_p}{R_s})^2

. The astronomer assumes the total flux

F

comes from a single star. However, in this binary scenario, the total flux is

F_{total} = F_1 + F_2 = 2F_1

. The change in flux during transit is

\Delta F = F_1 (\frac{R_p}{R_{s1}})^2

. The observed depth is

\frac{\Delta F}{F_{total}} = \frac{F_1 (R_p/R_{s1})^2}{2F_1} = \frac{1}{2}(\frac{R_p}{R_{s1}})^2

. The astronomer, assuming a single star, would calculate a radius

R_{p, measured}

such that

(\frac{R_{p, measured}}{R_{s1}})^2 = \frac{1}{2}(\frac{R_p}{R_{s1}})^2

. Solving for

R_{p, measured}

gives

R_{p, measured} = \frac{R_p}{\sqrt{2}}

. Thus, the radius is underestimated by a factor of

\sqrt{2}

Question 13

The orbit has a high eccentricity, and the observer's line of sight is aligned with the semi-major axis.
The orbit is circular, and its inclination is exactly 90 degrees relative to the plane of the sky.
The orbit has zero or very low eccentricity, and its inclination is close to 90 degrees. (correct answer)
The orbit is retrograde, and it has a moderate eccentricity that is averaged out by the observations.

i

Question 14

The primary advantage of detecting an exoplanet with both the transit and radial velocity methods, as opposed to just one, is that it allows for the direct calculation of the planet's:

surface temperature, by combining its distance from the star with its albedo.
orbital eccentricity, by comparing the shape of the transit and the velocity curve.
atmospheric composition, by analyzing the star's spectrum during the transit.
bulk density, by combining the radius from the transit and the mass from the radial velocity. (correct answer)

Explanation: The transit method measures the planet's radius (relative to its star). The radial velocity method measures the planet's minimum mass (m sin i). If a planet transits, we know the inclination (i) is very close to 90°, so sin i ≈ 1, and the RV method gives us the true mass. With both mass (m) and radius (R), the bulk density (

\rho = m/V

, where

V = \frac{4}{3}\pi R^3

) can be calculated. This is a crucial step in characterizing a planet as rocky, gaseous, or icy.

Question 15

The radial velocity method fundamentally relies on measuring:

the periodic dimming of the planet as it passes behind the star.
the Doppler shift of spectral lines in the light from the host star. (correct answer)
the gravitational lensing effect of the planet as it passes in front of a background star.
the Doppler shift of reflected light from the planet's atmosphere.

Explanation: The radial velocity method detects planets indirectly by observing their gravitational effect on their host star. As the planet orbits, it pulls the star in a small counter-orbit. This motion of the star towards and away from Earth causes the star's light to be periodically blueshifted and redshifted. This is observed as a Doppler shift in the star's spectral absorption or emission lines. The planet's own light is far too faint to be measured this way.

Question 16

Why is the probability of detecting a planet via the transit method significantly lower for planets with long orbital periods compared to those with short orbital periods?

Planets in long-period orbits are typically smaller and block less light, making their transits too shallow to detect.
The gravitational influence of planets in long-period orbits is too weak to maintain a stable, transiting alignment.
Stellar activity and noise are more likely to obscure the infrequent and long-duration transits from distant planets.
The required geometric alignment of the star, planet, and Earth is much less likely for a planet that is farther from its star. (correct answer)

Explanation: When evaluating planet detection methods, you need to consider both the physics of the detection technique and the geometric constraints involved. The transit method depends on a very specific alignment where the planet passes directly between its star and Earth, temporarily blocking a small fraction of the star's light. The correct answer is D because geometric probability decreases dramatically with orbital distance. For a transit to occur, the planet's orbital plane must be aligned within a very narrow range of our line of sight to the star. This critical angle gets smaller as the planet's orbit increases. A planet close to its star has a relatively wide "window" of acceptable orbital orientations that would produce transits visible from Earth. However, a planet farther out requires much more precise alignment - the orbital plane must be oriented within an increasingly narrow range to cross our sightline to the star. Let's examine why the other options miss the mark. Option A incorrectly assumes planet size correlates with orbital period - many long-period planets are actually gas giants that would create detectable transits. Option B misunderstands orbital mechanics; gravitational stability isn't the issue since many confirmed long-period planets maintain stable orbits and transit regularly. Option C focuses on observational challenges like stellar noise, but the fundamental limitation isn't technical - it's geometric. Remember this key principle: in astronomy, detection probability often comes down to geometric constraints. When you see questions about why certain detection methods favor specific types of objects, consider whether alignment, orientation, or viewing angle creates the primary limitation.

Question 17

The host star's rotation is causing a precession in the planet's orbital plane.
The gravitational influence of at least one other planet in the same system is perturbing the transiting planet's orbit. (correct answer)
The planet's orbit is highly eccentric, causing its orbital speed to vary significantly.
The presence of large, long-lived starspots on the host star is interfering with the measurement of the transit midpoint.

Question 18

The planet orbiting Star A has twice the radius of the planet orbiting Star B.
The product $m_p \sin i$ for Planet A is twice the value for Planet B. (correct answer)
The orbital inclination of Planet A is twice the orbital inclination of Planet B.
The planet orbiting Star B is twice as far from its star as the planet orbiting Star A.

Explanation: The amplitude (K) of the radial velocity curve is proportional to

m_p \sin i

m_p \sin i

. If the amplitude for Star A is twice that for Star B, then the minimum mass (

m_p \sin i

) of Planet A must be twice that of Planet B. We cannot separate the mass from the inclination with this information alone.

Question 19

A radial velocity signal is detected from a star, but its shape is not a simple sinusoid. Instead, it resembles a sawtooth pattern, rising sharply and then declining slowly. This pattern is characteristic of a planet in a(n):

highly inclined orbit viewed nearly face-on.
circular orbit that is precessing due to stellar oblateness.
retrograde orbit that opposes the star's natural rotation.
highly eccentric orbit, where the star's speed changes dramatically. (correct answer)

Explanation: When analyzing radial velocity data from exoplanet detection, the shape of the velocity curve tells you about the planet's orbital characteristics. A standard circular orbit produces a smooth sinusoidal pattern, but deviations from this reveal important orbital properties. The sawtooth pattern described—with sharp rises and slow declines—indicates dramatically varying orbital speeds. This happens in highly eccentric orbits where the planet follows Kepler's laws: it moves fastest at periapsis (closest approach) and slowest at apapsis (farthest point). The sharp velocity changes occur when the planet rapidly swings around the star at periapsis, while the gradual changes correspond to the slower motion at apapsis. This creates the characteristic asymmetric, sawtooth-like radial velocity signature, making D correct. A is wrong because orbital inclination affects the amplitude of the radial velocity signal, not its shape—a face-on orbit would actually produce little to no detectable radial velocity signal. B is incorrect because orbital precession would cause gradual shifts in the timing or orientation of the velocity curve over long periods, not the sharp asymmetric pattern described. C is wrong because retrograde motion affects the direction of orbital motion relative to stellar rotation, but doesn't change the fundamental relationship between orbital eccentricity and velocity variations. Remember: when you see asymmetric or "kinked" radial velocity curves on astronomy exams, think eccentricity. Circular orbits produce smooth sinusoids, while eccentric orbits create the dramatic speed variations that distort these curves into more complex shapes.

Question 20

The depth will be smaller and the duration will be shorter.
The depth will be smaller and the duration will be longer. (correct answer)
The depth will be larger and the duration will be shorter.
The depth will be larger and the duration will be longer.

Explanation: Transit depth is

\Delta F/F \approx (R_p/R_s)^2

. Since the planet's radius

R_p

is fixed and Star A is larger (larger

R_s

), the ratio

R_p/R_s

is smaller, resulting in a smaller (shallower) transit depth. For transit duration, Star A is more massive, so at the same orbital distance, the planet must orbit faster (from Kepler's laws,

v = \sqrt{GM/r}

). However, the path across the larger star's disk is also longer. The duration is roughly

2R_s/v

. Since

v

scales with

\sqrt{M_s}

and the path scales with

R_s

, the duration scales with

R_s / \sqrt{M_s}

. For main-sequence stars, radius scales roughly with mass (

R \propto M^{0.8}

). Therefore, duration scales roughly as

M^{0.8} / M^{0.5} = M^{0.3}

. Since Star A is more massive, the duration will be longer.