 # Artificial Satellites – an overview

Artificial satellites are  man – made object that orbit the earth .

## Introduction

Let us consider a several kilometers high tower on the surface of earth as shown in Fig. 1

A stone is thrown from the top of the tower with a horizontal direction with certain velocity, it may hit the surface of the earth at point A after following a parabolic path. Now the stone is thrown with a greater horizontal velocity , the stone will meet the earth at the point say B, a little away from the foot of the tower.

If we keep on increasing the velocity of the projection , it will hit the surface of the earth always at a point a little further away, till at a certain horizontal velocity of the projection of the stone, it will be in a state of free fall under gravity i.e., it will be attempting to fall to the earth but missing it all the times. The result will be that the stone will follow a circular path around the earth. As the stone is moving around the earth, it may be called an artificial satellite of the earth .

In Practical life , an artificial satellite is launched into space with the help of at least a two stage rocket . The satellite is placed at the tip of the rocket. The first stage of the rocket takes the satellite vertically up against the gravity to the desired height of its orbit above the surface of the earth. Then the first stage of the rocket detaches, and the second stage tilts it through 90°. The second stage is used to impart required horizontal velocity to the rocket to make the satellite revolve in a circular orbit around the earth. This horizontal velocity is called critical orbital velocity or simply orbital velocity and depends on the height of the orbit. In fact, the magnitude of the horizontal velocity imparted to the satellite determines the successful launching of the satellite as explained below :

• In case the horizontal velocity imparted is less than the critical orbital velocity, the satellite will follow a parabolic path and fall to the surface of the earth .
• In case the horizontal velocity imparted is precisely equal to the critical orbital velocity, the satellite will move in the desired circular orbit around the earth .
• In case the horizontal velocity imparted is greater than the critical orbital velocity but less than the escape velocity , the satellite will move in an elliptical orbit .
• In case the horizontal velocity imparted is greater than the escape velocity, the satellite will leave the gravitational field of the earth and escape into the space .

## Natural satellite and Artificial satellite

Artificial satellites are basically object or machine made by humans that are sent into space and revolve in a orbit around another heavenly body ( mostly the earth , but also Mars, the Moon, and Sun ). A natural satellite is any object (i.e. a planet, moon, comet, asteroid) as part of natural processes in the universe that orbits another heavenly body (stars, planets). So our moon is a natural satellite of earth. And earth is a natural satellite of our sun .

So basically we can say , A satellite is a body which is constantly revolving in an orbit around a comparatively much larger body . Artificial satellite are man mad while natural satellite are made by nature.

Generally artificial satellites is also known as earth satellite .

## Orbital velocity of Satellite

Satellites are the natural or artificial objects that revolve in a orbit around a planet under its gravitational attraction . For example Moon is a natural satellite of earth , and althose there are lot of artificial satellites pf earth , one of them is INSAT-1B .

Orbital velocity of a satellite is basically the velocity of satellite required to put the satellite into its orbit around the earth .
For revolution of satellite (around the earth ( as shown in Fig.3 ) the gravitational pull should be equal to the the required centripetal force .

mv²/r = GMm/r²

v = [GM/r]1/2

v = [gR²/(R+h)]1/2

v = R[g/(R+h)]1/2

[ As GM = gR² and r = R+h ]

## Time period of Satellite

Time period of a satellite is the the time taken by satellite in completing one complete revolution around the earth .

we can get , time period by formula T = Circumference of the orbit /orbital velocity of satellite

T = 2πr/v = 2πr[ r/GM ]1/2

= 2π [ r³/GM ]1/2

= 2π [ r³/gR² ]

= 2π [ (R+h)³/gR² ]1/2

T = 2π [ R/g ]1/2[1+(h/R)]3/2

[ As r = R+h ]

NOTE – Time period of satellite does not depends on the mass of satellite .

## Height of satellite

As we know, time period of satellite ,

T = 2π[ r³/GM ]1/2

= 2π[ (R+h)³/gR² ]1/2

By squaring and rearranging both sides

gR²T²/4π² = ( R+h )³

h = [ T²gR²/4π² ]1/3 – R

By knowing the value of time period we can calculate the height of satellite from the surface of the earth .

## Angular momentum of Satellite( L )

Let’s suppose there is a satellite of mass m and is revolving with linear speed v on the orbital path of radius r around the earth , its angular momentum can be given by the formula ,

L = mvr = mr[ GM/r ]1/2 = [ m²GMr]1/2

From the above equation, it is clear that angular momentum of a satellite depends on both the mass of the satellite (m) and the mass of the planet ( M ) . It also depends upon the radius of the orbit ( r ) of the satellite .

## Energy of an orbiting satellite

The total mechanical energy of a satellite revolving around the earth is the sum of its potential energy (U) and kinetic energy (K) .

The potential energy of a satellite is due to its position with respect to earth. It appears because of the gravitational pull acting on the satellite due to earth. If a satellite of mass m is revolving around the earth of mass M , radius R , with orbital velocity v, in
an orbit of radius r , then potential energy of the satellite is

U = – GMm/r

The kinetic energy of a satellite ,

K = mv²/2
[ Put v= (GM/r)1/2]

K = GMm/2r

Thus, total mechanical energy of a satellite is

E = U + K

E = -GMm/r + GMm/2r = – GMm/2r

E = – GMm/2r = – GMm/2(R+h) Fig. 4 Graph of variation of Kinetic energy, potential energy and total energy of satellite with distance r .

If the radius r of the orbit of the satellite decreases , then it follows that

(i) its kinetic energy will increase i.e., its orbital velocity will increase
(ii) its potential energy will decrease (as it becomes more negative) and
(iii) its total energy will decrease (as it also becomes more negative)

If the satellite is orbiting close to earth, then r = R. Now, total energy of the satellite is E = – GMm/2R

The above expression has been obtained considering the circular orbit of the satellite. When the orbit of satellite becomes elliptical, both the PE and KE of the satellite vary from point to point but the total energy remains constant and is negative, as in the case of circular orbit .

The total energy of a circularly orbiting satellite which is negative , the potential energy is negative and magnitude of potential energy is twice the magnitude of the positive kinetic energy. We know that when any body is at infinity distance the total energy of a body will be zero if or positive , zero in case of body has no velocity . Therefore , if total energy of a satellite becomes zero or positive the satellite will escape to infinity .

NOTE- Since a satellite is always at finite distance from the earth , its total energy can never be positive or zero .

## Binding energy of a satellite

Binding energy of a satellite is the energy required to remove the satellite from its orbit around the earth to infinity .
Binding energy is equal to negative value of total mechanical energy of a satellite in its orbit . Thus, binding energy is ,

Binding energy = GMm/2r

## Weather satellite

A weather satellite is a type of satellite that is basically used to monitor the weather and climate of earth as well as used to help with more accurate weather forecasting . Weather satellite can be polar or geostationary , both satellite systems have unique characteristics . Satellite data , having a global view , complements land-based systems such as radiosondes , weather radar , and surface observing systems .

In modern era location based information is much more important and it can be possible with Satellite Navigation system .

GPS(Global Positioning System) , the most utlised system. Consisted of up to 32 medium orbit satellites. Operational since 1978 and globally since 1994. Russia has its own GLONASS (Global Navigation Satellite System), Europe its GALILEO and China its BeiDou .

## Astronomical satellites

Astronomical satellites are satellites used for observation of Planets, galaxies and other outer space objects and to make star maps . They can be used to study mysterious phenomena such as black holes and quasars . they can be used to take pictures of the planets in the solar system .

## communications satellite

A communications satellite consists of basic equipment supporting a mission – the attitude control subsystem, a power supply subsystem, TT&C subsystem , propulsion subsystem, the thermal control subsystem, and structure subsystem .
These are used for Television , phone or internet transmission .
eg. The Indian National Satellite (INSAT) system is one of the largest domestic communication satellite systems in Asia-Pacific region with nine operational communication satellites placed in Geo-stationary orbit .

## Military satellites

These are artificial satellites used for military purposes like reconnaissance and surveillance, signals intelligence, communications, navigation, and meteorology . military satellites dominate space from 600 to 1,200 miles altitude .

At that time , India has nearly 15 military-application satellites , with the latest GSAT-7A , these satellite are dedicated to the air force but shared by the army .

## Earth Observation satellites

These are used to photograph and image the earth and help us to monitor and protect our environment, manage our resources , respond to global humanitarian disasters and enable sustainable development. They provide essential information on a vast number of areas, including , ice thickness , crop health , ocean salinity and air quality .
Low earth orbit are manly used so that a more detailed image can be produced .

## International Space Station (ISS)

The International Space Station (ISS) is basically a habitable space laboratory , ISS is the largest single structure put into space by humans .
At an altitude of 400 km , the ISS travels with a speed of 28,000 km/h and make a complete revolution of the Earth in every 92 minutes . Inside the ISS ( in a microgravity environment ) Scientists are able to perform many valuable experiments .

## Solved Examples based on satellites

• Problem 1 . If a satellite is revolving around the earth in a circular orbit with velocity half to the magnitude of escape velocity from earth , determine the height of the satellite above the earth’s surface .

Solution – At ‘r’ distance from center of earth , orbital velocity of the satellite is

vo = [ GM/r ]1/2

= [ GM/(R+h) ]1/2. ….( i )

given , vo = ve/2

vo = ( 2GM/R )1/2/2

vo = [ GM/2R ]1/2 ….( ii )

with the help of Equation ( i ) and ( ii ),
we get

h = R = 6400 km

• Problem 2. Find out the minimum energy needed to launch a satellite in a circular orbit from surface of a planet at an altitude of 2R , mass of planet is M and mass of satellite is m .

Solution –
Let the energy required is E .
then ,

E = ( Energy of satellite )final – ( energy of satellite )initial

finally satellite is at distance of r from center of earth ( here r = R+2R=3R )

final energy of satellite = – GMm/2r

Ef = – GMm/6R

Initially when satellite in on earth surface ( before lunching ) then it will be at rest .
So ,
Initial energy = Ei = – GMm/R

Mininum energy required ( E ) = Ef – Ei

E = [ – GMm/6R ] – [ – GMm/R ]

E = 5GMm/6R

• Problem 3 . By considering , the orbit of the earth round the sun is a circle . Determine the mass of the Sun if the distance between the sun and earth is 1.49 x 1011 m and value of G=6.66 x 10-11 Nm²/kg² .

Solution –
Orbital velocity of earth will be

V= [ GM/r ]1/2 ( Orbital speed )

Where M denotes the mass of sun and r denotes the orbit radius of earth .

We know that , time period of earth around the sun is T = 365 days ,

thus we have

T = 2πr/v

T = 2πr [ r/GM ]1/2

M = 4π²r³/GT²
put the value ,

M = [4×(3.14)²×(1.49×10¹¹)³]/[(365×24×3600)²×(6.66×10-11]

M = 1.972×10²² Kg