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 7760 km. Satellite B is to orbit at an altitude of 23600 km. The radius of Earth Rgis 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 12.6 kg? (d) By how much? (a) Number Units (b) Number Uni (d) Number Unii

Answer 1

(b) starting with

mv^2/R = GMm/R^2

v^2 = GM/R

v(A) = √(GM/((6370+7760)*10^3) = √(GM/(14130*10^3)) m/s

v(B) = √(GM/((6370+23600)*10^3) = √(GM/(29970*10^3)) m/s

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*14130*10^3) J

Ekin(B) = GMm/(2*29970*10^3) J

EkinB/EkinA = 0.471

(a) the potential energy of an object is

Epot = twice the negative value of the kinetic energy

Epot = - GMm/R

EpotA = -GMm/(14130*10^3)

EpotB = -GMm/(29970*10^3)

EpotB/EpotA = 0.471

(c) the total energy is Ekin + Epot

Etot(A) = GMm/(2*14130*10^3) - GMm/(14130*10^3)

EtotA = GMm(1/(2*14130*10^3) - 1/(14130*10^3))

EtotA = - GMm((1/(2*14130*10^3))

EtotB = - GMm(1/(2*29970*10^3))

--->since 1/(2*14130*10^3) is larger than 1(2*29970*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*29970*10^3)) - [ - GMm((1/(2*14130*10^3))]

∆Etot = GMm(1/(2*14130*10^3) - 1/(2*29970*10^3))

∆Etot = 6.67*10^-11*5.974*10^24*12.6((1/(2*14130*10^3) - 1/(2*29970*10^3))

∆Etot = 9.39 *10^7 J