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

**alexmahone** · September 24th 2011, 06:18 AM

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.

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

$a=0$

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

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

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

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

$r\approx 345,661\ km$

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

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

$\frac{v^2}{2}=\frac{GM_e}{r}+\frac{GM_m}{S-r}+C$

$v(345,661)=0$

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

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

$=-1.290$

$\frac{v^2}{2}=\frac{GM_e}{r}+\frac{GM_m}{S-r}-1.290$

$v(R)=v_0$

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

$v_0^2\approx 124.7564$

$v_0\approx 11.1\ km/s$

**CaptainBlack** · September 24th 2011, 06:50 AM

Originally Posted by alexmahone

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.

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

$a=0$

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

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

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

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

$r\approx 345,661\ km$

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

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

$\frac{v^2}{2}=\frac{GM_e}{r}+\frac{GM_m}{S-r}+C$

$v(345,661)=0$

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

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

$=-1.290$

$\frac{v^2}{2}=\frac{GM_e}{r}+\frac{GM_m}{S-r}-1.290$

$v(R)=v_0$

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

$v_0^2\approx 124.7564$

$v_0\approx 11.1\ km/s$

Why do you think you can treat this as a one dimensional problem?

CB

**alexmahone** · September 24th 2011, 06:53 AM

Originally Posted by CaptainBlack

Why do you think you can treat this as a one dimensional problem?

CB

The gravitational effect of the sun on the projectile is probably negligibly small due to the large distance between them.

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