Answer to Question #105125 in Classical Mechanics for Frenco

Question

1. Consider the mechanical energy of a body in geostationary orbit above the Earth's equator, at rGS=42000 km.

Consider the mechanical energy of the same body on Earth at the South pole, at re = 6400 km. For this problem, we consider the Earth to be spherical. (Remember, the object at the equator is in orbit, the object at the Pole is not in orbit.)

G = 6.67×10^−11 Nm^2 kg^−2, and the mass of the Earth is M = 5.97×10^24 kg

What is the difference in the mechanical energy per kilogram between the two?

E = ___ MJ.kg^−1(to two significant figures, don't use scientific notation)

Answer 1

Mechanical energy at the surface of the earth=

$\dfrac{GM}{R}=\dfrac{6.67×10^{−11} \times5.97×10^{24}}{(6400)\times10^3)}=\dfrac{39.8199\times10^{13}}{64\times10^5}$

$=\dfrac{39.82\times 10^8}{64}=6.22\times 10^7J/kg$

Mechanical energy per kg at height =$\dfrac{GM}{(R+h)} ​ + \dfrac{ v ^ 2}2 ​$

$=\dfrac{GM}{(R+h)}+\dfrac{GM}{(R+h)}=\dfrac{2GM}{(R+h)}$

$=\dfrac{2\times6.67×10^{−11} \times5.97×10^{24}}{(6400+42000)\times10^3)}$

$=\dfrac{2\times 39.8199\times 10^{13}}{(462\times 10^5)}$

$=\dfrac{79.6398\times10^{13}}{462\times 10^5}=0.172\times 10^8 J/kg$

$=1.72\times 10^7 J/kg$

Hence the required difference=$6.22×10 ^7 −1.72×10 ^7 =4.5×10 ^7 J/kg$