Question

What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?

Ignore the Earth's rotation.

Answer 1

Answer: 3.63 km/s

Step-by-step explanation:

The escape velocity equation for a craft launched from the Earth surface is:

`V_(e)=\sqrt{(2GM)/(R)}`

Where:

`V_(e)` is the escape velocity

`G=6.67(10)^(-11) Nm^(2)/kg^(2)` is the Universal Gravitational constant

`M=5.976(10)^(24)kg` is the mass of the Earth

`R=6371 km=6371000 m` is the Earth's radius

However, in this situation the craft would be launched at a height
`h=54000 km=54000000 m` over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

`V_(e)=\sqrt{(2GM)/(R+h)}`

`V_(e)=\sqrt{(2(6.67(10)^(-11) Nm^(2)/kg^(2))(5.976(10)^(24)kg))/(6371000 m+54000000 m)}`

Finally:

`V_(e)=3633.86 m/s \approx 3.63 km/s`