IOAA 2015, Теория, кратки задачи, български — Задача 11

1

$\frac{1}{2}mv_0^{\ 2} - \frac{GmM}{R} = 0 \rightarrow R = \frac{2GM}{v_0^{\ 2}}$

[20]

2

$\Delta m(r) = \rho4\pi r^2\Delta r = 4\pi\alpha r\Delta r \rightarrow M = 2\pi\alpha R^2$

$(dm(r) = \rho4\pi r^2dr = 4\pi\alpha rdr \rightarrow M = 2\pi\alpha R^2)$

[30]

3

$M = \alpha2\pi R^2 = \alpha2\pi\frac{4G^2M^2}{v_0^{\ 4}} \rightarrow M = \frac{v_0^{\ 4}}{8\pi\alpha G^2}$

[30]

4

$M = \frac{v_0^{\ 4}}{8\pi\alpha G^2} = \frac{(1.5\times10^{4}\ ms^{-1})^4}{8\pi(5\times10^{13}\ kgm^{-2})(6.67\times10^{-11}\mathrm{kg^{-1}m^3s^{-2}})^2} = 9.1\times10^{21}\ kg$

[20]