A satellite of mass 5510 kg orbits the Earth and has a period of 6510 s .   a.)Determine the radius of its circular orbit. b.)Determine the magnitude of the Earth's gravitational force on the satellite.

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A: mass of satellite (m) = 2600 kg altitude of satellite above the surface of earth (h) = 450 km

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A: 3) (a) T = 2π √(Re+h)3/√(gRe2)  (Re+h)3 = T2gRe2 / 4π2  (Re+h)3 = (24×3600)2×9.8×(6400×103)2 / 4π2…

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Q: question 43 please

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A: a) The expression used to find orbital speed of satellite is mv2r=GMmr2v=GMr=GMrE+r Substitute…

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A: Given data Orbital distance, r=1.05×1010mOrbital period,T=6.3days=5.44×105s

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A: Given: Radius of orbit of satellite above earth surface = 228 mi. = 366.93 km.

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A: The gravitational force be defined as,

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A: It is given that,  h=678 mi=(678×1609) m=1.091×106 m Radius of satellite's orbit is,…

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A: Given Data: The mass of Mars is, M=6.417×1023 kg The radius of Mars is, R=3390 km The rotation…

Q: 5. A satellite is in a circular orbit around the Earth at an altitude of h = 1, 250 km above the…

A: Given: Satellite is at height h=1250 km=1250×103 m=1.25×106 m Radius of earth RE=6.37×106 m Mass…

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A: The equation for the radial acceleration of the satellite is given by,

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Q: A 5.00 kg satellite is to be launch in geosynchronous orbit on Earth. (a) At what altitude above the…

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A: Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts…

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A: the centripetal acceleration is, ac=v2r Here, r~R the radius of earth. ac=g=9.80 m/s2.

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Q: A 100 kg satellite is placed into a near circular orbit with an average path radius of 6.671 x 10^6…

A: Given mass of satellite m= 100 kg Radius of satellite orbit = R = 6.671 × 106 m a)  To find,  period…

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A: Given Data Mass of the asteroid is Ma=2.0×10-4 ×Me    (Meis the mass of the Earth) Distance of the…

Q: NASA is expected to send a 2600-kg satellite 450 km above the earth's surface. (a) What is its…

A: Given : m = 2600 kg h = 450 km = 450 x 103 m M = 5.97 x 1024 kg

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Q: Unreasonable Results - (a) Based on Kepler’s laws and information on the orbital characteristics of…

A: (a) Here, r3T2=1.01×1019 km3/y2 When T=1.00 hour, r3=1.01×10191365×242r=1.01×10191365×2423r=5086 km

Q: An 76 kg satellite orbits a distant planet with a radius of 4500 km and a period of 280 min. From…

A: Write the formula for the force. F=ma Here, F is the force, m is the mass and a is the acceleration.

Q: A 530 kg satellite orbits at a distance from the Earth's center of about 7.9 Earth radii. What…

A: The expression for find gravitational force is Fg=GMmr2=GMm7.9re2

Q: (5). A satellite is in a circular orbit around the Earth at an altitude of h = 1, 250 km above the…

A: