Question

What is the escape speed from Jupiter?

Short answer

The escape speed from Jupiter is $59.5km/s.$

Step-by-step solution

Given information

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

Calculation

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.

Continuation of calculation

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

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

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

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

Equating potential energy to kinetic energy :

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

Continuation of calculation

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

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

Final answer

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