Okay, back-of-the-envelope type calculations here:
The "escape velocity" for a planet is the speed you need to achieve lift-off, i.e. to completely escape the gravity of a planet (or other body) and escape into space.
Obviously, if a human jumps up--using their usual body strength--and winds up shooting off into space forever, they're in trouble, so let's get a handle on how small an asteroid has to be before that can happen.
Say a person is strong enough, when they're standing on Earth, to jump a half metre into the air before they fall back down. We want to find out how fast they're moving when they leave the ground at the start of their jump, because that's a good benchmark for how hard they can jump (i.e. how fast they can accelerate their body mass) on an asteroid too.
The kinematics equation we'll need is:
(vf)^2 = (vi)^2 + 2ad
where:
vf is their final velocity,
vi is their initial velocity,
a is the acceleration due to gravity near the surface of the Earth, and
d is how far they have travelled.
So we have:
d=0.5 m (for jumping up half a metre),
a = -9.81 m/s^2 (acceleration due to gravity near the surface of Earth)
vf=0 m/s (because they stop moving upward once they reach the height of 0.5m)
After substituting in those values and making a cancellation that gives us:
vi^2 = 9.81 (m/s)^2
It'll be easier for us later to just leave vi squared like that. Sorry if it makes the notation confusing.
So the next thing we want to do is figure out how small an asteroid has to be before this rather-modest amount of human strength is enough to launch a person off the asteroid permanently.
The formula for the escape velocity of a body is:
ve^2 = 2GM/r
where:
ve = escape velocity, i.e. the velocity needed to achieve lift-off,
G = the universal gravitational constant for our universe, which is 6.67 x 10^(-11) m^3/kg/s^2
M = the mass of the asteroid, and
r = the radius of the asteroid
Now before we stick anything into this equation, let's first simplify things a little by getting rid of the mass M.
The mass of a planet is equal to its density times its volume, and the volume of a sphere is (4/3)*(Pi)*(r^3).
So M/r can be re-written as (density)*(4/3)*(Pi)*(r^2).
That gives us:
ve^2 = 2G*(density)*(4/3)*(Pi)*(r^2)
Old Mother Google tells me a typical asteroid has a density of 2 g/cm^3, which converts to 2000 kg/m^3. So now we have all the numbers we need to figure out how big an asteroid is when a human, who has the strength to jump 0.5 m into the air on Earth, is capable of launching themselves into space using that same amount of strength.
We want to set vi^2 in our first equation equal to ve^2 in the escape velocity equation. That gives:
(9.81) = (2) * (6.67x10^[-11]) * (2000) * (4/3) * (Pi) * (r^2)
Solving for r, which is the radius of the asteroid, we get:
r = 2962 m = 2.962 km
So any asteroid with a diameter of about (2 x 2.962km) ~ 6 km across is one that a normal human can launch themselves into space off of just by jumping really hard.
If you want to calculate how high up they can jump on LARGER asteroids, that's something we can work out too.
First, we figure out what the acceleration due to gravity is on that asteroid using the following expression:
a = GM/r^2 = G * (density) * (4/3) * (Pi) * r
Substituting in the values:
G = 6.67x10^(-11) m^3/kg/s^2
density = 2000 kg/m^3
we get:
a = 5.59x10^(-7) * r
where r is still the radius of the asteroid. The units for r will be metres, and the units for 'a' will be metres per second-squared.
To then figure out how high up the person can jump on the larger asteroid, you substitute that 'a' value into the equation we used above:
(vf)^2 = (vi)^2 + 2ad
where
a = the acceleration we just figured out, 5.59x10^(-7) * r,
(vf)^2 = 0, because they stop moving upward at the top of their trajectory,
(vi)^2 is the value we calculated above for a typical human, (vi)^2 = 9.81 (m/s)^2, and
d = the height they manage to jump to, which is the quantity we're trying to find.
This gives a simplified expression of:
9.81 (m/s)^2 = 2 * (5.59x10^[-7]) * r * d
Or, solving for d, the distance the person can jump upward:
d = 8,775,000 * (1/r)
So, for example, on a 10km wide asteroid, r=5km=5000m, and that means a person can jump:
d = 8,775,000 / 5000 = 1755m = 1.755 km high
Which is still really freaking high, and they will take ages to float back down again.
In other words, jumping around on asteroids is bad news for us big, strong, Earth-adapted humans.
So walking would be a delicate, tip-toe-don't-sneeze problem, the cards on the card table need to have weights on them, or every little breeze would knock them around, and drinks would probably work but everyone would be prone to splashing them everywhere, i.e. anyone who slams their glass down on the table will paint the ceiling with whatever was still in it.
That said, asteroids can be up to 500km across, so you can figure out whether that would be enough to keep people stuck to the ground:
diameter=500km => radius = 250km = 250,000m
So:
d = 8,755,000 / 250,000 = 35 metres
People can still bound around like grasshoppers, but they also can stay on the ground a bit more easily.