Asatellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the tangent of the surface of the earth. assuming the radius of the earth to be 6378 km and that is 5.976 * 10^6 kg, determine the eccentricity of the orbit.

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore

and radial component of given velocity is zero

we have

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

so

solvingt for

therefore eccentrcity of orbit is 0.22

answer:

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