Step 3 - Orbital Period and Distance
The orbital period, T, of a planet is the time it takes the planet to complete one full orbit around its star. If multiple transits of the same exoplanet are observed, then the time interval between consecutive transits – detected dips in the light curve – is a direct measure of the orbital period of the planet.
Based on the orbital period, T, we can derive the distance, d, between the planet and the star, using Kepler’s Third Law:
T^2 = (\frac{4\pi^2}{GM_s} )d^3
where G is the gravitational constant and {\text{M}_{\text{s}}} is the mass of the star.
Watch the video to learn more, complete your calculations and then check your solutions with our expert. When you’re ready to continue to the next step, return to this page and click “continue the investigation“.
Watch the video on exoplanet orbital period and distance:
Subtitles are available (automatically generated by YouTube) – select your language using the YouTube player controls.
Ready for the KELT-3b solution?
Have you solved the orbital period and distance of KELT-3b? Check below to see whether your results matched our expert’s solution for determining the orbital period and distance of KELT-3b.
Let’s now analyse KELT-3b data as an example. In this exercise you should pay close attention to the units.
- The gravitational constant in SI units is G = 6.67430 x 10^{-11} \text{m}^3 \text{kg}^{-1} \text{s}^{-2}
- The mass of the star KELT-3 is known: \text{M}_s = 1.96 \text{M}_\text{Sun}
- We need to convert its mass to SI units: \text{M}_s = 3.90 \text{x} 10^{30} \text{kg}
- From the model fit we have learned that the orbital period, T = 2.70339 days. Converting the orbital period to seconds: T = 233573 s.
We now have all the information needed to determine the distance between the star and exoplanet.
\text{d} = \sqrt[3]{\frac{\text{G}\text{M}_s}{4\pi^2}T^2} = \sqrt[3]{\frac{6.67430 \text{x} 10^{-11} \text{x} 3.90 \text{x} 10^{30}}{4\pi^2}233573^2} = 7.112 \text{x} 10^9 \text{m} = 0.048 au
Let’s now compare KELT-3b’s period and mean orbital distance to the planets in our Solar System:
Planet | Period (days) | Mean orbital distance (au) |
KELT-3b | 2.70339 | 0.048 |
Mercury | 87.97 | 0.4 |
| Earth | 365.25 | 1 |
| Neptune | 60266.25 | 30 |
Table 1: Comparison of the period and mean orbital distance for KELT-3b and planets in the Solar System.
KELT-3b has a much shorter orbital period than Mercury, the closest planet to the Sun in our Solar System, and is orbiting much closer to its host star. The transit photometry method finds planets in such close-in orbits more easily than it finds planets in much larger orbits like those of the outer planets in our Solar System.
When will the next transit of your exoplanet be? How does the orbital distance calculated using Kepler’s Third Law compare to the result from the best model fit value?
Step 3 Complete!
Have you analysed the Cheops data and determined the orbital period and distance of your exoplanet using Kepler’s Third Law? If yes, you can continue your investigation into the exoplanet’s properties with Step 4 – the temperature and habitability of an exoplanet!
