Asked by christian
A 1,160 kg satellite orbits Earth with a tangential speed of 7,446 m/s. If the satellite experiences a centripetal force of 8,955 N, what is the height of the satellite above the surface of Earth? Recall that Earth’s radius is 6.38 × 106 m and Earth’s mass is 5.97 × 1024 kg.
3.71 × 1028 m
8.02 × 105 m
7.20 × 105 m
9.67 × 1028 m
All Answers 1
Answered by
GPT-5 mini
AI
Use F = m v^2 / r to find the orbital radius r:
r = m v^2 / F = (1160 kg)(7446 m/s)^2 / 8955 N
≈ (1160)(55,442,916) / 8955
≈ 7.186 × 10^6 m.
Height above Earth's surface:
h = r − R_earth = 7.186 × 10^6 − 6.38 × 10^6 ≈ 8.02 × 10^5 m.
Answer: 8.02 × 10^5 m.
r = m v^2 / F = (1160 kg)(7446 m/s)^2 / 8955 N
≈ (1160)(55,442,916) / 8955
≈ 7.186 × 10^6 m.
Height above Earth's surface:
h = r − R_earth = 7.186 × 10^6 − 6.38 × 10^6 ≈ 8.02 × 10^5 m.
Answer: 8.02 × 10^5 m.
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