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Jan. 31, 2023•0 likes•3 views

Jan. 31, 2023•0 likes•3 views

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Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 946.0 N on the earth weighs 920.0 N at the north pole of Planet X and only 848.0 N at its equator. The distance from the north pole to the equator is 1.889Ã—10 4 km , measured along the surface of Planet X. a) How long is the day on Planet X? b) If a 4.800Ã—10 4 kg satellite is placed in a circular orbit 4000 km above the surface of Planet X, what will be its orbital period? Solution m = 946N / 9.8m/sÂ² = 96.53 kg ? mass of astronaut F = GmM/rÂ² 920 N = 6.674*10 -11 NÂ·mÂ²/kgÂ² * 96.53kg * M / (1.889*10 7 m)Â² so M = 5.09566*10 25 kg The centripetal acceleration at the equator is a = (920N - 848N) / 96.53kg = 0.7458 m/sÂ² = ?Â²r = ?Â² * (1.889e7m) ? = 1.987e-4 rad/s T = 2?/? = 34 610 s = 9 h 36 min 50 s The mass of the satellite is immaterial. For \"orbit\", centripetal acceleration = gravitational acceleration, or ?Â²r = GM/rÂ² rearranges to ? = ?(GM/rÂ³) where G = Newton\'s gravitational constant = 6.674e?11 NÂ·mÂ²/kgÂ² and M = mass of X = 5.09566*10 25 kg from above. So ? = ?(6.674*10 -11 NÂ·mÂ²/kgÂ² * 5.09566*10 25 kg / (1.889*10 7 m + 4*10 6 m)Â³ ) = 5.325e-4 rad/s T = 2?/? = 11799.27 s .

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- 1. Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 946.0 N on the earth weighs 920.0 N at the north pole of Planet X and only 848.0 N at its equator. The distance from the north pole to the equator is 1.889Ã—10 4 km , measured along the surface of Planet X. a) How long is the day on Planet X? b) If a 4.800Ã—10 4 kg satellite is placed in a circular orbit 4000 km above the surface of Planet X, what will be its orbital period? Solution m = 946N / 9.8m/sÂ² = 96.53 kg ? mass of astronaut F = GmM/rÂ² 920 N = 6.674*10 -11 NÂ·mÂ²/kgÂ² * 96.53kg * M / (1.889*10 7 m)Â² so M = 5.09566*10 25 kg The centripetal acceleration at the equator is a = (920N - 848N) / 96.53kg = 0.7458 m/sÂ² = ?Â²r = ?Â² * (1.889e7m) ? = 1.987e-4 rad/s T = 2?/? = 34 610 s = 9 h 36 min 50 s The mass of the satellite is immaterial. For "orbit", centripetal acceleration = gravitational acceleration, or ?Â²r = GM/rÂ² rearranges to ? = ?(GM/rÂ³) where G = Newton's gravitational constant = 6.674e?11 NÂ·mÂ²/kgÂ² and M = mass of X = 5.09566*10 25 kg from above. So ? = ?(6.674*10 -11 NÂ·mÂ²/kgÂ² * 5.09566*10 25 kg / (1.889*10 7 m + 4*10 6 m)Â³ ) = 5.325e-4 rad/s T = 2?/? = 11799.27 s