# Roughly, how fast do rockets travel?

I know that different rockets are built with different a range of different sizes or purposes, but there must be some kind of range of typical speeds for those things when leaving the Earth.

So how fast are rockets or any other human made ship when leaving the Earth, after the engines shut down?

• I've tried to clarify the wording of your question a bit to make some things clearer. The speed is constantly changing while the engines are still running, so I added "...after the engines shut down" to it. You might consider trying to do a little investigation yourself. For example how fast does an object move in low Earth orbit? (LEO) roughly? How far is the moon, and how long did it take the Apollo missions to reach the moon once they left Earth orbit? – uhoh Aug 20 '17 at 12:12
• Speed depends on your frame of reference. The Earth is moving at some 30km/sec along its orbital path around the Sun, which is orbiting the galactic center at some 220 km/sec. To achieve a low Earth orbit, a rocket must accelerate its payload by about 7.8km/sec; Apollo had to accelerate to about 10.4 km/sec to get to the Moon, and Voyager 1 has attained speeds of about 17km/sec on its journey through the solar system and beyond. – Anthony X Aug 20 '17 at 14:20
• You'll probably need to think what you mean by"leave the earth" - do you mean go into orbit? Which orbit? Do you mean to leave orbit? – Rory Alsop Aug 20 '17 at 14:28
• @Rory Alsop I meant to leave orbit, aiming for a inter stellar trip. – Matthew Aug 20 '17 at 14:31
• In that case: 25000mph is escape velocity. – Rory Alsop Aug 20 '17 at 14:40

Your ballpark estimate, a velocity you can compare everything else to is: 8 km/s. This is the orbital speed in Low Earth Orbit (LEO); you'll see this number quite frequently. It's something to get started with. You need that much of "horizontal" speed, on top of whatever's needed to lift the rocket above the atmosphere to enter the orbit.

Circular orbit velocity: $v_o = \sqrt{{GM} \over r}$. Escape velocity $v_e = \sqrt{{2GM} \over r}$. ${v_e \over v_o} = \sqrt{2}$. So, multiply LEO speed by $\sqrt{2}$ (resulting in 11.3km/s) for escape speed of Earth. That multiplier applies to any body and any circular orbit, giving the speed needed for leaving given body. The speed for interplanetary transfers when escaping Earth will usually be higher - usually only slightly so (for Mars and Venus transfers), so that the probe reaches the target instead of escaping Earth gravity and ending up nowhere in particular.

Then more advanced orbital mechanics kicks in, varying the speed wildly, regardless of what you measure the speed relative to. Earth moves around the Sun at ~30km/s so once you escape Earth gravity, this is the baseline to which you add (or subtract) your travel speed.

In strongly eccentric elliptical orbits, the speed will vary depending on where in the orbit you are: near the body it will be close to escape speed; at apoapsis you'll crawl at scarce meters per second, taking weeks until gravity pulls you back in for something of like half an hour dashing past the planet. Low gravity bodies, like comets or asteroids will have very slow orbits, you could enter orbit of Cruithne, a 10km asteroid, by running.

High circular orbits are much slower than low orbits; Pluto's about 1km/s (vs Earth's 30km/s). Moon is also 1km/s (relative to Earth; vs LEO 8km/s.) I'd recommend playing Kerbal Space Program to understand how the speed varies in orbital mechanics - e.g. how your rocket blasting at 11km/s past Earth's upper atmosphere ends up crawling at speed of a bicycle once it approaches the edges of Earth sphere of influence.

But your ballparks are 8km/s for entering LEO, 1km/s for Moon, or trans-Neptunian objects, 30km/s for solar orbit around Earth altitude. And a scarce meter per second or less for orbiting asteroids and the likes.

Remember, multiply circular orbit by $\sqrt{2}$ for escape speed; to escape Earth, $30\sqrt{2} = 42$ km/s to escape the solar system.

• Nice answer! Where does the $\sqrt{2}$ come from? Adding a link to that would be helpful for both the OP and future readers to see how it's derived and to double-check the validity. – uhoh Aug 21 '17 at 0:54
• @uhoh: Circular orbit velocity: $v_o = \sqrt{{GM} \over r}$. Escape velocity $v_e = \sqrt{{2GM} \over r}$. ${v_e \over v_o} = \sqrt{2}$. I think deriving the equations for these two velocities is beyond the scope of this question. – SF. Aug 21 '17 at 10:17
• Comments are considered temporary. I'm suggesting if you refer to an equation in the answer, just add the link in the answer. I try to do it every single time I refer to something. This gives both the OP and future readers maximal chance of following things up in case they want to learn more. – uhoh Aug 21 '17 at 10:21