Answer: 3.63 km/s
Explanation:
The escape velocity equation for a craft launched from the Earth surface is:
[tex]V_{e}=\sqrt{\frac{2GM}{R}}[/tex]
Where:
[tex]V_{e}[/tex] is the escape velocity
[tex]G=6.67(10)^{-11} Nm^{2}/kg^{2}[/tex] is the Universal Gravitational constant
[tex]M=5.976(10)^{24}kg[/tex] is the mass of the Earth
[tex]R=6371 km=6371000 m[/tex] is the Earth's radius
However, in this situation the craft would be launched at a height [tex]h=54000 km=54000000 m[/tex] over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:
[tex]V_{e}=\sqrt{\frac{2GM}{R+h}}[/tex]
[tex]V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}[/tex]
Finally:
[tex]V_{e}=3633.86 m/s \approx 3.63 km/s[/tex]
