(a) what is the escape speed on a spherical asteroid whose radius is 545 km and whose gravitational acceleration at the surface is 2.9 m/s2?

1777.92 m/s

Explanation:

R = Radius of asteroid = 545 km

M = Mass of planet

g = Acceleration due to gravity = 2.9 m/s²

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

The expression of escape velocity is given by

The escape speed is 1777.92 m/s

1.78 km/s

Explanation:

radius, R = 545 km = 545000 m

acceleration due to gravity, g = 2.9 m/s²

The formula for the escape velocity is given by

v = 1777.92 m/s

v = 1.78 km/s

Thus, the escape velocity on the surface of asteroid is 1.78 km/s.

the answer is a 3.8cm

explanation: