Mean solar day is the time interval between two successive noon when sun passes through zenith point (meridian).
Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian).
By drawing appropriate diagram showing the earth's spin and orbital motion, show that mean solar day is 4 min longer than the sidereal day. In other words, distant stars would rise 4 min early every successive day.
Consider the diagram below, the earth moves from the point P to Q in one solar day.
Every day the earth advances in the orbit by approximately $$1^{\circ}$$. Then, it will have to rotate by $$361^{\circ}$$ (which we define as 1 day) to have the sun at zenith point again.
$$\because 361^{\circ}$$ corresponds to 24 h.
$$\therefore 1^{\circ}$$ corresponds to $$\frac{24}{361} \times 1=0.066 \mathrm{~h}=3.99 \mathrm{~min} \approx 4 \mathrm{~min}$$
Hence, distant stars would rise 4 min early every successive day.
Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the mid-point of the line joining their centres be in stable equilibrium or unstable equilibrium? Give reason for your answer.
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.
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).
$$\begin{aligned} & \text { Potential energy of the object at the surface of the earth }=-\frac{G M m}{R} \\ & \text { PE of the object at a height equal to the radius of the earth }=-\frac{G M m}{2 R} \\ & \begin{aligned} \therefore \quad \text { Gain in PE of the object } & =\frac{-G M m}{2 R}-\left(-\frac{G M m}{R}\right) \\ & =\frac{-G M m+2 G M m}{2 R}=+\frac{G M m}{2 R} \\ & =\frac{g R^2 \times m}{2 R}=\frac{1}{2} m g R \quad (\because GM=gR^2) \end{aligned} \end{aligned}$$