Physics question concerning gravity. please help!

since (1/2)mv^2=mgh, kinetic energy is equal to potential energy

so v^2=2gh

draw the free body diagram and the force on the body is:

mdv/dt=-mR^2(/x+R)^2

so dv/dt=(dv/dx)(dx/dt)=(vdv/dx)

vdv/dx=-gR^2/(x+R^2)

v^2/2=gR^2/(x+R)+c

at x=0, v=0

so c=v0^2/2-gR

or v^2/2=-gR+2gR^2/(x+R)

at x=infinity, ve^2/2=-gR

or ve^2=-2gR

or ve=sqrt(2gR)

v=at+ve

v=gt+sqrt(2gR)

(1/2)mv^2=gt+sqrt(2gR)

but( 1/2)mv^2=mgh

so v^2=2gh

or (1/2)m2gh=gt+sqrt(2gR)

mgh=gt+sqrt(2gR)

gt=mgh-sqrt(2gR)

t=mh-sqrt(2R/g)

x=(1/2)gt^2+v0t

h=(1/2)gt^2-sqrt(2gR)t

h=(1/2)g(mh-sqrt(2R/g)^2
-sqrt(2gR)(mh-sqrt(2R/g))

2h=g(m^2h^2-2mhsqrt(2R/g)+2R/g)
-2mhsqrt(2gR)+2sqrt(4R^2)

2h=gm^2h^2-2mhsqrt(2R/g)
+2R/g-2mhsqrt(2gR)+4R

now mg=GMm/R^2

or g=GM/R^2

where mass=density * volume

so M=3500 (4/3) pi R^3

so g=G (3500)(4/3) pi R^3/R^2

or g=14660GR

and G=6.673*10^-11 Nm^2/kg^2

so g=(14660*6.673*10^-11)R=(9.78*10^-7)R

sub into the distance equation,

2h=gm^2h^2-2mhsqrt(2R/g)+2R/g
-2mhsqrt(2gR)+4R

2h=(9.78*10^-7)m^2h^2R
-2mhsqrt(2/9.78*10^-7)
+2/9.78*10^-7-2mhsqrt(2(9.78*10^-7))R+4…

2h=(9.78*10^-7)m^2h^2R-2.86*10^3mh
+2.04*10^6-2.8*10^-3mhR+4R

but now we know mass=density*volume

so m=3500 (4/3) pi R^3=14.66*10^3R^3

sub into the equation,

2h=(9.78*10^-7)(14.66*10^3)^2h^2R^7
-2.86*10^3(14.66*10^3)hR
+2.04*10^6-(2.8*10^-3)(14.66*10^3)hR^4+…

2h=210h^2R^7-41hR^4+4R
-4.19*10^7hR+2.04*10^6

210h^2R^7-41hR^4-(4-4.19*10^7h)R
+2.04*10^6-2h=0

i can't solve this using maple 15. all i can do is plot R against h using maple 15....

h plateau at 10 m, R at 5m............

so the size of the asteroid probably is 5m and if he jump more than 10m, he will jump into space!!!