## Task 2 - 010912

In ballistics, the term "mean precision" $p_{m}$ is frequently used. If $p_{m}$ is taken as the radius of a circle, then about 20 percent of all hits lie within this circle. All hits are captured by a circle with a radius that is approximately $4 \frac{1}{2}$ times larger. Western military experts currently estimate a mean precision (for rockets) of $p_{m}=0.5$ percent of the range. Later, they aim to achieve values of $p_{m}=0.1$ percent and in the distant future even $p_{m}=0.05$ percent.

a) How large would the radius of the 20 percent circle and that of the circle containing all hits be for these values if the range is $12500 \mathrm{~km}$?

b) What mean precision $p_{m}$ did the Soviet Union achieve, considering that the radius of the circle containing all hits in the tests conducted in October 1961 was less than $1 \mathrm{~km}$?
a) The value $p_{m}$ gives the percentage of the ratio of the radius $r_{20}$ of the $20\%$ circle to the range. Since for the radius $r_{100}$ of the circle in which all hits land, $r_{100}=4 \frac{1}{2} r_{20}$ holds, the circles for the given $p_{m}$ have the following radii:

| $p_{m}$ | $0.5\%$ | $0.1\%$ | $0.05\%$ |
| :---: | :---: | :---: | :---: |
| $r_{20}$ | $62.5 \mathrm{~km}$ | $12.5 \mathrm{~km}$ | $6.25 \mathrm{~km}$ |
| $r_{100}$ | $281.25 \mathrm{~km}$ | $56.25 \mathrm{~km}$ | $28.125 \mathrm{~km}$ |

b) If $r_{100}<1 \mathrm{~km}$, then $r_{20}=\frac{2}{9} r_{100}<\frac{2}{9} \mathrm{~km}=0.222 \mathrm{~km}$. Thus, the ratio of $r_{20}$ to the range is less than $\frac{0.222 \mathrm{~km}}{12500 \mathrm{~km}} \approx 0.000018$, so $p_{m}<0.0018\%$.