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inysia [295]
2 years ago
9

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

Physics
1 answer:
Natasha_Volkova [10]2 years ago
7 0

Answer: 3.63 km/s

Explanation:

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

V_{e}=\sqrt{\frac{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{\frac{2GM}{R+h}}

V_{e}=\sqrt{\frac{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

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