A satellite orbits the Earth with a period of 24 hours. The radius of Earth is 6400 km and the mass of Earth is 6.0 x 1024 kg. Determine the height of the satellite above the Earth’s surface.
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67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) (gravitational constant) - \( M_{\text{Earth}} = 6.0 \times 10^{24} \, \text{kg} \) (mass of Earth) - \( T = 24 \times 3600 \, \text{s} \) (period of rotation in seconds) ** Show more…
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