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.
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