question_answer 63)

                  A satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2 R where \[R=6400\,\,km\] is the radius of the earth. Find eccentricity of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius 6R?                 [\[G=6.67\times {{10}^{-11}}\,SI\] units and \[M=6\times {{10}^{24}}\,kg\]]

Answer:

                  (i) We know \[{{r}_{p}}=a(1-e)=a-ae\]                 and \[{{r}_{a}}=a(1+e)=a+ae\]                 \[\therefore \] \[e=\,\frac{{{r}_{a}}-{{r}_{p}}}{{{r}_{a}}+{{r}_{p}}}=\frac{4R}{8R}=0.5\]                 (ii) According to the law of conservation o, angular momentum,                 \[m{{v}_{p}}{{r}_{p}}=m{{\upsilon }_{a}}{{r}_{a}}\]                 \[\therefore \] \[\frac{{{\upsilon }_{p}}}{{{\upsilon }_{a}}}=\,\frac{{{r}_{a}}}{{{r}_{p}}}=\frac{6R}{2R}=3\]                 or \[{{\upsilon }_{p}}=\,3{{\upsilon }_{a}}\]                                               ?. (i)                 Now, total energy of satellite at perihelion                 \[=\frac{1}{2}\,m\,\upsilon _{p}^{2}\,-\,\frac{GMm}{{{r}_{p}}}\]                 Total energy of satellite at aphelion                 \[=\frac{1}{2}\,mv_{a}^{2}\,-\frac{GMm}{{{r}_{a}}}\]                 According to the law of conservation of energy.                 \[\frac{1}{2}m\,\nu _{p}^{2}\,-\,\frac{Gmm}{{{r}_{p}}}\,=\frac{1}{2}\,m\,v_{a}^{2}-\,\frac{GMm}{{{r}_{a}}}\]                 or \[\frac{\upsilon _{p}^{2}}{2}-\,\frac{\upsilon _{a}^{2}}{2}=\,-GM\,\left( \frac{1}{{{r}_{a}}}-\,\frac{1}{{{r}_{p}}} \right)\]                 \[\left( v_{p}^{2}-v_{a}^{2} \right)=\,-2GM\,\left( \frac{1}{{{r}_{a}}}-\frac{1}{{{r}_{p}}} \right)\]                 \[=-\,2GM\,\frac{({{r}_{p}}-{{r}_{a}})}{{{r}_{a}}{{r}_{p}}}\]                 or \[8\,\,\upsilon _{a}^{2}=\,-2GM\,\,\frac{({{r}_{p}}-{{r}_{a}})}{{{r}_{a}}{{r}_{p}}}\]                 or \[\upsilon _{a}^{2}\,=\,-\frac{GM}{4}\,\frac{({{r}_{p}}-{{r}_{a}})}{{{r}_{a}}{{r}_{p}}}\]                 \[=\,\frac{-GM}{4}\,\,\times \,\left( \frac{-1}{3R} \right)\]                 \[=\,\frac{-GM}{12\,R}\,\,\,=\,\frac{6.67\,\,\times \,{{10}^{-11}}\times 6\,\times {{10}^{24}}}{12\times \,6.4\,\times \,{{10}^{6}}}\]                 \[=5.21\times {{10}^{6}}\]                 \[\therefore \] \[{{\upsilon }_{a}}=2.28\times \,{{10}^{3}}\,m\,{{s}^{-1}}=2.28\,\,km\,{{s}^{-1}}\]                 From eqn. (i), \[{{\upsilon }_{p}}=\,3\times \,2.28\]             \[=\,6.48\,km\,{{s}^{-1}}\]                 (iii) We knowm, velocity of a satellite in an orbit of radius r is given by                 \[\upsilon =\frac{\sqrt{GM}}{r}\]                 \[\therefore \] for \[r=6R,\,\upsilon \,=\sqrt{\frac{GM}{6R}}\]                 \[=\,\sqrt{\frac{6.67\,\times 1\,{{0}^{-11}}\times 6\,\times {{10}^{24}}}{6\times \,6.4\times \,{{10}^{6}}}}\]                 \[=3.23\text{ }km\text{ }{{s}^{1}}\]                 Hence, the satellite has to given additional speed by an amount \[=(3.23-2.28)=0.95\,km\,{{s}^{-1}}\] by throwing or firing rockets from the satellite.