What is the escape velocity of Jupiter in km/s?
Answer
59.5 km/s
Answer
59.5 km/s
The escape velocity of Jupiter is 59.5 kilometers per second. This is the speed an object would need at Jupiter’s reference radius to escape the planet’s gravitational pull without further propulsion, ignoring atmospheric drag and other bodies.
Escape velocity depends on mass and distance from the center. Jupiter is vastly more massive than Earth, so its gravity is much stronger. Its escape velocity is more than five times Earth’s value of about 11.2 kilometers per second. The number is calculated from Jupiter’s mass and radius rather than measured from a rocket launch.
Because Jupiter is a gas giant, it has no solid surface like Earth. Astronomers therefore specify a reference level, commonly the visible cloud-top or standard-radius region, when quoting the value. The 59.5 km/s figure should not be confused with Jupiter’s surface gravity, which is about 24.79 m/s², or with the speed of a spacecraft already far from the planet. NASA’s Juno spacecraft used carefully planned high-speed trajectories and gravity assists while exploring the Jovian system.
Source: Wikipedia · fact-checked Aug. 2026