Question

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to orbit at an altitude of 7530 km. Satellite B is to orbit at an altitude of 24200 km. The radius of Earth R_Eis 6370 km. (a) What is the ratio of the potential energy of satellite B to that of satellite A, in orbit? (b) What is the ratio of the kinetic energy of satellite B to that of satellite A, in orbit? (c) Which satellite (answer A or B) has the greater total energy if each has a mass of 28.5 kg? (d) By how much?

Answer 1

mv²/R = GMm/R²
v² = GM/R
v(A) = ?(GM/((6370+7530) x 10³) = ?(GM/(13900 x 10³)) m/s
v(B) = ?(GM/((6370+24200) x 10³) = ?(GM/(30570 x 10³)) m/s

the same formula gives the expression for the kinetic energy of an orbiting object:
mv²/R = GMm/R²
mv²= GMm/R
1/2 x mv² = GMm/(2R) = E_k
E_k(A) = GMm/(2 x 13900 x 10³) J
E_k(B) = GMm/(2 x 30570 x 10³) J
E_k(B)/E_k(A) = 0.454

the potential energy of an object is
E_p = twice the negative value of the kinetic energy
E_p = - GMm/R
E_pA = -GMm/(13900 x 10³)
E_potB = -GMm/(30570 x 10³)
E_p(B)/E_p(A) = 0.454

the total energy is E_k + E_p
E_t(A) = GMm/(2 x 13900 x 10³) - GMm/(13900 x 10³)
E_t(A) = GMm(1/(2 x 13900 x 10³) - 1/(13900 x 10³))
E_t(A) = - GMm((1/(2 x 13900 x 10³))
E_t(B) = - GMm(1/(2 x 30570 x 10³))
--->since 1/(2 x 13900 x 10³) is larger than 1(2 x 30570 x 10³) it follows, that the absolute value
of E_t(A) is lager than that of E_t(B), but since the total energies are negative, it follows that
E_t(B) is larger than E_t(A).

For the difference, subtract E_t(A) from E_t(B)
E_t(B) - E_t(A) = - GMm((1/(2 x 30570 x 10³)) - [ - GMm((1/(2 x 13900 x 10³))]
?E_t = GMm(1/(2 x 13900 x 10³) - 1/(2 x 30570 x 10³))
?E_t = 6.67 x 10^-11 x 5.974 x 10²⁴ x 28.5 x ((1/(2 x 13900 x 10³) - 1/(2 x 30570 x 10³))
?E_t = 2.227 x10⁸ J