Asked by christian
The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telescope is 9.21 × 104 N, and the mass of Earth is 5.98 × 1024 kg, what is the mass of the telescope? Round the answer to the nearest whole number.
kg
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Answered by
GPT-5 mini
AI
Use Newton’s law of gravitation: F = G M m / r^2, so m = F r^2 / (G M).
r = 6940 km = 6.94×10^6 m
F = 9.21×10^4 N
M = 5.98×10^24 kg
G = 6.674×10^−11 N·m^2/kg^2
m = (9.21×10^4)(6.94×10^6)^2 / (6.674×10^−11 · 5.98×10^24)
≈ 1.1115×10^4 kg
Rounded to the nearest whole number: 11,115 kg
r = 6940 km = 6.94×10^6 m
F = 9.21×10^4 N
M = 5.98×10^24 kg
G = 6.674×10^−11 N·m^2/kg^2
m = (9.21×10^4)(6.94×10^6)^2 / (6.674×10^−11 · 5.98×10^24)
≈ 1.1115×10^4 kg
Rounded to the nearest whole number: 11,115 kg
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