What size an object in space would have to be in order to have just enough gravity to hold a 200lb person to it's surface, yet so little gravity that one could jump clean off of it?
Is there a equation one could use to figure this out and what would this phenomenon be called?
2007-08-14 14:16:02 · 5 answers · asked by bender_xr217 7 in Science & Mathematics ➔ Physics
I know that some people around here think that a wikipedia link is lame - but here it is.
You are talking about calculating the "escape velocity" - or rather the smallest mass on which a person can stand and exert sufficient force against to reach escape velocity at the end of the leg thrust.
Calculating it will take into account the mass of your planet - the presumed density of the material from which the planet is made - the force you can generate with your legs.
Good luck in finding an answer - sounds like a fun project. I am not so sure that it is really easy to figure out - as there are so many variables that affect each other - on the other hand, maybe there is an elegant solution that cancels out alot of the factors.
Hey, other answers - a little help here??? - I know that there are people on this site just chomping at the bit to get their hands on an interesting problem to solve - this sounds like one.
2007-08-14 15:00:04 · answer #1 · answered by Anonymous · 2⤊ 0⤋
"Jump clean off" means to reach escape velocity. There is a simple formula that relates escape velocity, surface gravity, and planet's diameter. it is:
Ve = sqrt(g_s·D)
where "Ve" is escape velocity, "g_s" is the the gravitational acceleration at the planet's surface, and D is the planet's diameter. [Note: the formulas given by others are also correct: but this form is more userful for our purposes here.]
So let's assume that we want the planet's surface gravity to be the same as the earth's, but we want the escape velocity to be "human jump" speed--let's say 5 mph.
That gives us:
D = (5 mph)² / (9.8 m/sec²) = 20 inches.
Okay, a little too small. So let's reduce the surface gravity requirement, and say it's only 0.01% of earth gravity (that means the person will weigh about a third of an ounce). In that case, the planet's diameter comes out to be about 3 miles.
2007-08-15 13:21:47 · answer #2 · answered by RickB 7 · 1⤊ 0⤋
These are all approximations, as the human body is not a fixed point in space, but rather a flailing, floppy blob of liquid.
High-jump world record: Javier Sotomayor: 2.45m
Jump velocity: sqrt(2*g*h) = 6.93m/s
Escape velocity = sqrt(2*G*M/r) = 6.93
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M/r = 7.2 x 10^11 kg/m
In other words, the ratio of an objects mass (kg) must be less than or equal to 7.2x10^11 times its radius (m) in order for a person to be able to jump at escape velocity.
Deimos, for example (as given above) has a ratio of (1.8*10^15) / (5500) = 3.3 x 10^11 kg/m.
2007-08-15 13:17:57 · answer #3 · answered by MooseBoys 6 · 1⤊ 0⤋
The moons of Mars Phobos and Deimos have escape velocities 11m/s and 7m/s. The latter corresponds to 2.5m high jump on on Earth, ie pretty close to capability of humans and probably within the reach of grasshoppers.
(Humans can run as fast as 12m/s, but cannot not jump with such speed)
2007-08-15 13:16:26 · answer #4 · answered by Alexander 6 · 1⤊ 0⤋
yeah, i've also heard that the moons of mars are so small that they might fit this description. they may be just big enough to hold a person if they jumped, but they are small enough that you could throw a baseball into orbit from the ground. then even i could hit better than barry bonds without cheating like he did
2007-08-15 13:34:25 · answer #5 · answered by brandon 5 · 1⤊ 0⤋