Expression for Orbital velocity of Satellite and Time Period

Orbital Velocity of Satellite:

When any satellite moves about the planet in a particular orbit then the velocity of the satellite is called the orbital velocity of the satellite.

Expression for Orbital Velocity of Satellite:

Let us consider:

The mass of the satellite = $m$

The mass of planet= $M$

The radius of Planet =$R$

The satellite is moving about the planet at height=$h$

The satellite is moving about the planet with orbital velocity=$v_{\circ}$

The distance from the center of the planet to satellite=$r$

The force of gravitation between the planet and the satellite

$F=G \frac{M \: m}{r^{2}} \qquad(1)$

This force work act as a centripetal force to revolve the satellite around the planet i.e.

$F=\frac{m v_{\circ}^{2}}{r} \qquad(2)$

From the above equation $(1)$ and equation $(2)$, we get

$\frac{m v_{\circ}^{2}}{r} = G \frac{M \: m}{r^{2}}$

$v_{\circ}= \sqrt{\frac{G \: M }{r}}$

Where $r=R+h$ then

$v_{\circ}= \sqrt{\frac{G \: M }{R+h}} \qquad(3)$

This is the equation of the orbital velocity of the satellite.

If any satellite revolves around the earth then the orbital velocity

Orbital Velocity of Satellite moving around the Earth

$v_{\circ}= \sqrt{\frac{G \: M_{e} }{R_{e}+h}}$

$v_{\circ}= \sqrt{\frac{gR_{e}^{2}}{R_{e}+h}} \qquad \left( \because G \: M_{e} = gR_{e}^{2} \right)$

$v_{\circ}= R_{e} \sqrt{\frac{g}{R_{e}+h}} $

This is the equation of the orbital velocity of a satellite revolving around the earth.

If the satellite is orbiting very close to the surface of the earth ( i.e $h=0$) then the orbital velocity

$v_{\circ}= R_{e} \sqrt{\frac{g}{R_{e}+0}} $

$v_{\circ}= \sqrt{g R_{e}} $

Now subtitute the value of radius of earth (i.e $R_{e}=6.4 \times 10^{6} \: m$) and gravitational accelertaion ($g=9.8 \:m/sec^{2}$) then orbital velocity

$v_{\circ}=7.92 \: Km/sec$

The time period of Revolving Satellite:

The time taken by satellite to complete on revolution around the planet is called the time period of the satellite.

Let us consider the time period of the revolving satellite is $T$ Then

$T= \frac{Distance \: covered \: by \: Satellite \: in \: one \: revolution}{Orbital \: Velocity}$

$T= \frac{2 \pi r}{v_{\circ}}$

$T= \frac{2 \pi \left( R+h \right)}{v_{\circ}}$

Now subtitute the value of orbital velocity $v_{\circ}$ from equation $(3)$ in above equation then

$T= \frac{2 \pi \left( R+h \right)}{\sqrt{\frac{G \: M }{R+h}}}$

$T= 2 \pi \sqrt{ \frac{ \left( R+h \right)^{3}}{G M}}$

This is the equation of the time period of the revolution of satellites.

If a satellite revolves around the earth then the time period

$T= 2 \pi \sqrt{ \frac{ \left( R_{e}+h \right)^{3}}{G M_{e}}}$

$T= 2 \pi \sqrt{ \frac{ \left( R_{e}+h \right)^{3}}{g R_{e}^{2}}} \qquad \left( \because G \: M_{e} = gR_{e}^{2} \right)$

This is the equation of the time period of revolution of satellite revolving around the earth.

If the satellite revolves very nearly around the earth (i.e $h=0$) then the time period of the satellite from the above equation

$T= 2 \pi \sqrt{ \frac{ \left( R_{e}+0 \right)^{3}}{g R_{e}^{2}}} $

$T= 2 \pi \sqrt{ \frac{ R_{e}}{g}} $

$T= 84.6 \: min $

Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig

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Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

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Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ $\frac{d^{2} \psi(x)}{d x

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