Asked by christian
The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. Given that Earth’s mass is 5.97 × 1024 kg and its radius is 6.38 × 106 m, what is HST’s tangential speed? Make sure to use correct significant figures.
m/s
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Answered by
GPT-5 mini
AI
Use v = sqrt(GM/(R + h)).
Given: G = 6.67430×10^-11 m^3 kg^-1 s^-2, M = 5.97×10^24 kg, R = 6.38×10^6 m, h = 569 km = 5.69×10^5 m.
r = R + h = 6.38×10^6 + 5.69×10^5 = 6.949×10^6 m.
GM = 6.67430×10^-11 · 5.97×10^24 ≈ 3.9846×10^14 m^3/s^2.
v = sqrt(GM/r) = sqrt(3.9846×10^14 / 6.949×10^6) ≈ 7.57×10^3 m/s.
With correct significant figures: v = 7.57 × 10^3 m/s.
Given: G = 6.67430×10^-11 m^3 kg^-1 s^-2, M = 5.97×10^24 kg, R = 6.38×10^6 m, h = 569 km = 5.69×10^5 m.
r = R + h = 6.38×10^6 + 5.69×10^5 = 6.949×10^6 m.
GM = 6.67430×10^-11 · 5.97×10^24 ≈ 3.9846×10^14 m^3/s^2.
v = sqrt(GM/r) = sqrt(3.9846×10^14 / 6.949×10^6) ≈ 7.57×10^3 m/s.
With correct significant figures: v = 7.57 × 10^3 m/s.
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