Apollo Lunar Mission

During the Apollo lunar explorations of the late 1960s and early 1970s, the main section of the spaceship remained in orbit about the Moon with one astronaut in it while the other two astronauts descended to the surface in the landing module. Assume the mass of this section is 5200 kg.
a) If the main section orbited about 52 mi above the lunar surface, determine the gravitational potential energy.
b) Determine the total energy.
c) Determine the energy needed to “escape” the Moon for the main section of the lunar exploration mission in orbit.

a). $52 miles = 52*1609 =83668 m$
To calculate the potential energy one must know the value of g at the surface of the moon.
$g_{moon} = k* Mmoon/Rmoon^2$ where $k$ is the gravitational constant
$g_{moon} = 6.67*10^{-11} *7.3477*10^{22}/1737100^2 =1.62 m/s^2$
The altitude of 52 miles is small compared to the Moon radius, hence $g_{moon}$ is the same.
The potential energy is $E_p = m*g_{moon}*h =5200*1.62*83668 =704819232 J =0.7 GJ$
b)
the speed of the module at the altitude of 52 miles comes from equating the centrifugal force with the gravitational force.
$m*g_{moon} = m*v^2/R$
$v^2 = g_{moon} *(R_{moon}+h) = 1.62*(1737100+83668) =2949644.16 m^2/s^2$
the kinetic energy is $Ec = m*v^2/2 =5200*2949644.16/2 =7669074816 J = 7.67 GJ$
The total energy is $E = Ep+Ec =7.67+0.7 =8.37 GJ$
c)
The energy needed to escape the moon can be computed in two ways
1. it is the kinetic energy needed to orbit the moon at altitude 0 meter (hence one must compute the speed necessary for stable trajectory at 0 meters – this comes from equating the centrifugal force with the gravitational force at 0 meters altitude)
$m*g_{moon} = m*v^2/R$
$v^2 = g_{moon} *R_{moon} = 1.62*1737100 =2814102 m^2/s^2$
The energy is therefore
$E = m*V^2/2 =5200*2814102/2 =7316665200 J = 7.32 GJ$
2.
It is the integral of the potential of the field from the distance $R_{moon}$ to infinity, where the potential of the field is defined as
$\Gamma(R) = k* M_{moon}/R$, for $R > R_{moon}$
$Energy = \int (m*\Gamma) dR = kmM_{moon}\int _{R_{moon}}^\infty(1/R)*dR$


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