97x1024 kg rearth638x106 m g667x1011 nm2kg2 solution

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Unformatted text preview: ed 5.58 km/s. a. Find the distance of the satellite’s orbit from the surface of the Earth. Assume that the Earth is a perfect sphere. b. Calculate the radial acceleration of the satellite. [MEarth=5.97x1024 kg, REarth=6.38x106 m, G=6.67x10 ­11 Nm2/kg2] Solution: a) For a circular orbit, mv 2 GM E m GM (6.67x10 !11 )(5.97x10 24 ) = , or r = 2 E = = 1.28x10 7 m. This is the distance 2 32 r r v (5.58x10 ) from the center of the Earth, so the distance from the surface is r ! RE = 6.4 x10 6 m, which is about RE. b) The radial acceleration in just the centripetal acceleratio...
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