What is the angle between the equatorial plane and the orbital plane of
(a) polar satellite?
(b) geostationary satellite?
Consider the diagram where plane of geostationary and polar satellite are shown.
Clearly
(a) Angle between the equatorial plane and orbital plane of a polar satellite is $$90^{\circ}$$.
(b) Angle between equatorial plane and orbital plane of a geostationary satellite is $$0^{\circ}$$.
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.