Question

Space junk of mass $m$ leaves its orbit and heads towards an annular disc-shaped meteor at rest relative to earth along its axis at a distance $8R$ with a velocity $v$. The surface mass density of the meteor is $\sigma$ and its inner and outer radii are $R$ and $4R$. Find the velocity with which the junk passes the centre of the meteor. (Thickness of the meteor is negligible).

Answer 1

Consider an elemental ring of radius $r$ and width $dr$.
We have potential energy of the system $(ring+junk)$
$d{E}_1=\frac{Gmdm}{(x^2+r^2{)}^{1/2}}=-\frac{Gm2\pi rdr\sigma }{(x^2+r^2{)}^{1/2}}$
$E_1=-2\pi Gm\sigma {\int }_{R_1}^{R_2}\frac{rdr}{(x^2+r^2{)}^{1/2}}$
Substituting, $x^2+r^2=t$
$2rdr=dt;rdr=dt/2$
$E_1=-2\pi Gm\sigma \int \frac{dt}{2(t{)}^{1/2}}=-2\pi Gm\sigma |(x^2+r^2{)}^{1/2}{|}_{R_1}^{R_2}$
$=-2\pi Gm\sigma \left[\right(x^2+R_2^2{)}^{1/2}-(x^2+R_1^2{)}^{1/2}]......\left(i\right)$
At the centre $E_2=-2\pi Gm\sigma (R_2-R_1)......\left(ii\right)$
The initial kinetic energy of the junk $E_3=\frac{1}{2}m{v}^2$
Let the velocity of the junk at the centre be $v^{\prime }$ then $E_4=\frac{1}{2}M{v}^{\prime 2}$
Conserving energy, $E_1+E_3=E_2+E_4$
$\Rightarrow -2\pi Gm\sigma \left[\right(x^2+R_2^2{)}^{1/2}-(x^2+R_1^2{)}^{1/2}]+\frac{1}{2}m{v}^2=-2\pi Gm\sigma (R_2-R_1)+\frac{1}{2}m{v}^2$
$\Rightarrow -2\pi G\sigma \left[\right(64R^2+16R^2{)}^{1/2}-(64R^2+R^2{)}^{1/2}]+\frac{1}{2}{v}^2=-2\pi G\sigma \left(3R\right)+\frac{1}{2}{v}^{\prime 2}$
$\Rightarrow v^{\prime }={\left\{4\pi G\sigma \right[3-\sqrt{5}(4-\sqrt{13})]R+v^2\}}^{1/2}$.