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A: Mass of Rom, mR = 5.7×1026 kg Mass of Galla, mG = 2×1030 kgGravitational force, Fg = 8.8×1023…
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A: Solution:-
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A: Given: The radius of the earth is re = 6400 km
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A: Newton's Law of Universal Gravitation The gravitational force between two objects of mass m1 and m2…
Q: The gas-giant planet Rom (mass 5.7⨯1026 kg) goes around the star Galla (mass 2.0⨯1030 kg) in a…
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Q: A satellite in orbit above Earth's equator is traveling at an orbital speed of 7.45 KM/s. a) How…
A: Part A. Orbital speed of satellite = Mass of the Earth = M = If the distance from the Earth center…
Q: An Earth satellite has an orbital period of 5.3 h. What is its orbital radius? (ME = 5.98 x 1024 kg)
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Q: Calculate the period of a satellite orbiting a planet with the mass 8.7 x 1022 kg given the height…
A: Mass of the planet, M=8.7×1022kg Height to the satellite's orbit, h=300km=300×103m Radius of the…
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A: GivenFor a uniform, spherical, non rotating planet with radius Rs = 4.66 x 103 km, let gs, and gh be…
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A: Given, Mass of the earth =ME The radius of the earth = RE
Q: The radius of the Earth may be taken to be 6,371 km. Assume a satellite is in orbit at an altitude…
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Q: What will be the B.E. of a satellite of mass 80 kg.revolving in an orbit close to the earth's…
A: We need to compute-B.E. of satellite revolving close to earth=?B.E. of satellite at a height 1600 km…
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A: given that, The mean radius of earth from the Sun is 0.67 times the mars mean distance of mars from…
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Q: Time period of a revolution of a satellite is T, then its K. E. proportional to: (a) T-1 (b) T-2 (c)…
A: To find-Relation of K.E. with T=?Given-Periodic time=T
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Q: What is the escape speed from a planet of mass M = 3.1 x 1023 kg and radius R = 2.6 x 106 m? Write…
A: Given data: Mass of planet (M) = 3.1×1023 kg Radius (R) = 2.6×106 m Universal Gravitational…
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Q: Calcuiate the escape velocity of a body at a height 1600 km above the surface of the earth. Radius…
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Q: A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits…
A: The expression from the Kepler's third law is, T12T22=R13R23...............(1) Here, T1 is the time…
What is the circular velocity of an Earth satellite 1000 km above Earth’s surface? (Note: Earth’s average radius is 6370 km. Hint: Convert all quantities to m, kg, s.)
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