Question

An astronaut has a centripetal acceleration of 2.34 x 102 m/s2. What is the velocity of the astronaut at 6.78 x 106 meters above the surface of Earth?

Answer 1

Answer on Question #62482 – Physics – Molecular Physics – Thermodynamics

Question:

An astronaut has a centripetal acceleration of $2.34 \times 10^{2} \frac{m}{s^{2}}$ . What is the velocity of the astronaut at $6.78 \times 10^{6}$ meters above the surface of Earth?

Answer:

$$a = \omega^ {2} (R _ {\oplus} + h) = \frac {v ^ {2}}{R _ {\oplus} + h} \Rightarrow v = \sqrt {a (R _ {\oplus} + h)} = 5 5. 4 7 \cdot 1 0 ^ {3} \frac {m}{s};$$

Velocity of the astronaut at $6.78 \times 10^{6}$ meters above the surface of Earth is $55.47 \cdot 10^{3} \frac{m}{s}$ .

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