Show the nature of the following graph for a satellite orbitting the earth.
(a) KE versus orbital radius $$R$$
(b) PE versus orbital radius $$R$$
(c) TE versus orbital radius $$R$$
Consider the diagram, where a satellite of mass $m$, moving around the earth in a circular orbit of radius $R$.
Orbital speed of the satellite orbitting the earth is given by $v_0=\sqrt{\frac{G M}{R}}$ where, $M$ and $R$ are the mass and radius of the earth.
(a) $\therefore$ KE of a satellite of mass $m, E_K=\frac{1}{2} m v_o^2=\frac{1}{2} m \times \frac{G M}{R}$
$\therefore \quad E_K \propto \frac{1}{R}$
It means the KE decreases exponentially with radius.
The graph for KE versus orbital radius $R$ is shown in figure.
(b) Potential energy of a satellite $E_p=-\frac{G M m}{R}$
$$E_P \propto-\frac{1}{R}$$
The graph for PE versus orbital radius $R$ is shown in figure.
(c) Total energy of the satellite $E=E_K+E_P=\frac{G M m}{2 R}-\frac{G M m}{R}$
$$=-\frac{G M m}{2 R}$$
The graph for total energy versus orbital radius $R$ is shown in the figure.
Shown are several curves [fig. (a), (b), (c), (d), (e), (f)]. Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).
The trajectory of a particle under gravitational force of the earth will be a conic section (for motion outside the earth) with the centre of the earth as a focus. Ony (c) meets this requirement.
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.