91Ó°ÊÓ

Farthest point distance (earth's surface and satellite)$=360\text{km}$

Closest point distance (earth's surface and satellite)$=180\text{km}$

Radius of earth$=6.37×{10}^6\text{m}$

As the distance between the farthest point and the closest point is given, find the distance between the satellite and the earth’s center. If the average is taken of both distances, find the semi-major axis.

From the formula of apogee and perigee distance, find the eccentricity of the orbit.

Formulae are as follows:

$\mathbf{\text{apogee}}\mathbf{=}{\mathbf{R}}_{\mathbf{a}}\mathbf{=}\mathbf{a}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{e}\mathbf{)}$

$\mathbf{\text{perigee}}\mathbf{=}{\mathit{R}}_{\mathbf{P}}\mathbf{=}\mathit{a}\mathbf{(}\mathbf{1}\mathbf{−}\mathbf{e}\mathbf{)}$

Farthest point distance (earth’s surface and satellite)$=360\text{ k³¾}$

Farthest point distance (earth's center and satellite)

$\begin{array}{c}{\text{R}}_{\text{a}}=360\text{ k³¾+Radiusofearth}\\ =\text{360}×{\text{10}}^{\text{3}}\text{″¾}+\text{6.37}×{\text{10}}^{\text{6}}\text{″¾}\\ =6.73×{\text{10}}^{\text{6}}\text{″¾}\end{array}$

Closest point distance (earth's surface and satellite)$\text{=180km}$

Closest point distance (earth's center and satellite)

$\begin{array}{c}{\text{R}}_{\mathrm{P}}=\text{180km}+\text{Radiusofearth}\\ =\text{180}×{\text{10}}^{\text{3}}\text{″¾}+6.37×{10}^6\text{ }\mathrm{m}\\ =6.55×{10}^{\text{6}}\text{m}\end{array}$

From diagram,

$\text{Semimajoraxis}=\mathrm{a}=\frac{{\mathrm{R}}_{\mathrm{a}}+{\mathrm{R}}_{\mathrm{p}}}{2}$

$\begin{array}{c}\text{Semimajoraxis}=\mathrm{a}=\frac{(6.73×{10}^6\text{ }\mathrm{m})+(6.55×{10}^6\text{ }\mathrm{m})}{2}\\ =6.64×{10}^{\text{6}}\text{m}\end{array}$

Hence, the semi-major axis of the satellite is$6.64×{10}^6\text{m}$ .

Now,

${\mathrm{R}}_{\mathrm{a}}=\mathrm{a}(1+\mathrm{e})$

${\mathrm{R}}_{\mathrm{P}}=\mathrm{a}(1−\mathrm{e})$

Adding the above two equations,

$\begin{array}{c}{\mathrm{R}}_{\mathrm{a}}+{\mathrm{R}}_{\mathrm{p}}=\left(\mathrm{a}(1+\mathrm{e})\right)+\left(\mathrm{a}(1−\mathrm{e})\right)\\ =\mathrm{a}+\mathrm{ae}+\mathrm{a}−\mathrm{ae}\\ =2\mathrm{a}\end{array}$

Now,

${\mathrm{R}}_{\mathrm{a}}=\mathrm{a}(1+\mathrm{e})$

${\mathrm{R}}_{\mathrm{P}}=\mathrm{a}(1−\mathrm{e})$

Subtracting the above two equations,

$\begin{array}{c}{\mathrm{R}}_{\mathrm{a}}−{\mathrm{R}}_{\mathrm{p}}=\left(\mathrm{a}(1+\mathrm{e})\right)−\left(\mathrm{a}(1−\mathrm{e})\right)\\ =\mathrm{a}+\mathrm{ae}−\mathrm{a}+\mathrm{ae}\\ =2\mathrm{ae}\end{array}$

Dividing the two equations $({\mathrm{R}}_{\mathrm{a}}−{\mathrm{R}}_{\mathrm{p}})\text{ }\mathrm{and}\text{ }({\mathrm{R}}_{\mathrm{a}}+{\mathrm{R}}_{\mathrm{p}}),$

$\begin{array}{c}\frac{({\mathrm{R}}_{\mathrm{a}}−{\mathrm{R}}_{\mathrm{p}})}{({\mathrm{R}}_{\mathrm{a}}+{\mathrm{R}}_{\mathrm{p}})}=\frac{2\mathrm{ae}}{2\mathrm{a}}\\ \mathrm{e}=\frac{({\mathrm{R}}_{\mathrm{a}}−{\mathrm{R}}_{\mathrm{p}})}{({\mathrm{R}}_{\mathrm{a}}+{\mathrm{R}}_{\mathrm{p}})}\\ \mathrm{e}=\frac{((6.73×{10}^6\text{ }\mathrm{m})−(6.55×{10}^6\text{ }\mathrm{m}))}{((6.73×{10}^6\text{ }\mathrm{m})+(6.55×{10}^6\text{ }\mathrm{m}))}\\ \mathrm{e}=0.0136\end{array}$

Hence, theeccentricity of the orbit of the satellite is$0.0136$.

Therefore, using the farthest and closest distance of the satellite from the center of the earth, the semi-major axis, and eccentricity of the orbiting satellite can be found.