Two Earth satellites, A and B , each of mass m , are to be launched into circula

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 5140 km. Satellite B is to orbit at an altitude of 18800 km. The radius of Earth REis 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 23.3 kg? (d) By how much?

Explanation / Answer

starting with
mv^2/R = GMm/R^2
v^2 = GM/R
v(A) = (GM/((6370+5140)*10^3) = (GM/(11510*10^3)) m/s
v(B) = (GM/((6370+18800)*10^3) = (GM/(25170*19^3)) m/s

b) the same formula gives the expression for the kinetic energy of an orbiting object:
mv^2/R = GMm/R^2
mv^2= GMm/R
1/2*mv^2 = GMm/(2R) = Ekin
Ekin(A) = GMm/(2*11510*10^3) J
Ekin(B) = GMm/(2*25170*10^3) J
EkinB/EkinA = 0.4573 approx

a) the potential energy of an object is
Epot = twice the negative value of the kinetic energy
Epot = - GMm/R
EpotA = -GMm/(11510*10^3)
EpotB = -GMm/(25170*10^3)
EpotB/EpotA = 0.4573 approx


c) the total energy is Ekin + Epot
Etot(A) = GMm/(2*11510*10^3) - GMm/(11510*10^3)
EtotA = GMm(1/(2*11510*10^3) - 1/(11510*10^3))
EtotA = - GMm((1/(2*11510*10^3))
EtotB = - GMm(1/(2*25170*10^3))
--->since 1/(2*11510*10^3) is larger than 1(2*25170*10^3) it follows, that the absolute value
of EtotA is lager than that of EtotB, but since the total energies are negative, it follows that
EtotB is larger than EtotA.

d) For the difference, subtract EtotA from EtotB
EtotB - EtotA = - GMm((1/(2*25170*10^3)) - [ - GMm((1/(2*11510*10^3))]
Etot = GMm(1/(2*11510*10^3) - 1/(2*25170*10^3))
Etot = 6.67*10^(-11)*5.972*10^24*23.3[(1/(2*11510*10^3) - 1/(2*25170*10^3)] taking mass of earth= 5.972* 10^24 kg
Etot = 2.1880 x 10^8 joule

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