Problem 2 A small satellite is in a circular orbit around the Earth at a height

Question

Problem 2 A small satellite is in a circular orbit around the Earth at a height of 320 km above the Earth's surface. The radius of the Earth is RE 6.38x106 m, the Earth's mass is ME 5.98x1024 kg, and Newton's gravitational constant is G 6.67x10-11 N-m2/kg2 (a) Find the speed v and the orbital period T of the satellite in this circular orbit. (b) Find the escape velocity vesc from the Earth for the satellite in this orbit. (c) Find the frequency of oscillation of small radial displacements from this circular orbit. (d) Find the total energy of the satellite in its orbit. Express your answer as e, i.e., in units of energy per unit mass. Problem 3 Assume the satellite in Problem 2 is in an elliptical, rather than circular, orbit. Its minimum height above the Earth's surface is 320 km while its maximum height is 1000 km. (a) Find the eccentricity of this orbit. (b) Find the satellite's speed at perigee and at apogee. (c) Find the total energy of the satellite in its orbit. Express your answer as e, i.e., in units of energy per unit mass.

Explanation / Answer

(A) Rp = 6371 + 320 = 6691 km

Ra = 6371 + 1000 = 7371 km

e = (Ra - Rp)/(Ra + Rp) = 0.048

(B) a = (Rp + Ra)/2 = 7031 km = 7.031 x 10^6 m

GM = (6.67 x 10^-11)(5.976 x 10^24)= 3.986 x 10^14

v = sqrt[ G M (2/r - 1/a)]

rp = 6.691 x 10^6 m

putting values,

vp = 7903 m/s .....Speed at perigee

at apogee,

ra = 7.371 x 10^6 m

va = 7174 m/s

(C) total energy = PE + KE

= - G M m / r + m v^2 /2

= m [ (-6.67 x 10^-11 x5.976 x 10^24 / (6.691 x 10^6)) + (7903^2 / 2)]

= - 2.834 x 10^7 m

energy per unit mass = - 2.834 x 10^7 J/kg

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