Consider the diagram below, in which six point masses are placed at six vertices $A, B, C, D, E$ and $F$

$$\begin{aligned} A C & =A G+G C=2 A G \\ & =2 l \cos 30^{\circ}=2 l \sqrt{3} / 2 \\ & =\sqrt{3} l=A E \\ A D & =A H+H J+J D \\ & =l \sin 30^{\circ}+l+l \sin 30^{\circ}=2 l \end{aligned}$$

$$ \text { Force on mass } m \text { at } A \text { due to mass } m \text { at } B \text { is, } f_1=\frac{G m m}{l^2} \text { along } A B \text {. } $$

$$\begin{aligned} &\text { Force on mass } m \text { at } A \text { due to mass } m \text { at } C \text { is, } f_2=\frac{G m \times m}{(\sqrt{3} l)^2}=\frac{G m^2}{3 l^2} \text { along } A C \text {. }\quad [\because A C=\sqrt{3 l}] \end{aligned}$$

Force on mass $m$ at $A$ due to mass $m$ at $D$ is, $f_3=\frac{G m \times m}{(2 l)^2}=\frac{G m^2}{4 l^2}$ along $A D . [\because A D=2 l]$

Force on mass $m$ at $A$ due to mass $m$ at $E$ is, $f_4=\frac{G m \times m}{(\sqrt{3} l)^2}=\frac{G m^2}{3 l^2}$ along $A E$.

Force on mass $m$ at $A$ due to mass $m$ at $F$ is, $f_5=\frac{G m \times m}{l^2}=\frac{G m^2}{l^2}$ along $A F$.

Resultant force due to $f_1$ and $f_5$ is, $F_1=\sqrt{f_1^2+f_5^2+2 f_1 f_5 \cos 120^{\circ}}=\frac{G m^2}{l^2}$ along $A D$. \quad \left[\because \text { Angle between } f_1 \text { and } f_5=120^{\circ}\right]

Resultant force due to $f_2$ and $f_4$ is, $F_2=\sqrt{f_2^2+f_4^2+2 f_2 f_4 \cos 60^{\circ}}$

$$=\frac{\sqrt{3} G m^2}{3 l^2}=\frac{G m^2}{\sqrt{3} l^2} \text { along } A D$$

So, net force along $A D=F_1+F_2+F_3=\frac{G m^2}{l^2}+\frac{G m^2}{\sqrt{3} l^2}+\frac{G m^2}{4 l^2}=\frac{G m^2}{l^2}\left(1+\frac{1}{\sqrt{3}}+\frac{1}{4}\right)$.

Consider the adjacent diagram

$$\begin{aligned} \text{Given,}\quad\text { mass of the earth } M & =6 \times 10^{24} \mathrm{~kg} \\ \text { Radius of the earth, } R & =6400 \mathrm{~km}=6.4 \times 10^6 \mathrm{~m} \\ \text { Time period } T & =24 \mathrm{~h} \\ & =24 \times 60 \times 60=86400 \mathrm{~s} \\ G & =6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2 \end{aligned}$$

$$\begin{aligned} &\text { (a) Time period }\\ &T=2 \pi \sqrt{\frac{(R+h)^3}{\mathrm{GM}}} \quad\left[\because v_0=\sqrt{\frac{\mathrm{G} M}{R+h}} \text { and } T=\frac{2 \pi(R+h)}{v_0}\right] \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad T^2 & =4 \pi^2 \frac{(R+h)^3}{G M} \Rightarrow(R+h)^3=\frac{T^2 G M}{4 \pi^2} \\ \Rightarrow \quad R+h & =\left(\frac{T^2 G M}{4 \pi^2}\right)^{1 / 3} \Rightarrow h=\left(\frac{T^2 G M}{4 \pi^2}\right)^{1 / 3}-R \\ \Rightarrow \quad h & =\left[\frac{(24 \times 60 \times 60)^2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{4 \times(3.14)^2}\right]^{1 / 3}-6.4 \times 10^6 \\ & =4.23 \times 10^7-6.4 \times 10^6 \\ & =(42.3-6.4) \times 10^6 \\ & =35.9 \times 10^6 \mathrm{~m} \\ & =3.59 \times 10^7 \mathrm{~m} \end{aligned}$$

$$\text { (b) If satellite is at height } h \text { from the earth's surface, then according to the diagram. }$$

$$\begin{aligned} \cos \theta & =\frac{R}{R+h}=\frac{1}{\left(1+\frac{h}{R}\right)}=\frac{1}{\left(1+\frac{3.59 \times 10^7}{6.4 \times 10^6}\right)} \\ & =\frac{1}{1+5.61}=\frac{1}{6.61}=0.1513=\cos 81^{\prime} 18^{\prime} \\ \theta & =81^{\prime} 18^{\prime} \\ \therefore \quad 2 \theta & =2 \times\left(81^{\prime} 18^{\prime}\right)=162^{\circ} 36^{\prime} \end{aligned}$$

If $n$ is the number of satellites needed to cover entire the earth, then

$$n=\frac{360^{\circ}}{2 \theta}=\frac{360^{\circ}}{162^{\circ} 36^{\prime}}=2.31$$

$\therefore \quad$ Minimum 3 satellites are required to cover entire the earth.

Consider the diagram. Let $m$ be the mass of the earth, $v_p, v_a$ be the velocity of the earth at perigee and apogee respectively. Similarly, $\omega_p$ and $\omega_a$ are corresponding angular velocities.

Angular momentum and areal velocity are constant as the earth orbits the sun.

At perigee, $r_p^2 \omega_p=r_a^2 \omega_a$ at apogee $$\quad$$ .... (i)

If $a$ is the semi-major axis of the earth's orbit, then $r_p=a(1-e)$ and $r_a=a(1+e)\quad \text{... (ii)}$

$$ \begin{array}{ll} \therefore \frac{\omega_p}{\omega_a}=\left(\frac{1+e}{1-e}\right)^2, e=0.0167 & \text { [from Eqs. (i) and (ii)] } \\ \therefore\frac{\omega_p}{\omega_a}=1.0691 & \end{array}$$

Let $\omega$ be angular speed which is geometric mean of $\omega_p$ and $\omega_a$ and corresponds to mean solar day,

$$\begin{array}{ll} \therefore & \left(\frac{\omega_p}{\omega}\right)\left(\frac{\omega}{\omega_a}\right)=1.0691 \\ \therefore & \frac{\omega_p}{\omega}=\frac{\omega}{\omega_a}=1.034 \end{array}$$

If $\omega$ corresponds to $1^{\circ}$ per day (mean angular speed), then $\omega_p=1.034^{\circ}$ per day and $\omega_{\mathrm{a}}=0.967^{\circ}$ per day. Since, $361^{\circ}=24$, mean solar day, we get $361.034^{\circ}$ which corresponds to $24 \mathrm{~h}, 8.14^{\prime \prime}$ ( $8.1^{\prime \prime}$ longer) and $360.967^{\circ}$ corresponds to 23 h 59 min $52^{\prime \prime}$ ( $7.9^{\prime \prime}$ smaller).

This does not explain the actual variation of the length of the day during the year.

Given,

$r_p=$ radius of perihelion $=2 R$

$$r_a=\text { radius of apnelion }=6 R$$

Hence, we can write

$$ \begin{aligned} & r_a=a(1+e)=6 R \quad \text{... (i)}\\ & r_p=a(1-e)=2 R \quad \text{... (ii)} \end{aligned}$$

Solving Eqs. (i) and (ii), we get eccentricity, $e=\frac{1}{2}$

By conservation of angular momentum, angular momentum at perigee $=$ angular momentum at apogee

$$\begin{array}{ll} \therefore & m v_p r_p=m v_a r_a \\ \therefore & \frac{v_a}{v_p}=\frac{1}{3} \end{array}$$

where $m$ is mass of the satellite. Applying conservation of energy, energy at perigee $=$ energy at apogee

$$\frac{1}{2} m v_p^2-\frac{G M m}{r_p}=\frac{1}{2} m v_a^2-\frac{G M m}{r_a}$$

where $M$ is the mass of the earth.

$$\begin{aligned} &\begin{aligned} \therefore\quad v_p^2\left(1-\frac{1}{9}\right) & =-2 G M\left(\frac{1}{r_a}-\frac{1}{r_p}\right)=2 G M\left(\frac{1}{r_p}-\frac{1}{r_a}\right) \quad\left(\text { By putting } v_a=\frac{v_p}{3}\right) \\ v_p & =\frac{\left[2 G M\left(\frac{1}{r_p}-\frac{1}{r_a}\right)\right]^{1 / 2}}{\left[1-\left(V_a / V_p\right)^2\right]^{1 / 2}}=\left[\frac{\frac{2 G M}{R}\left(\frac{1}{2}-\frac{1}{6}\right)}{\left(1-\frac{1}{9}\right)}\right]^{1 / 2} \\ & =\left(\frac{2 / 3}{8 / 9} \frac{G M}{R}\right)^{1 / 2}=\sqrt{\frac{3}{4}} \frac{G M}{R}=6.85 \mathrm{~km} / \mathrm{s} \\ v_p & =6.85 \mathrm{~km} / \mathrm{s}, v_a=2.28 \mathrm{~km} / \mathrm{s} \end{aligned} \end{aligned}$$

For circular orbit of radius $r$,

$$\begin{aligned} v_c & =\text { orbital velocity }=\sqrt{\frac{G M}{r}} \\ \text{For}\quad r & =6 R, v_c=\sqrt{\frac{G M}{6 R}}=3.23 \mathrm{~km} / \mathrm{s} \end{aligned}$$

Hence, to transfer to a circular orbit at apogee, we have to boost the velocity by $\Delta=(3.23-2.28)=0.95 \mathrm{~km} / \mathrm{s}$. This can be done by suitably firing rockets from the satellite.