Answer
(a) The satellite's orbital speed is 6.95 m/s
(b) The escape speed from the asteroid is 12.3 m/s
Work Step by Step
(a) Let $M_a$ be the asteroid's mass. Let $R$ be the asteroid's radius. Note that the satellite's orbital radius is $R+5.0~km$. We can use the orbital radius to find the orbital speed.
$v = \sqrt{\frac{G~M_a}{R+5.0~km}}$
$v = \sqrt{\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.0\times 10^{16}~kg)}{1.38\times 10^4~m}}$
$v = 6.95~m/s$
The satellite's orbital speed is 6.95 m/s.
(b) We can find the escape speed from the asteroid.
$v_{esc} = \sqrt{\frac{2~G~M_a}{R}}$
$v_{esc} = \sqrt{\frac{(2)(6.67\times 10^{-11}~m^3/kg~s^2)(1.0\times 10^{16}~kg)}{8.8\times 10^3~m}}$
$v_{esc} = 12.3~m/s$
The escape speed from the asteroid is 12.3 m/s.
