A) A satellite m = 500 kg orbits the earth at a distance d = 201 km, above the s

Question

A) A satellite m = 500 kg orbits the earth at a distance d = 201 km, above the surface of the planet. The radius of the earth is re = 6.38 × 106 m and the gravitational constant G = 6.67 × 10-11 N m2/kg2 and the Earth's mass is me = 5.98 × 1024 kg. What is the speed of the satellite in m/s?

C) Suppose you are told that a satellite orbiting the Earth has a orbital period of 2.1 hours. Using the orbital characteristics of the Moon (RM = 3.84 × 105km and TM = 0.7481 y), use Kepler's laws to calculate the orbital radius for the satellite in km. Is there anything unreasonable?

r3(km3/y2) 1.01x1019 3.35x1024 3.35x1024 Period TO) 0.07481 1.000 11.86 0.00485 (1.77 d) 3.19x 10 0.00972 (3.55 d) 3.20x 10-1 0.0196 (7.16 d) 3.19x 10 0.0457 (16.19 d) | 3.20× 1021 Satellite Parent Earth Sun Average orbital radius(km) 3.84x105 1.496x108 7.783x108 4.22x 10 Moon Earth upiter Jupitero Europa 6.71x10 Ganymede 1.07x10 Callisto 1.88x106 ©theexpertta.com

Explanation / Answer

A) velocity = G*M / r

Here M is mass of earth, r is the distance of satellite from the center.

velocity = 6.67E-11*5.98E-24 / 6380000 + 201000

Velocity = 3.988E14 / 6581000

Velocity = 60598693.21

Velocity = 7784.5 m/s

B) orbital radius = 3T2G*M / 4*pi*pi

Putting all the values, we get

Orbital radius = 3 5.579E25 = 3.83E5 km

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