Answer to Question #105228 in Classical Mechanics for Jeff

Question

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

How much more mechanical energy per kilogram does an object on the ground at the Equator than on the ground at the Pole?
E =E= ___ MJ.kg^−1
(2 sig figs, do not use scientific notation)

Answer 1

As per the given question,

Orbital distance from the surface of the earth (h)=42000 km

Radius of the earth at south pole(R) =64000km

$G=6.67×10^{−11} Nm^2 /kg^2$

mass of the Earth is $M=5.97×10^{24} kg$

a)

Minimum velocity required to move in the orbit(v),

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

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

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

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

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

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

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

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

Mechanical energy at the surface of the earth=$\dfrac{GM}{R}=\dfrac{6.67×10^{−11} \times5.97×10^{24}}{(6400)\times10^3)}=\dfrac{39.8199\times10^{13}}{64\times10^5}$

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

Hence the required difference = $=6.22\times 10^7-1.72\times 10^7=4.5\times 10^7$ J/kg