Satellite orbit

\[\frac{{{r_{\max }}}}{{{r_{\min }}}} = \frac{{1 + e}}{{1 – e}} = \frac{{1 + 0.0014}}{{1 – 0.0014}} = 1.0028\]

\[ \Rightarrow {r_{\max }} = {r_{\min }}\left( {1.0028} \right)\]

\[{r_{\min }} = {\rm{radius}}\,{\rm{of}}\,{\rm{earth}}\,{\rm{ + }}\,{\rm{altitude}}\,{\rm{above}}\,{\rm{earth’s}}\,{\rm{surface}}\]

\[ \Rightarrow {{\rm{r}}_{\min }} = 6.4 \times {10^6} + 0.4 \times {10^6} = 6.8 \times {10^6}\,m\]

\[\Rightarrow {r_{\max }} = 6.8 \times {10^6}\left( {1.0028} \right) = 681904\,m = 6.81904 \times {10^6}\,m\]

Therefore satellite’s altitude at apogee is

\[6.81904 \times {10^6} – 6.4 \times {10^6} = 419040\,m = 419.04\,km\]

Time period of the satellite is given as

\[{T^2} = \frac{{4{\pi ^2}}}{{{k^2}}}{a^3}\]

\[{k^2} = G{M_{earth}} = 3.986 \times {10^{14}}\,{m^3}/{s^2}\]

\[ \Rightarrow a = {\rm{semi}}\,{\rm{major}}\,{\rm{axis}}\,{\rm{ = }}\frac{{{r_{\max }} + {r_{\min }}}}{2} = \frac{{6.81904 \times {{10}^6} + 6.8 \times {{10}^6}}}{2} = 6.80952 \times {10^6}\,m\]

\[{T^2} = \frac{{4{\pi ^2}}}{{3.986 \times {{10}^{14}}}} \times {\left( {6.80952 \times {{10}^6}} \right)^3}\]

\[ \Rightarrow T = 5592.242\,s = 1.5534\,hr\]

\[{V_{perigee}} = {r_{\min }}\dot \theta \]

\[\dot \theta = \frac{h}{{{{\left( {{r_{\min }}} \right)}^2}}}\]

\[{h^2} = {r_{\min }}\left( {1 + e} \right){k^2}\]

\[ \Rightarrow {h^2} = \left( {6.8 \times {{10}^6}} \right)\left( {1 + 0.0014} \right)3.986 \times {10^{14}}\]

\[ \Rightarrow h = 5.20987 \times {10^{10}}\]

Therefore,

\[\dot \theta = \frac{{5.20987 \times {{10}^{10}}}}{{{{\left( {6.8 \times {{10}^6}} \right)}^2}}} = 0.0011267\,rad/s\]

\[{V_{perigee}} = \left( {6.8 \times {{10}^6} \times 0.0011267} \right) = 7661.56\,m/s = 7.66156\,km/s\]