There is a moon orbiting an Earth-like planet. The mass of the moon is 5.37 × 1022 kg, the center-to-center separation of the planet and the moon is 5.34 × 105 km, the orbital period of the moon is 21.6 days, and the radius of the moon is 1060 km. What is the angular momentum of the moon about the planet? Answer in units of kg m2 /s.

There is a moon orbiting an Earth-like planet. The mass of the moon is 5.37 ×...
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.030×105 km2.030×105 km and the planet has a radius of 4125 km4125 km and a mass of 7.15×1022 kg7.15×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N⋅m2/kg26.67×10−11N·m2/kg2.
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.270 × 105 km and the planet has a radius of 3.150 × 103 km and a mass of 6.55 x 1022 kg, how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67 x 10lN-m2/kg2 T-0.88125...
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.195×105 km and the planet has a radius of 4.400×103 km and a mass of 7.60×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N⋅m2/kg2.
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.435×105 km2.435×105 km and the planet has a radius of 4075 km4075 km and a mass of 5.05×1022 kg5.05×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg26.67×10−11N·m2/kg2. T = ? DAYS
A planet of mass ?=8.45×1024 kg is orbiting in a circular path a star of mass ?=3.85×1029 kg . The radius of the orbit is ?=3.85×107 km . What is the orbital period (in Earth days) of the planet ?planet ? ?planet=
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.285×105 km and the planet has a radius of 3.650×103 km and a mass of 4.30×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2
An undiscovered planet, many light-years from Earth, has one moon which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.00000 × 105 km and the planet has a radius of 3225 km and a mass of 6.85 × 1022 kg, how long (in days) does it take the moon to make one revolution around the planet. The gravitational constant is 6.67 × 10-11 N·m2/kg2.
An undiscovered planet, many light-years from Earth, has one moon which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.10500 × 105 km and the planet has a radius of 4.350 × 103 km and a mass of 7.75 × 1022 kg, how long (in days) does it take the moon to make one revolution around the planet. The gravitational constant is 6.67 × 10-11 N·m2/kg2.
An undiscovered planet, many light-years from Earth, has one moon which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.06000 × 105 km and the planet has a radius of 3275 km and a mass of 4.60 × 1022 kg, how long (in days) does it take the moon to make one revolution around the planet. The gravitational constant is 6.67 × 10-11 N·m2/kg2.
What is the angular momentum of the moon about the Earth? The mass of the moon is 7.35 x 10^22 kg, the center-to-center separation of the Earth and the moon is 3.84 x 10^5 km, and the orbital period of the moon is 27.3 days. How do you get 3.84 x 10^8? v = 2 x pi x r/T L = m x 2 x pi x r x r/ T L = (7.35 x 10^22 x 2 x 3.14...