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
[tex]g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}[/tex]
The expression of escape velocity is given by
[tex]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[/tex]
The escape speed is 1777.92 m/s
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
[tex]v=\sqrt{2gR}[/tex]
[tex]v=\sqrt{2\times 545000\times 2.9}[/tex]
v = 1777.92 m/s
v = 1.78 km/s
Thus, the escape velocity on the surface of asteroid is 1.78 km/s.
