(a) What is the minimum speed a spacecraft must have when…

(a) What is the minimum speed a spacecraft must have when leaving the Earth (at a height of 200 km) to reach the Moon? (b) If the spacecraft's mass is $m = 800\ \mathrm{t}$, what minimum kinetic energy (in kWh) is needed at launch?

Step 1: Idea

Between the Earth and the Moon there is a point A where the two pulls are equal. Before A the Earth's pull wins and slows the craft down; beyond A the Moon's pull wins and draws the craft in. So the craft only needs enough energy to reach A, with speed practically zero there.

Step 2: Neutral point

Let $x$ be the distance of A from the Earth's centre and $d$ the Earth–Moon distance. With the mass ratio $k = m_M/m_Z = 0.0118$ and the distance at perigee $d = 363\,000\ \mathrm{km} \approx 57R$ ($R$ = Earth's radius):

\[ \gamma\frac{m\,m_Z}{x^2} = \gamma\frac{m\,m_M}{(d - x)^2}, \]

\[ \frac{d - x}{x} = \sqrt k. \]

\[ x = \frac{d}{1 + \sqrt k}, \]

\[ x = \frac{57R}{1.109} \approx 51.4R. \]

Point A is about 90% of the way to the Moon, because the Moon is so much lighter.

Step 3: (a) Minimum speed

Energy conservation between the start (distance $r_0$ from the Earth's centre) and A, where the speed is zero. We neglect the Moon's pull at the start and its (small) contribution along the way; it lowers the result by only about 0.01 km/s:

\[ \tfrac12mv^2 - \gamma\frac{m\,m_Z}{r_0} = -\gamma\frac{m\,m_Z}{x}, \]

\[ \tfrac12v^2 = gR^2\left(\frac{1}{r_0} - \frac{1}{51.4R}\right). \]

Here $\gamma m_Z = gR^2$. From the surface ($r_0 = R$) and from a height $h = 200\ \mathrm{km}$ ($r_0 = R + h$): $r_0 = R: \ v = \sqrt{2gR\cdot\tfrac{50.4}{51.4}}$.

\[ r_0 = R + h: \ v = \sqrt{2gR\left(\tfrac{R}{R + h} - \tfrac{1}{51.4}\right)}. \]

\[\begin{gathered} v \approx 11.07\ \mathrm{km/s}\ (\text{surface}) \\ v \approx 10.90\ \mathrm{km/s}\ (h = 200\ \mathrm{km}) \end{gathered}\]

Both are only slightly below the escape velocity (11.2 km/s from the surface): reaching the Moon costs almost as much as leaving the Earth for good.

Step 4: (b) Energy for 800 t

\[ E = \tfrac12mv^2 = \tfrac12 \cdot 8 \cdot 10^5 \cdot (1.107 \cdot 10^4)^2, \]

\[ E \approx 4.9 \cdot 10^{13}\ \mathrm{J}. \]

Since $1\ \mathrm{kWh} = 3.6 \cdot 10^6\ \mathrm{J}$: \[\begin{gathered} E \approx 1.36 \cdot 10^7\ \mathrm{kWh} = 13.6\ \mathrm{GWh} \\ (13.2\ \mathrm{GWh}\ \text{from } 200\ \mathrm{km}) \end{gathered}\]

That is roughly the output of a large power station (1 GW) running for 14 hours, and it is only the final kinetic energy; a real rocket needs far more chemical energy.

Check

$v$ is just below the escape speed $\sqrt{2gR} = 11.2\ \mathrm{km/s}$, since A is far out where the Earth's potential is small $\checkmark$

$ 4.9 \cdot 10^{13}/3.6 \cdot 10^6 $ $= 1.36 \cdot 10^7\ \mathrm{kWh}\ \checkmark $