Jumping off a comet cliff:

The Cliffs of Comet Churyumov–Gerasimenko, a cliff is 1000 meters high. If you were on top and jumped off, how long would it take and how fast would you be going at the bottom of the cliff? Gravity there is about 1/100,000 of Earth.

Originally posted by sonhouse
The Cliffs of Comet Churyumov–Gerasimenko, a cliff is 1000 meters high. If you were on top and jumped off, how long would it take and how fast would you be going at the bottom of the cliff? Gravity there is about 1/100,000 of Earth.

a = g/100,000 ~ 10/100,000
v = sqrt(2as) = sqrt(2*10/100,000*1000) = sqrt(1/5) ~ 0.45 m/s
s = ½at²
t = sqrt(2*s/a) = sqrt(2*1000/(10/100,000)) = sqrt(20,000,000) ~ 4500 s ~ 1 hr 20 minutes.

Originally posted by DeepThought
a = g/100,000 ~ 10/100,000
v = sqrt(2as) = sqrt(2*10/100,000*1000) = sqrt(1/5) ~ 0.45 m/s
s = ½at²
t = sqrt(2*s/a) = sqrt(2*1000/(10/100,000)) = sqrt(20,000,000) ~ 4500 s ~ 1 hr 20 minutes.

yep, that's what I got. If you jump off you better make sure you have enough O2 to reach bottom!

Originally posted by sonhouse
yep, that's what I got. If you jump off you better make sure you have enough O2 to reach bottom!

Don't jump of, push off. Treat it like navigating in zero G.

I was going to say, if you "jumped" off, you probably wouldn't land at all.

Originally posted by forkedknight
I was going to say, if you "jumped" off, you probably wouldn't land at all.

lets say a person has a 1 ft vertical leap hear on earth.

Let T = Kinetic Energy
U = Potential Energy (Gravitational)

Using Conservation of Energy (Neglecting non conservative forces)

T = U

1/2*m*v² = m*g*h

v = Sqrt (2*g*h)

g = 32.2 ft/s²
h = 1 ft

v = 8 ft/s

The escape velocity of the comet according to Wiki is estimated at 3 ft/s. So I would suspect that "jumping" as you and Mr Whitehead pointed out would be a bad idea!

Originally posted by joe shmo
lets say a person has a 1 ft vertical leap hear on earth.

Let T = Kinetic Energy
U = Potential Energy (Gravitational)

Using Conservation of Energy (Neglecting non conservative forces)

T = U

1/2*m*v² = m*g*h

v = Sqrt (2*g*h)

g = 32.2 ft/s²
h = 1 ft

v = 8 ft/s

The escape velocity of the comet according to Wiki is estimated at 3 ft/s. So I would suspect that "jumping" as you and Mr Whitehead pointed out would be a bad idea!

yeah, you could probably push off with one hand and reach escape velocity of that comet. So you would have to 'jump' very carefully! If you had some kind of propulsion on your EVA suit, you could examine the sides of the cliff in some detail, you could hover for an hour with little expenditure of thrust!

Considering the extremely low gravity, it was a real trick of navigation to get the main craft to just orbit the thing! Orbiting at what, one km per hour?

Originally posted by sonhouse
yeah, you could probably push off with one hand and reach escape velocity of that comet. So you would have to 'jump' very carefully! If you had some kind of propulsion on your EVA suit, you could examine the sides of the cliff in some detail, you could hover for an hour with little expenditure of thrust!

Considering the extremely low gravity, it was a re ...[text shortened]... of navigation to get the main craft to just orbit the thing! Orbiting at what, one km per hour?

The action in the movie Gravity ( George Clooney and Sandra Bullock) is a perfect example of what can happen in such a case. Awesome movie, by the way.