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 7640 km. SatelliteB is to orbit at an altitude of 19100 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 14.2 kg? (d) By how much?

I cannot seem to get the answer for D. I have the answer for A and B at 0.0551 and C i have that satellite B is greater.

Explanation / Answer

Re = radius of the earth = 6370 km
Ra = radius of the A orbit = Re + 7640 = 14010 km
Rb = radius of the B orbit. = Re + 19100 = 26740 km
M = mass of the earth = 5.9742x10^24 kg
m = mass of each satellite 14.2 kg
G = gravitational constant = 6.673x10^(-11) m^3 kg^-1 s^-2

(a) PEa = -GMm/Ra
PEb = -GMm/Rb

PEb/ PEa = Ra/Rb = 0.524

(b)Centripetal force is m*v^2/r
This must equal the gravitaional force = GMm/r^2
GM/r^2 = v^2/r
v^2 = GM/r
KE = (1/2)m*GM/r = (1/2)GMm/r

KEb/KEa = Ra/Rb = 0.524

(c) You can just use the difference in energies between the orbits.
KE + PE = (1/2)m(GM/r) - GMm/r
KE + PE = -(1/2)GMm/r
Ea = -(1/2)GMm/Ra
Eb = -(1/2)GMm/Rb

As Rb > Ra, so Eb > Ea (because B has less -ve energy so)

d) difference = Eb - Ea = -(1/2)GMm/Rb + (1/2)GMm/Ra
= (1/2)GMm[1/Ra - 1/Rb] = 96180736 J =96 kJ

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