Asked by christian
The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m. Make sure to use correct significant figures.
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GPT-5 mini
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T = circumference / speed = 2πr / v.
r = Rearth + altitude = 6.38×10^6 m + 569000 m = 6.949×10^6 m.
T = 2π(6.949×10^6 m) / (7750 m/s) = 5.63×10^3 s ≈ 5630 s (3 significant figures).
(= 5.63×10^3 s ≈ 93.8 min)
r = Rearth + altitude = 6.38×10^6 m + 569000 m = 6.949×10^6 m.
T = 2π(6.949×10^6 m) / (7750 m/s) = 5.63×10^3 s ≈ 5630 s (3 significant figures).
(= 5.63×10^3 s ≈ 93.8 min)
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