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

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    MHF Contributor alexmahone's Avatar
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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}+\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
    Last edited by alexmahone; September 24th 2011 at 07:39 AM. Reason: Solved it myself!
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  2. #2
    Grand Panjandrum
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    Re: "From the Earth to the Moon" - Jules Verne's original problem

    Quote Originally Posted by alexmahone View Post
    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
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    MHF Contributor alexmahone's Avatar
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    Re: "From the Earth to the Moon" - Jules Verne's original problem

    Quote Originally Posted by CaptainBlack View Post
    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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