An object of mass $m$ is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance $R$ to $2 R$ from the centre of the earth. What is the gain in its potential energy?
Consider the diagram where an object of mass m is raised from the surface of the earth to a distance (height) equal to the radius of the earth (R).
$$\begin{aligned} & \text { Potential energy of the object at the surface of the earth }=-\frac{G M m}{R} \\ & \text { PE of the object at a height equal to the radius of the earth }=-\frac{G M m}{2 R} \\ & \begin{aligned} \therefore \quad \text { Gain in PE of the object } & =\frac{-G M m}{2 R}-\left(-\frac{G M m}{R}\right) \\ & =\frac{-G M m+2 G M m}{2 R}=+\frac{G M m}{2 R} \\ & =\frac{g R^2 \times m}{2 R}=\frac{1}{2} m g R \quad (\because GM=gR^2) \end{aligned} \end{aligned}$$
A mass $m$ is placed at $P$ a distance $h$ along the normal through the centre 0 of a thin circular ring of mass $M$ and radius $r$ (figure).
If the mass is moved further away such that $O P$ becomes $2 h$, by what factor the force of gravitation will decrease, if $h=r$ ?
Consider the diagram, in which a system consisting of a ring and a point mass is shown.
Gravitational force acting on an object of mass $m$, placed at point $P$ at a distance $h$ along the normal through the centre of a circular ring of mass $M$ and radius $r$ is given by
$$F=\frac{\text { GMmh }}{\left(r^2+h^2\right)^{3 / 2}}\quad$$ (along PO) ...(i)
$$\begin{aligned} &\text { When mass is displaced upto distance } 2 h \text {, then }\\ &\begin{aligned} F^{\prime} & =\frac{G M m \times 2 h}{\left[r^2+(2 h)^2\right]^{3 / 2}} \quad [\because h=2r]\\ & =\frac{2 G M m h}{\left(r^2+4 h^2\right)^{3 / 2}} \quad \text{... (ii)} \end{aligned} \end{aligned}$$
$\begin{aligned} & \text { When } h=r \text {, then from Eq. (i) } \\ & \qquad \begin{array}{ll}F & =\frac{G M m \times r}{\left(r^2+r^2\right)^{3 / 2}} \Rightarrow F=\frac{G M m}{2 \sqrt{2 r^2}} \\ \text { and } & F^{\prime}=\frac{2 G M m r}{\left(r^2+4 r^2\right)^{3 / 2}}=\frac{2 G M m}{5 \sqrt{5 r^2}} \quad \text { [From Eq. (ii) substituting } h=r \text { ] } \\ \therefore & \frac{F^{\prime}}{F}=\frac{4 \sqrt{2}}{5 \sqrt{5}} \\ \Rightarrow & F^{\prime}=\frac{4 \sqrt{2}}{5 \sqrt{5}} F\end{array}\end{aligned}$
A star like the sun has several bodies moving around it at different distances. Consider that all of them are moving in circular orbits. Let $r$ be the distance of the body from the centre of the star and let its linear velocity be $v$, angular velocity $\omega$, kinetic energy $K$, gravitational potential energy $U$, total energy $E$ and angular momentum $l$. As the radius $r$ of the orbit increases, determine which of the above quantities increase and which ones decrease.
$$\begin{array}{ll} \text { Linear velocity of the body } & v=\sqrt{\frac{G M}{r}} \\ \Rightarrow & v \propto \frac{1}{\sqrt{r}} \end{array}$$
Therefore, when $r$ increases, $v$ decreases.
Angular velocity of the body $\quad \omega=\frac{2 \pi}{T}$
According to Kepler's law of period,
$$T^2 \propto r^3 \Rightarrow T=k r^{3 / 2}$$
where $k$ is a constant
$$\therefore \quad \omega=\frac{2 \pi}{k r^{3 / 2}} \Rightarrow \omega \propto \frac{1}{r^{3 / 2}} \quad\left(\because \omega=\frac{2 \pi}{T}\right)$$
Therefore, when $r$ increases, $\omega$ decreases.
Kinetic energy of the body
$K=\frac{1}{2} m v^2=\frac{1}{2} m \times \frac{G M}{r}=\frac{G M m}{2 r}$
$\therefore \quad K \propto \frac{1}{r}$
Therefore, when $r$ increases, KE decreases. Gravitational potential energy of the body,
$$U=-\frac{G M m}{r} \Rightarrow U \propto-\frac{1}{r}$$
Therefore, when $r$ increases, PE becomes less negative i.e., increases.
Total energy of the body
$$E=\mathrm{KE}+\mathrm{PE}=\frac{\mathrm{GMm}}{2 r}+\left(-\frac{\mathrm{GMm}}{r}\right)=-\frac{\mathrm{GMm}}{2 r}$$
Therefore, when $r$ increases, total energy becomes less negative, i.e., increases.
Angular momentum of the body $L=m v r=m r \sqrt{\frac{G M}{r}}=m \sqrt{G M r}$ $\therefore \quad L \propto \sqrt{r}$
Therefore, when $r$ increases, angular momentum $L$ increases.
Six point masses of mass $m$ each are at the vertices of a regular hexagon of side $l$. Calculate the force on any of the masses.
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 $A D=F_1+F_2+F_3=\frac{G m^2}{l^2}+\frac{G m^2}{\sqrt{3} l^2}+\frac{G m^2}{4 l^2}=\frac{G m^2}{l^2}\left(1+\frac{1}{\sqrt{3}}+\frac{1}{4}\right)$.
A satellite is to be placed in equatorial geostationary orbit around the earth for communication.
(a) Calculate height of such a satellite.
(b) Find out the minimum number of satellites that are needed to cover entire earth, so that atleast one satellite is visible from any point on the equator.
$$\left[M=6 \times 10^{24} \mathrm{~kg}, R=6400 \mathrm{~km}, T=24 \mathrm{~h}, G=6.67 \times 10^{-11} \mathrm{SI} \text { unit }\right]$$
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.