Question

(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?

Answer 1

Answer:

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

$v=\sqrt{2gR}$

$v=\sqrt{2\times 545000\times 2.9}$

v = 1777.92 m/s

v = 1.78 km/s

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

Answer 2

Answer:

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

$g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}$

The expression of escape velocity is given by

$v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s$

The escape speed is 1777.92 m/s