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A space shuttle S is sent to service the Hubble Space Telescope H which is in a circular orbit 6.0 x 105m above the Earth's surface. The space shuttle finally reaches the same orbit as f and its thrust rockets are shut down. Given that the radius of the Earth is 6400km.
a. Calculate the speed of the space-shuttle S in the orbit.
b. Calculate the time take for the space-shuttle S to revolve one revolution in the orbit​

Respuesta :

Given:
- Radius of Earth, \( r_{\text{Earth}} = 6400 \times 10^3 \, \text{m}\)
- Altitude of the orbit, \( h = 6.0 \times 10^5 \, \text{m}\)
- Gravitational constant, \( G = 6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2\)
- Mass of Earth, \( M_{\text{Earth}} = 5.972 \times 10^{24} \, \text{kg}\)

a. Speed of the space shuttle in the orbit:
\[ v = \sqrt{\frac{G \cdot M_{\text{Earth}}}{r_{\text{Earth}} + h}} \]

\[ v = \sqrt{\frac{(6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2) \cdot (5.972 \times 10^{24} \, \text{kg})}{6400 \times 10^3 \, \text{m} + 6.0 \times 10^5 \, \text{m}}} \]

\[ v ≈ 7.75 \times 10^3 \, \text{m/s} \]

b. Time taken for one revolution:
\[ T = \frac{2\pi \cdot (r_{\text{Earth}} + h)}{v} \]

\[ T ≈ \frac{2\pi \cdot (6400 \times 10^3 + 6.0 \times 10^5)}{7.75 \times 10^3} \]

\[ T ≈ 5370 \, \text{s} \]

So,
a. The speed of the space shuttle in the orbit is approximately \(7.75 \times 10^3 \, \text{m/s}\).
b. The time taken for the space shuttle to complete one revolution in the orbit is approximately \(5370 \, \text{s}\).