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The free-fall acceleration on the surface of Jupiter is about two and one half times that on the surface of the Earth. The radius of Jupiter is about 11.0 RE (RE = Earth's radius = 6.4 106 m). Find the ratio of their average densities, ρJupiter/ρEarth.
12 years ago

Answers

AARON
1/6.4106
10 years ago
ALLAH
AKBAR
8 years ago
Alexander Vega Soto
5/22
4 years ago

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