Oblicz okres obiegu satelity poruszającego się na wysokości 600km nad powierzchnią Ziemi.

[latex]h=0,6cdot10^{6} [m]\\R=6,37cdot10^{6} [m]\\m_{1}=5,98cdot10^{24} [kg][/latex] Promień orbity satelity: [latex]r=R+h=6,97cdot10^{6} [m][/latex] Siła grawitacji pełni rolę siły dośrodkowej: [latex]F_{d}=F_{g}\\frac{m_{2}v^{2}}{r}=frac{Gm_{1}m_{2}}{r^{2}}\\v^{2}r=Gm_{1} | v=frac{2pi r}{T}\\frac{4pi^{2}r^{3}}{T^{2}}=Gm_{1}\\T=sqrt{frac{4pi^{2}r^{3}}{Gm_{1}}}=sqrt{frac{4cdot(3,14)^{2}cdot(6,97cdot10^{6})^{3}}{6,67cdot10^{-11}cdot5,98cdot10^{24}}}approx5,79cdot10^{3} [s][/latex]

[latex]dane:\h = 600 km = 0,6*10^{6} m\R_{z} = 6370 km = 6,37*10^{6} m\r = R_{z}+h = 6,36*10^{6}m+0,6*10^{6}m = 6,96*10^{6} m\g = 10frac{m}{s^{2}}\szukane:\T = ?[/latex] [latex]F_{g} = F_{r}\\frac{GMm}{r^{2}} = frac{mv^{2}}{r} |*frac{r}{m}\\frac{GM}{r} = v^{2}\\ale\\v = frac{2 pi r}{T} = frac{2 pi (R_{z}+h)}{T}\i\g = frac{GM}{R_{z}^{2}} -> GM = gR__{z}}^{2}\\frac{gR_{z}^{2}}{R_{z}+h}=[frac{2 pi (R_{z}+h)}{T}]^{2}[/latex] [latex]frac{gr_{z}^{2}}{R_{z}+h} = frac{4 pi ^{2}(R_{z}+h)^{2}}{T^{2}}\\T^{2}gR_{z}^{2} = 4 pi ^{2}(R_{z}+h)^{3} /:gR{z}^{2}}\\T^{2} = frac{4 pi ^{2}(R_{z}+h)^{3}}{gR_{z}^{2}}\\T = frac{2 pi (R_{z}+h)}{R_{z}}*sqrt{frac{R_{z}+h}{g}}\\T = frac{2*3,14*6,96*10^{6}}{6,37*10^{6}}*sqrt{frac{6,96*10{6}}{10}} s\\T approx5670 s = 1h 34,5min[/latex]