(4)

Let the planet is moving above the earth surface at $R$ distance.

$T=\frac{2\pi R}{v}$                                 ...(i)

As, the centripetal force of satellite is due to gravitational force of earth.

$\frac{GMm}{R^2}=\frac{m{v}^2}{R}\Rightarrow \frac{GM}{R}=v^2 \text{or} v=\sqrt{\frac{GM}{R}}$

Using (i), $T=\frac{2\pi R\sqrt{R}}{\sqrt{GM}}$

$T^2=\frac{4{\pi }^2{R}^3}{G(d\times \frac{4}{3}\pi R^3)}=\frac{3\pi }{Gd}. \text{Hence,} \frac{3\pi }{Gd}=T^2$

(1)

Time period of Geostationary satellite is,

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

$\therefore$    $\frac{T_1^2}{T_2^2}=\frac{a_1^3}{a_2^3}\Rightarrow \frac{{{\displaystyle (}{\displaystyle 24}{\displaystyle )}}^2}{T_2^2}=\frac{{{\displaystyle (}{\displaystyle 7}{\displaystyle R}{\displaystyle {}_E}{\displaystyle )}}^3}{{{\displaystyle (}{\displaystyle 3}{\displaystyle .}{\displaystyle 5}{\displaystyle R}{\displaystyle {}_E}{\displaystyle )}}^3}$

$\Rightarrow T_2^2=\frac{{{\displaystyle (}{\displaystyle 24}{\displaystyle )}}^2\times {(3.5)}^3}{{\left(7\right)}^3}\Rightarrow T_2=\frac{\sqrt{{{\displaystyle (}{\displaystyle 24}{\displaystyle )}}^2}}{\sqrt{8}}=6\sqrt{2}$ $h.$

(3)

The orbital speed of the satellite is, $v_0=R\sqrt{\frac{g}{(R+h)}}$

where $R$ is the earth’s radius, $g$ is the acceleration due to gravity on earth’s surface, and $h$ is the height above the surface of Earth.

Here, $R=6.38\times {10}^6\text{m}, g=9.8{\text{ms}}^{-2}, h=0.25\times {10}^6\text{m}$

$\therefore$   $v_0=$$(6.38\times {10}^6\text{m})$$\sqrt{\frac{(9.8 {\text{ms}}^{-2})}{(6.38\times {10}^6\text{m}+0.25\times {10}^6\text{m})}}$

              $=7.76\times {10}^3{\text{ms}}^{-1}=7.76\text{km}$ $s^{-1}$

(2)

The gravitational force on the satellite S acts towards the centre of the earth, so the acceleration of the satellite S is always directed towards the centre of the earth.