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A satellite 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.

Respuesta :

Answer:

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[tex] \theta angle = 0[/tex]

and radial component of given velocity is zero

we have[tex] h = r_o v_r_o = 6378+600 =6.97*10^6 m[/tex]

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

we know that

[tex]\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)[/tex]

[tex]GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s[/tex]

so

[tex]\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)[/tex]

solvingt for [tex] \epsilon)[/tex]

[tex]\epsilon = 0.22)[/tex]

therefore eccentrcity of orbit is 0.22

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