91Ó°ÊÓ

We know, mass of the planet = $1.898×{10}^{27}kg.$, radius of the planet =$71492km.$

Escape speed is only possible at the condition when the kinetic energy exceeds the gravitational potential energy at the surface.

The kinetic energy is given by : $K.E=\frac{1}{2}m{v}^2.$

Where, 'v' is the escape velocity.

'm' is the mass of a planet.

The potential energy is given by : $P.E=\frac{GMm}{R}.$

Where, 'G' is the gravitational constant$=6.67×{10}^{-11}{\mathrm{m}}^3/\mathrm{kg}·{\mathrm{s}}^2.$

'M' is the mass of the planet $=1.898×{10}^{27}\mathrm{kg}.$

'R' is the radius of the planet $=71492\mathrm{km}.$

Equating potential energy to kinetic energy :

$$\begin{gathered} \frac{1}{2}m{v}^2=\frac{GMm}{R} \\ v=\sqrt{\frac{2GM}{R}}. \end{gathered}$$

Substituting the values of$G,M,R$we get,

$$\begin{gathered} v=\sqrt{\frac{2GM}{R}} \\ v=\sqrt{\frac{2×(6.67×{10}^{-11}{m}^3/kg{s}^2)(1.898×{10}^{27}kg)}{(71.492×{10}^{3}m)}} \\ v=5.95×{10}^4\mathrm{m}/\mathrm{s} \\ v=59.5\mathrm{km}/\mathrm{s}. \end{gathered}$$

Escape speed from Jupiter is $59.5\mathrm{km}/\mathrm{s}.$