PS3-1 - g E we have g E g M = M E M M w R M R E W 2...

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Problem Set 3: Problem 1 . Problem: The mass of Earth is M E =5 . 97 · 10 24 kg and its average radius is R E =6 . 372 · 10 3 km. The mass of Mars is M M =6 . 42 · 10 23 kg and its average radius is R M =3 . 397 · 10 3 km. If James Bond ’smassis90kg ,wha tish iswe igh tonMars , W 007 , in kiloNewtons? Solution: We must first compute the gravitational acceleration on Mars, g M . In general, the gravitational acceleration at a planet’s surface is g = GM R 2 where
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Unformatted text preview: g E , we have g E g M = M E M M w R M R E W 2 Substituting the given values for planetary mass and radius, we find g E g M = 5 . 97 · 10 24 kg 6 . 42 · 10 23 kg w 3 . 397 · 10 3 km 6 . 372 · 10 3 km W 2 = 2 . 6429 Therefore, the gravitational acceleration on Mars is g M = 9 . 807 m/sec 2 2 . 6429 = 3 . 71 m sec 2 Thus, denoting Bond’s mass by m , his weight on Mars would be W 007 = mg M = (90 kg ) D 3 . 71 m/sec 2 i = 333 . 9 N ≈ 1 3 kN...
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