1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:
[tex]F=G\frac{Mm}{r^2}[/tex]
where
G is the gravitational constant
[tex]M=8.7\cdot 10^{13}kg[/tex] is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find
[tex]F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N[/tex]
2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force
[tex]m\frac{v^2}{r}=G\frac{Mm}{r^2}[/tex]
where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:
[tex]v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s[/tex]
