A satellite is launched and attains a velocity of $30400 \mathrm{km}/\mathrm{hr}$ relative to the centre of the earth at a height of 320 km from the earth’s surface. It has been guided into a path that is parallel to the
earth’s surface at burnout. Choose the correct options.

Given that the velocity of the satellite is

$\mathrm{v}=30400\times \frac{5}{18}\mathrm{m}/\mathrm{s}=8444.44\mathrm{m}/\mathrm{s}$

Let us calculate the orbital velocity $v_0$ for the satellite to move along a circular orbit of height 320 km.

$\because v_0=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}=\sqrt{\frac{{\mathrm{gR}}^2}{\mathrm{r}}}$

$=\sqrt{\frac{9.8\times {(6.4\times {10}^6)}^2}{672\times {10}^4}}=7728.73\mathrm{m}/\mathrm{s}$

Also, the escape velocity for the satellite in this position is given by

$v_{\mathrm{e}}=\sqrt{2}{v}_0=10930.08\mathrm{m}/\mathrm{s}$

Evidently, $7728.73<8444.44<10930.08$

$\text{i.e.,}{v}_0<v<v_e$

So, the satellite moves along an elliptical orbit, with the position of launch as the perigee. (figure).

(b) $\mathrm{Let} {\mathrm{r}}_{\max }$  be the maximum distance of the satellite from the earth’s centre, and $v$ ’ the corresponding velocity.

From, conservation law of angular momentum and energy between perigee and apogee, we have $8444.44(672\times {10}^4)=v^{\mathrm{\prime }}\left({\mathrm{r}}_{max}\right)$

and $\frac{1}{2}(8444.44{)}^2-\frac{{\mathrm{gR}}^2}{672\times {10}^4}=\frac{1}{2}{v}^{\text{'}2}-\frac{{\mathrm{gR}}^2}{{\mathrm{r}}_{max}}$

Eliminating $v$ ’ from the above two equations, we have

$3565.43\times {10}^4\left[1-{\left(\frac{672}{{\mathrm{r}}^{\mathrm{\prime }}}\right)}^2\right]$

$=\frac{9.8\times {(640\times {10}^4)}^2}{{10}^4}[\frac{1}{672}-\frac{1}{\mathrm{r}},]$

where $r^{\mathrm{\prime }}=\frac{r_{max}}{{10}^4}\mathrm{m}$

$\Rightarrow 3565.43\times {10}^4\left(\frac{{\mathrm{r}}^{\text{'}2}-(672{)}^2}{{\mathrm{r}}^{\text{'}2}}\right)$

$=4014.08\times {10}^7\left[\frac{{\mathrm{r}}^{\mathrm{\prime }}-672}{672{\mathrm{r}}^{\mathrm{\prime }}}\right]$

$\Rightarrow r^{\mathrm{\prime }}+672=1.675r^{\mathrm{\prime }}\Rightarrow r^{\mathrm{\prime }}=\frac{672}{0.675}=995\mathrm{m}$

$\therefore {\mathrm{r}}_{max}=995\times {10}^4\mathrm{m}$

The maximum height of the satellite

$=r_{\text{max}}-\mathrm{R}=9950-6400=3550\mathrm{km}$

(c) The semi-major axis of the elliptical path will be

$\mathrm{a}=\frac{{\mathrm{r}}_{max}+{\mathrm{r}}_{min}}{2}=\left(\frac{995+672}{2}\right)\times {10}^4\mathrm{m}$

$=833.5\times {10}^4\mathrm{m}$

From Kepler’s III rd law, of planetary motion,

$\mathrm{T}=2\pi \sqrt{\frac{{\mathrm{a}}^3}{\mathrm{GM}}}\Rightarrow \mathrm{T}=2\pi \sqrt{\frac{{\mathrm{a}}^3}{{\mathrm{gR}}^2}}$

$\Rightarrow \mathrm{T}=2\pi \sqrt{\frac{{(833.5\times {10}^4)}^3}{9.8\times {(64\times {10}^5)}^2}}$

$=2\times 3.14\times 1201.06$

$= 7546.5 sec = 2.09hrs$

(d) The escape velocity is given by

$v_{\mathrm{e}}=\sqrt{2}{v}_0=10930.09\mathrm{m}/\mathrm{s}$