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Math Help - minimum mass of planet

  1. #1
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    minimum mass of planet

    Certain neutron stars (extremely dense stars) are believed to be rotating at about 0.6 rev/s . If such a star has a radius of 12.5 km , what must be its minimum mass so that objects on its surface will be attracted to the star and not thrown off by the rapid rotation?

    i can't figure this problem out for the life of me.

    i think it has something to do with the force of gravity=centripetal acceleration...
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  2. #2
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    Quote Originally Posted by Chadly724 View Post
    Certain neutron stars (extremely dense stars) are believed to be rotating at about 0.6 rev/s . If such a star has a radius of 12.5 km , what must be its minimum mass so that objects on its surface will be attracted to the star and not thrown off by the rapid rotation?

    i can't figure this problem out for the life of me.

    i think it has something to do with the force of gravity=centripetal acceleration...
    The radial acceleration required to keep a particle of mass m on the equator is:

    <br />
a=r \dot \theta^2<br />

    the gravitational force on the paticle is:

    <br />
F_G=\frac{GmM}{r^2}<br />

    so for the the particle to be just bound the required condition is:

    <br />
F_G=\frac{GmM}{r^2}=ma=m\,r \dot \theta^2<br />

    where M is the mass of the Neutron Star.

    So:

    <br />
\frac{G\,M}{r^2}=r \dot \theta^2<br />

    So the limmiting mass is:

    <br />
M=\frac{r^3 \dot \theta^2}{G}<br />

    where  \dot \theta=0.6\times 2\times \pi \rm{\ radians\ per \ second}, r=12500 \rm{\ m}, and G=6.67 \,10^{-11}\rm{\ N\ m^2\ / kg^2}.

    Which appears to give: M=4.16\ 10^{23}\ \rm{kg}, which is \sim 2\times 10^{-7}\ \rm{M_{\bigodot}} or \sim 0.07 M_{\bigoplus}

    RonL
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  3. #3
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    now i get it...

    i was using the wrong equation for Fg because i didn't have a little r.
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  4. #4
    Forum Admin topsquark's Avatar
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    I suppose I should mention that we technically need to be using General Relativity to calculate the force at surface of the neutron star. I'd be happy to provide the force law if someone is interested. (I'd have to look it up and I'm feeling lazy at the moment.)

    -Dan
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    Quote Originally Posted by topsquark View Post
    I suppose I should mention that we technically need to be using General Relativity to calculate the force at surface of the neutron star. I'd be happy to provide the force law if someone is interested. (I'd have to look it up and I'm feeling lazy at the moment.)

    -Dan
    I'm not sure that we are talking neutron star with that radius and mass \sim 0.07 M_{\bigoplus}
    (not that I have calculated the density exactly but it looks nowhere near the required density to me,
    which is ~10^18 kg/m^3)

    RonL
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    I'm not sure that we are talking neutron star with that radius and mass \sim 0.07 M_{\bigoplus}
    (not that I have calculated the density exactly but it looks nowhere near the required density to me,
    which is ~10^18 kg/m^3)

    RonL
    You are correct, the density is about 4 orders of magnitude too small to be a neutron star. This puts it in the white dwarf category. However, relativistic effects should be observable from such.

    I just did the calculation (if I did it right, I hate these unit conversions!) for the white dwarf and I get that the acceleration only differs from Newtonian by a factor of about 0.00182%, whereas for the neutron star the relativistic correction fairly well overwhelms the result by a factor of some 91000%.

    So I guess the relativistic correction IS fairly small for this case after all.

    -Dan
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