Gravitation

Given, radius of earth, $R_e=6400 \mathrm{~km}$

Distance from centre of earth, $r=2000 \mathrm{~km}$

Distance from the surface of earth,

$\begin{aligned} d & =R_e-r \\ & =6400-2000 \\ & =4400 \mathrm{~km}=4.4 \times 10^6 \mathrm{~m} \end{aligned}$

If $g^{\prime}$ be the gravitation field below the surface of earth at a depth $d$, then

$\begin{aligned} g^{\prime} & =g\left(1-\frac{d}{R_{\mathrm{e}}}\right)=9.8\left(1-\frac{4.4 \times 10^6}{6.4 \times 10^6}\right) \quad\left[\because g=9.8 \mathrm{~m} / \mathrm{s}^2\right] \\ & =9.8 \times 0.3125=3.06 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$

From Kepler's law,

$\begin{array}{ll} & \frac{T_A^2}{T_B^2}=\frac{r_A^3}{r_B^3} \Rightarrow \frac{1^2}{8^2}=\frac{\left(10^4\right)^3}{r_B^3} \\ \Rightarrow & r_B^3=64 \times\left(10^4\right)^3 \\ \therefore & r_B=4 \times 10^4 \mathrm{~km} \end{array}$

Speed of satellite $A$,

$v_A=\frac{2 \pi r_A}{T_A}=\frac{2 \pi \times 10^4}{1}$

$=2 \pi \times 10^4 \mathrm{~km} / \mathrm{h}$

Speed of satellite B,

$\begin{aligned} & v_B=\frac{2 \pi r_B}{T_B}\\ & =\frac{2 \pi \times 4 \times 10^4}{8} \\ & =\pi \times 10^4 \mathrm{~km} / \mathrm{h} \end{aligned}$

The speed of $B$ relative to $A$ when they are closed.

$\begin{aligned} v_{B A} & =v_A-v_B \\ & =2 \pi \times 10^4-\pi \times 10^4 \\ & =\pi \times 10^4 \mathrm{~km} / \mathrm{h} \end{aligned}$

As, force $=\frac{G M m}{r^{3 / 2}}=m \omega^2 r$

$\begin{aligned} & \Rightarrow \quad \frac{G M m}{r^{3 / 2}}=\frac{4 \pi^2 m r}{T^2} \quad\left[\because T=\frac{2 \pi}{\omega}\right] \\ & \Rightarrow \quad T^2=\left(\frac{\Delta \pi^2}{G M}\right) \cdot r^{5 / 2} \\ & \Rightarrow \quad T^2 \propto r^{5 / 2} \\ \end{aligned}$

Given, $h=\frac{R_e}{2}$

Acceleration due to gravity at altitude $h$ is given by

$\begin{aligned} g^{\prime} & =\frac{g}{\left(1+\frac{h}{R_e}\right)^2}=\frac{g}{\left(1+\frac{R_e / 2}{R_e}\right)^2} \\ & =\frac{g}{\left(1+\frac{1}{2}\right)^2}=\frac{g}{\left(\frac{3}{2}\right)^2}=\frac{4 g}{9} \quad \text{.... (i)} \end{aligned}$

Weight of the body at the earth's surface,

$w=m g=63 \mathrm{~N}$ .... (ii)

Weight of the body at altitude, $h=\frac{R_e}{2}$

$w^{\prime}=m g^{\prime}=\frac{4}{9} m g$ ..... (iii)

Using Eq. (ii), we get

$w^{\prime}=\frac{4}{9} \times 63=28 \mathrm{~N}$

The orbital velocity of space ship in circular

$v_0=\sqrt{\frac{G M}{r}}$

If space ship is very close to earth surface, r = radius of earth = R

$\begin{aligned} & \therefore \quad v_0=\sqrt{\frac{G M}{R}}=\sqrt{\frac{R^2 g}{R}}=\sqrt{R g} \\ &\left(\text { since } g=\frac{G M}{R^2}\right) \end{aligned}$

$\begin{aligned} \text { Here, } \quad R & =6400 \mathrm{~km}=6.4 \times 10^6 \mathrm{~m} \\ g & =9 \cdot 8 \mathrm{~m} / \mathrm{s}^2 \\ \therefore \quad v_0 & =\sqrt{6 \cdot 4 \times 10^6 \times 9 \cdot 8} \\ & =7.92 \times 10^3 \mathrm{~m} / \mathrm{s} \\ & =7.92 \mathrm{~km} / \mathrm{s} \end{aligned}$

The escape velocity of space ship

$\begin{aligned} v_e & =\sqrt{(2 R g)}=\sqrt{2} v_0=\sqrt{2} \times 7 \cdot 92 \\ & =11 \cdot 20 \mathrm{~km} / \mathrm{s} \end{aligned}$

$\therefore$ Additional velocity required

$=11 \cdot 20-7 \cdot 92=3 \cdot 28 \mathrm{~km} / \mathrm{s}$

Therefore, velocity $3.28 \mathrm{~km} / \mathrm{s}$ must be added to orbital velocity to spaceship to overcome the gravitational pull.

From conservation of energy,

$\frac{1}{2} m v_1^2-\frac{G M_e m}{R}=\frac{1}{2} m v_2^2-\frac{G M_e m}{R+H}$

Here, $v_2=0$ (At maximum height)

$\therefore \quad \frac{1}{2} m v_1-\frac{G M_e m}{R}=-\frac{G M_e m}{R+H}$

Solving, we get

$H=\frac{R}{\frac{2 g R}{v_1^2}-1}=\frac{R}{\frac{v_e^2}{v_1^2}-1}$

Where $v_e$ is escape velocity

Here $v=\frac{1}{3} v_e \Rightarrow \frac{v_e}{v}=3$

$\therefore \quad H=\frac{R}{9-1}=\frac{R}{8}$

If $r_1$ and $r_2$ are radii of orbits of satellites $S_1$ and $S_2$ and $T_1, T_2$ are their respective time periods of revolution, then from Kepler's third law,

$\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3} \Rightarrow r_2=\left(\frac{T_2}{T_1}\right)^{2 / 3} \times r_1$

$\begin{aligned} \text { Here, } \quad T_1 & =1 \mathrm{~h} \\ T_2 & =8 \mathrm{~h} \\ r_1 & =10^4 \mathrm{~km} \\ \Rightarrow \quad\left(\frac{8 h}{1 h}\right)^{2 / 3} \times 10^4 & =4 \times 10^4 \mathrm{~km} \end{aligned}$

The speed of satellite $S_2$ relative to $S_1$

$\begin{aligned} \left|v_2-v_1\right| & =\frac{2 \pi r_2}{T_2}-\frac{2 \pi r_1}{T_1} \\ & =2 \pi\left[\frac{r_2}{T_2}-\frac{r_1}{T_1}\right] \\ & =2 \pi\left[\frac{4 \times 10^4}{8}-\frac{10^4}{1}\right] \mathrm{km} / \mathrm{h} \\ & =2 \pi \times \frac{1}{2} \times 10^4 \\ & =\pi \times 10^4 \mathrm{~km} / \mathrm{h} \end{aligned}$