Question

A 969-kg satellite orbits the Earth at a constant altitude of 103-km. (a) How much energy...

A 969-kg satellite orbits the Earth at a constant altitude of 103-km.

(a) How much energy must be added to the system to move the satellite into a circular orbit with altitude 208 km?

(b) What is the change in the system's kinetic energy?

(c) What is the change in the system's potential energy?

Homework Answers

Answer #1

here,

the mass of satellite , m = 969 kg

initial altitude of satellite , A1 = 103 km = 103 * 10^3 m

a)

the final altitude of satellite , A2 = 208 km

A2 = 208 * 10^3 m

the energy must be added to the system , E = (- G * m * M /(2 *(R + A2))) - ((- G * m * M /(2 *(R + A1))))

E = 6.67 * 10^-11 * 5.98 * 10^24 * 969 /2 * ( - 1/(6.37 * 10^6 + 208 * 10^3) + 1/(6.37 * 10^6 + 103 * 10^3)) J

E = 4.77 * 10^8 J

b)

the change in the system's kinetic energy , dKE = KEf - KEi

dKE = (G * m * M/(2*(R + A2)) - G * m * M/(2*(R + A1)))

dKE = 6.67 * 10^-11 * 969 * 5.98 * 10^24 /2 * ( 1/(6.37 * 10^6 + 208 * 10^3) - 1/(6.37 * 10^6 + 103 * 10^3)) J

dKE = - 4.77 * 10^8 J

c)

the change in potential energy , dPE = E - dKE

dPE = 4.77 * 10^8 J - ( -4.77 * 10^8) J

dPE = 9.54 * 10^8 J

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