Let the mass and radius of each identical heavy sphere be $$M$$ and $$R$$ respectively. An object of mass $$m$$ be placed at the mid-point $$P$$ of the line joining their centres.

Force acting on the object placed at the mid-point,

$$F_1=F_2=\frac{G M m}{(5 R)^2}$$

The direction of forces are opposite, therefore net force acting on the object is zero.

To check the stability of the equilibrium, we displace the object through a small distance $$x$$ towards sphere $$A$$.

Now, force acting towards sphere $$A, F_1^{\prime}=\frac{G M m}{(5 R-x)^2}$$

Force acting towards sphere $$B, F_2^{\prime}=\frac{G M m}{(5 R+x)^2}$$

As $$F_1^{\prime}>F_2{ }^{\prime}$$, therefore a resultant force $$\left(F_1{ }^{\prime}-F_2{ }^{\prime}\right)$$ acts on the object towards sphere $$A$$, therefore object start to move towards sphere $$A$$ and hence equilibrium is unstable.

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.

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.

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).

$$\begin{aligned} & \text { Potential energy of the object at the surface of the earth }=-\frac{G M m}{R} \\ & \text { PE of the object at a height equal to the radius of the earth }=-\frac{G M m}{2 R} \\ & \begin{aligned} \therefore \quad \text { Gain in PE of the object } & =\frac{-G M m}{2 R}-\left(-\frac{G M m}{R}\right) \\ & =\frac{-G M m+2 G M m}{2 R}=+\frac{G M m}{2 R} \\ & =\frac{g R^2 \times m}{2 R}=\frac{1}{2} m g R \quad (\because GM=gR^2) \end{aligned} \end{aligned}$$

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)

$$\begin{aligned} &\text { When mass is displaced upto distance } 2 h \text {, then }\\ &\begin{aligned} F^{\prime} & =\frac{G M m \times 2 h}{\left[r^2+(2 h)^2\right]^{3 / 2}} \quad [\because h=2r]\\ & =\frac{2 G M m h}{\left(r^2+4 h^2\right)^{3 / 2}} \quad \text{... (ii)} \end{aligned} \end{aligned}$$

$\begin{aligned} & \text { When } h=r \text {, then from Eq. (i) } \\ & \qquad \begin{array}{ll}F & =\frac{G M m \times r}{\left(r^2+r^2\right)^{3 / 2}} \Rightarrow F=\frac{G M m}{2 \sqrt{2 r^2}} \\ \text { and } & F^{\prime}=\frac{2 G M m r}{\left(r^2+4 r^2\right)^{3 / 2}}=\frac{2 G M m}{5 \sqrt{5 r^2}} \quad \text { [From Eq. (ii) substituting } h=r \text { ] } \\ \therefore & \frac{F^{\prime}}{F}=\frac{4 \sqrt{2}}{5 \sqrt{5}} \\ \Rightarrow & F^{\prime}=\frac{4 \sqrt{2}}{5 \sqrt{5}} F\end{array}\end{aligned}$