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"From the Earth to the Moon" - Jules Verne's original problem

In Jules Verne's original problem, find the minimal launch velocity

v_0

that suffices for the projectile to make it "From the Earth to the Moon". (To reach the moon, the projectile must only just pass the point between the moon and earth where its net acceleration vanishes.)

G \approx 6.726*10^{-20}\ km^3/kg.s^2

M_e=5.975*10^{24}\ kg

M_m=7.35*10^{22}\ kg

S is the distance between the centers of the earth and the moon and is = 384,000 km

R is the radius of the earth and is = 6,378 km

Let r be the distance of the projectile from the center of the earth.

a=-\frac{GM_e}{r^2}+\frac{GM_m}{(S-r)^2}

a=0

-\frac{GM_e}{r^2}+\frac{GM_m}{(S-r)^2}=0

\frac{r^2}{(S-r)^2}=\frac{M_e}{M_m}=\frac{5.975*10^{24}}{7.35*10^{22}}

\frac{r}{384,000-r}\approx 9.016

\frac{r}{384,000}=\frac{9.016}{10.016}

r\approx 345,661\ km

v\frac{dv}{dr}=-\frac{GM_e}{r^2}+\frac{GM_m}{(S-r)^2}

\int vdv=\int \left[-\frac{GM_e}{r^2}+\frac{GM_m}{(S-r)^2}\right]dr

\frac{v^2}{2}=\frac{GM_e}{r}+GM_m/(S-r)+C

v(345,661)=0

\frac{GM_e}{345,661}+\frac{GM_m}{384,400-345,661}+C=0

C=-6.726*10^{-20}\left(\frac{5.975*10^{24}}{345,661}+\frac{7.35*10^{22}}{38,739}\right)

=-1.290

\frac{v^2}{2}=\frac{GM_e}{r}+GM_m/(S-r)-1.290

v(R)=v_0

\frac{v_0^2}{2}=6.726*10^{-20}\left(\frac{5.975*10^{24}}{6,378}+\frac{7.35*10^{22}}{384,400-6,378}\right)-1.290

v_0^2\approx 124.7564

v_0\approx 11.1\ km/s

(edited 12 years ago)

Reply 1

M_E_X

e: units, sorry. I'll look through again.

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"From the Earth to the Moon" - Jules Verne's original problem

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