Odpowiedź:
[tex]d_P = \frac{M_P}{V_P}\\\\V_I = \sqrt{\frac{GM}{R} }\\R_P = 0,3 R_Z\\\\V_{IZ} = \sqrt{\frac{GM_Z}{R_Z} } = V_{IP} = \sqrt{\frac{GM_P}{R_P} }\\\frac{M_Z}{R_Z} = \frac{M_P}{R_P} \\\\M_P = M_Z \frac{R_P}{R_Z} = M_Z \frac{0,3 R_Z}{R_Z} = 0,3 M_Z\\V_P =\frac{4}{3} \pi R_P^3 = \frac{4}{3} \pi (0,3R_Z)^3 \\\\d_P = \frac{0,3 M_Z}{\frac{4}{3} \pi (0,3R_Z)^3} =\frac{3 M_Z}{4 \pi *0,3^2 R_Z^3} = 6,1369 * 10^{4} \frac{kg}{m^3}[/tex]
Wyjaśnienie: