Answer:
a) v =6.84 m/s
b) Vesc = 9.67 m/s
Explanation:
let v be the orbital speed of the satelite, Vesc be the escape speed from the asteroid, Fg be the force of gravity and Fc be the centripetal force.
a) At each point in the orbit :
Fc = Fg
m×v^2/r = GmM/(r^2)
r is the same and m is the same:
v^2 = GM/r
= [(6.67408×10^-11)×(1.10×10^16)]/(9.90×10^3 + 5.80×10^3)
= 46.76
where v = \sqrt{46.76 } = 6.8 m/s
therefore, the orbital speed of the satelite is 6.8 m/s.
b) the escape velocity is given by :
Vesc = \sqrt{2×G×M/r }
= \sqrt{2×(6.67408×10^-11)×(1.10×10^16)/(9.90×10^3 + 5.80×10^3) }
= 9.67 m/s.
therefore, for an object to escape the gravitationsl effects of the asteroid and escape it's atmosphere it will have to be moving with a speed of 9.67 m/s.
