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If someone without any propulsion jumped from a stationary ship in space, how fast would they travel?

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    $\begingroup$ You should be a little bit more detailed about what exactly you would like to know, as there are many possible ways that one could answer this. $\endgroup$ – tmwilson26 Oct 27 '15 at 18:08
  • $\begingroup$ What is "stationary...in space" supposed to mean? Stationary relative to what? And can you quantify "jumped"? (I mean, Jumped how hard?) $\endgroup$ – Solomon Slow Oct 27 '15 at 20:19
  • $\begingroup$ Sorry to be cryptic. Imagine you are in a space station, orbiting the earth. You jump out the door. how fast would you, (the average human), be traveling? Is 9 mph a good estimate? $\endgroup$ – JDSavage Oct 28 '15 at 19:36
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The highest a good athlete can jump on the earth's surface, where the acceleration due to gravity is about 32 feet per second squared, is about 3 feet. If the initial velocity from the jump is $v$, then the athlete's height at time $t$ is $vt-16t^2$. This is maximized at $t=v/32$, where the height is $v^2/64$. This gives $v^2/64=3$, or (roughly) $v=14$. So the athlete can achieve an initial velocity of about $14$ feet per second --- call it 9 miles per hour --- and of course in space you'd maintain this velocity pretty much forever.

(If you can jump either more or less than 3 feet off the ground, then, of course, correct the arithmetic accordingly to get your own speed in space.)re

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  • $\begingroup$ The world record for the highest standing jump is an impressive 6'3", giving 20 feet per second. $\endgroup$ – pela Oct 27 '15 at 19:56
  • $\begingroup$ @pela That is a little bit of an overestimate. Really good high jumpers manage to move their bodies over the bar while their center of mass passes under the bar. It helps to be very limber indeed. And yes it seems to be a small fractional effect but the square make for twice the fractional effect. $\endgroup$ – dmckee --- ex-moderator kitten Oct 27 '15 at 21:36
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    $\begingroup$ And the center of mass starts 3 or 4 feet above the ground... $\endgroup$ – DJohnM Oct 27 '15 at 23:33
  • $\begingroup$ @dmckee. In high jump, you're right that the CoM doesn't go as high. I hadn't considered that. But this was a box jump, where the CoM is maybe a foot or more over the edge. But as DJohnM says, the CoM starts several feet above the ground. $\endgroup$ – pela Oct 28 '15 at 14:42
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Velocities always have to be expressed with respect to some reference. When we talk about velocities on Earth, it's usually understood that we mean with respect to the surface of the Earth. If an astronaut on the ISS steps outside of the station, her velocity wrt. to ISS is (close to) zero, but wrt. to Earth's surface, it's 7.7 km/s, because this is the speed of the ISS wrt. Earth's surface. Wrt. Sun's surface, she would travel roughly 30 km/s, because this is Earth's speed around the Sun. Similarly, her speed wrt. the center of the Milky Way would be ~250 km/s, and so on.

Along the same lines, when you say "a stationary ship in space", you need to specify "stationary wrt. what?". It will always have some non-zero velocity wrt. something.

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  • $\begingroup$ I thought it was pretty clear that the question was asking for the jumper's velocity relative to the ship. $\endgroup$ – WillO Oct 27 '15 at 18:12
  • $\begingroup$ @WillO: I wouldn't say it's pretty clear, but +1 for your interpretation and calculations. $\endgroup$ – pela Oct 27 '15 at 19:55

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