An earth satellite is observed at perigee to be 250 km above earth’s surface, an

Question

An earth satellite is observed at perigee to be 250 km above earth’s surface, and traveling at 8500 m/s. Determine numerical values for the eccentricity of its orbit. Show your work. If any of the quantities cannot be determined, clearly state what additional information would be required, and show how you would calculate the quantity if you had that additional information.

(Note: 250 km is the distance above the surface of earth, not the distance from the center of earth. You may look up
values of G, the mass of earth, and the radius of earth.)

Explanation / Answer

We are given the satellite’s height hmin = 250 km; therefore the distance from the earth’s center is rmin = Re + hmin=6650 km. For any known satellite, we can approximate µ as m, since the satellite’s mass is a tiny fraction of the earth’s. Therefore, the angular momentum is l = mvmaxrmin, and the parameter c is c = l2/?µ=(vmaxrmin)2/GMe (where we have used the fact that ? = GMem). With the given numbers, we get c = 7960 km. Then, since rmin = c/(1 + o), we have o = (c - rmin)/rmin = 0.197. Similarly, rmax = c/(1 - o)=9910 km, so hmax = rmax = Re = 3510 km.

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