Skip to main content

Total energy of an orbiting Satellite and its Binding Energy

Definition: The total energy of a satellite revolving around the planet is the sum of kinetic energy (i.e. due to orbital motion) and potential energy (i.e. due to the gravitational potential energy of the satellite).

Derivation of the total energy of the satellite revolving around the planet:

Let us consider

The mass of the satellite = $m$

The mass of the planet = $M$

The satellite revolving around a planet at distance from the center of the planet = $r$

The radius of planet =$R$

The potential energy of the satellite is

$U=-\frac{G \: M \: m}{r} \qquad(1)$

The kinetic energy of the satellite is

$K=\frac{1}{2}m v_{e}^{2}$

$K=\frac{1}{2} m \left( \sqrt{\frac{G \: M}{r}} \right)^{2} \qquad \left( \because v_{e}^{2}=\sqrt{\frac{G \: M}{r}} \right)$

$K=\frac{1}{2} \left( \frac{G \: M \: m}{r} \right) \qquad(2)$

The total energy of the satellite is

$E= K+U$

Now substitute the value of the kinetic energy and potential energy in the above equation from equation $(1)$ and equation $(2)$

$E= \frac{1}{2} \left( \frac{G \: M \: m}{r} \right)+ \left( -\frac{G \: M \: m}{r} \right)$

$E= -\frac{1}{2} \left( \frac{G \: M \: m}{r} \right)$

$E= -\frac{1}{2} \left( \frac{G \: M \: m}{R+h} \right) \left( \because r=R+h \right)$

This is the expression for the total energy of the satellite revolving around the planet. The above expression shows that the total of a revolving satellite is negative.

The total energy of the orbiting satellite around the Earth:

Put $M=M_{e}$ and $R=R_{e}$ then

$E= -\frac{1}{2} \left( \frac{G \: M_{e} \: m}{R_{e}+h} \right) \left( \because r=R_{e}+h \right)$

$E= -\frac{1}{2} \left( \frac{g \: R_{e}^{2} \: m}{R_{e}+h} \right) \left( \because GM_{e}=g R_{e}^{2} \right)$

If the satellite revolves around near the earth (i.e. $h=0$) then the total energy of the satellite

$E= -\frac{1}{2} \left( \frac{g \: R_{e}^{2} \: m}{R_{e}+0} \right)$

$E= -\frac{1}{2} \left( \frac{g \: R_{e}^{2} \: m}{R_{e}} \right)$

$E= -\frac{1}{2} \left( g \: R_{e} \: m \right)$

Binding Energy of the Satellite:

The minimum amount of energy required to free the revolving satellite around the planet from its orbit is called the binding energy of the revolving satellite.

We know that the total energy of the revolving satellite is

$E= -\frac{1}{2} \left( \frac{G \: M \: m}{R+h} \right)$

The total energy of the satellite at infinity is zero. When an equal positive amount of energy of the total energy of the satellite is given to the satellite, the total energy of the satellite is become zero and the planet leaves its orbit. So this total positive energy is called the binding energy of the satellite. i.e.

$E= +\frac{1}{2} \left( \frac{G \: M \: m}{R+h} \right)$

The binding energy of the orbiting satellite around the earth

$E= +\frac{1}{2} \left( \frac{G \: M_{e} \: m}{R_{e}+h} \right)$

If a satellite revolves around near earth ($h=0$) then binding energy

$E= +\frac{1}{2} \left( \frac{G \: M_{e} \: m}{R_{e}} \right)$
Labels:

Comments

Post a Comment

If you have any doubt. Please let me know.

Popular Posts

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
Post a Comment
click here »

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
Post a Comment
click here »

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
Post a Comment
click here »