Question

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.10×10^16 kg and a radius of 9.90 km . Part A
What is the speed of a satellite orbiting 5.80 km above the surface?
Express your answer with the appropriate units
Part B
What is the escape speed from the asteroid?
Express your answer with the appropriate units.

Answer 1

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.