(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 working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.