Physics question concerning gravity. please help!

On Earth, a person can jump vertically and rise to a height, h. What is the radius of the largest spherical asteroid from which this person can escape by jumping straight upward? Assume that each cubic meter of the asteroid has a mass of 3500 kg.

The answer is: squareroot(1.00*10^7m*h)

An explanation would be awesome. Thanks so much in advance!

Let:
Me = mass of earth
Re = radius of earth
Ma = mass of asteroid
Ra = radius of asteroid

The person on earth can jump to a height "h". That means the energy that the person can expend in a jump is:
E = mgh

Since "mg" is the strength of the pull of gravity, we can rewrite this as:
E = (G(Me)m/Re²)h

Now imagine that this energy E is also just enough to let the person escape from the asteroid. That means his initial velocity is the same as the asteroid's escape velocity Ve. That is:
E = ½m(Ve)²

But escape velocity is also given by
Ve² = 2G(Ma)/Ra

So substitute:
E = ½m(2G(Ma)/Ra) = mG(Ma)/Ra

Combine this with our other expression for E to get:
(G(Me)m/Re²)h = mG(Ma)/Ra

Simplify (divide by mG)
h(Me)/Re² = Ma/Ra

Now let "ρ" ("rho") be the asteroid's density, which is given. That means:
ρ = mass/volume = Ma / (4/3 πRa³)

Rewrite as:
(4/3)πRa²ρ = Ma/Ra

Substute into previous equation containing Ma/Ra:
h(Me)/Re² = (4/3)πRa²ρ

Solve for "Ra":
Ra = sqrt[(3Me/(4πρRe²)) * h]

Finally, look up and plug in values for Me and Re; and plug in "3500 kg/m^3" for "ρ" (I tried it and got the expected answer.)

I start with h = v²/2g.

I solve for v; v = √2gh.

(Remember that it is his mass the person is accelerating, not his weight so his speed of launch shouldn't be affected by the weaker gravitational pull on the asteroid. If I get the right answer, or near nuff, then we'll see if I'm right.)