Shortest orbital period for a planet
- Who
- PSR J1719−1438 b
- What
- 2:10:37 hour(s):minute(s):second(s)
- Where
- Not Applicable
- When
- 25 August 2011
The planet with the shortest orbital period is PSR J1719−1438 b, which orbits the millisecond pulsar PSR J1719−1438 every 2 hours 10 minutes 37 seconds. This extra-solar planet, or exoplanet, is located around 4,000 light-years from Earth in the direction of the Serpens constellation. It was detected by astronomers studying variations in the spin of its parent pulsar on 25 August 2011.
Analysis of PSR J1719−1438 suggests that the planet orbiting it is around 40% of the size of Jupiter, but with around the same mass. That means it has an extraordinary density of 23 tonnes per cubic metre, and that its core is likely composed of crystalline carbon (similar to diamond). It orbits at a distance of 600,000 km (0.004 astronomical units), meaning that both the pulsar and its orbiting planet would fit inside the diameter of our Sun.
A pulsar is the collapsed core left behind after a supernova explosion, known as a neutron star. More specifically it is a type of neutron star distinguished by an extraordinarily powerful magnetic field and rapid rotation. These two traits mean that it generates a distinctive "pulse" – created by the magnetic poles sweeping across the sky – that can be detected by radio telescopes on Earth. PSR J1719−1438 has a mass 1.4 times greater than our own Sun, but is only around 38 km (23.6 mi) wide.