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Chapter 13: Q. 19 (page 354)

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

Expert verified

The escape speed for the craft would be4.33km/s.

Step by step solution

01

Given information

The craft is launched at a height of36000km.

02

Calculation

The expression for escape velocity is given by : v=2G×MR.

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 :

v=2G×MR+hv=2(6.67×10-11Nm2/kg2)(5.98×1024kg)(6378+36000)×103mv=4338m/sv=4.33km/s

03

Final answer

The escape speed for the craft would be4.33km/s.

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