43.1k views
3 votes
What is the escape speed(in km/s) from an Earth-like planet with mass 6.5e+24 kg and radius 70.0 x 10^5 m? Use the gravitational constant G = 6.67 x 10^-11 m^3kg^-1 s^-2

User Tlanigan
by
4.8k points

1 Answer

0 votes

Given,

The mass of the planet, M=6.5×10²⁴ kg

The radius of the planet, r=70.0×10⁵ m

The gravitational constant, G=6.67×10⁻¹¹ m³kg⁻¹s⁻²

The escape velocity is the minimum initial velocity that an object needs to have in order to escape the gravitational pull of a planet.

The escape velocity is given by,


v=\sqrt[]{(2GM)/(r)}

On substituting the known values,


\begin{gathered} v=\sqrt[]{(2*6.67*10^(-11)*6.5*10^(24))/(70.0*10^5)} \\ =11129.75\text{ m/s} \\ =11.13\text{ km/s} \end{gathered}

Thus the escape velocity of the planet is 11129.75 m/s or 11.13 km/s

User Michael McCabe
by
3.5k points