Question

A proposed space elevator would consist of a cable stretching from the earth’s surface to a satellite, orbiting far in space, that would keep the cable taut. A motorized climber could slowly carry rockets to the top, where they could be launched away from the earth using much less energy. What would be the escape speed for a craft launched from a space elevator at a height of $36,000km$? Ignore the earth’s rotation.

Short answer

The escape speed for the craft would be$4.33km/s.$

Step-by-step solution

Given information

The craft is launched at a height of$36000km.$

Calculation

The expression for escape velocity is given by : $v=\sqrt{\frac{2G\times M}{R}}.$

Where, 'v' is escape velocity.

'G' is the universal gravitational constant.

'M' is the mass of the earth.

'R' is the radius, here 'R' is (R + h).

Substituting the values of G, M, R, h in the above equation. And solving for 'V' we get :

$$\begin{gathered} v=\sqrt{\frac{2G\times M}{R+h}} \\ v=\sqrt{\frac{2(6.67\times {10}^{-11}N{m}^2/k{g}^2)(5.98\times {10}^{24}kg)}{(6378+36000)\times {10}^{3}m}} \\ v=4338\mathrm{m}/\mathrm{s} \\ v=4.33\mathrm{km}/\mathrm{s} \end{gathered}$$

Final answer

The escape speed for the craft would be$4.33km/s.$