In this article, we shall study about satellites, their types and uses of artificial satellites

Satellites:

Any object that revolves around a given planet in circular orbit under the influence of planet’s gravitational force is called as a satellite.

Types of Satellites:

Satellites are of two types. viz natural satellites and artificial satellites (man-made objects orbiting around the earth/planet). The moon is the natural satellite of the earth. The Earth, the Venus, and Jupiter are the natural satellites of the sun. INSAT-B, INSAT-IC are artificial satellites of the earth.

Projection of Satellite:

To launch a satellite in an orbit around the earth multistage rocket is used.  The launching involves two steps. In the first step, the satellite is taken to the desired height and then in the second step, it is projected horizontally with the calculated speed in a definite direction. If the velocity is proper it starts revolving in a stable circular orbit.

To understand the launching of satellite let us consider a simple case of the use of a two-stage rocket.  The satellite is kept at the tip of the two-stage rocket.

Initially, the first stage of the rocket is ignited on the ground so that the rocket is raised to the desired height, the first stage is detached. Then the rocket is rotated by remote control to point it in the horizontal direction. Then the second stage is ignited so the rocket gets push in the horizontal direction and acquires certain horizontal velocity (Vh).  When the fuel is completely burnt second stage also gets detached and the satellite starts orbiting around the earth.

A Satellite is Placed Outside Earth’s Atmosphere:

The satellite orbiting very close to the earth’s surface has an orbital speed of about 8 km/s. As the height of the satellite from the surface of the earth increases its orbital velocity decreases. If the satellite is placed in the atmosphere, due to the high velocity of satellite and friction between the atmosphere and the satellite, large heat will be produced and the satellite will get burnt.

Types of Artificial Satellites:

Geostationary or Geosynchronous or Communication Satellite:

A communication satellite is an artificial satellite which revolves around the earth in a circular orbit in the equatorial plane such that,

• its direction of motion is the same as the direction of rotation of the ‘earth about its axis.
• its period is the same as the period of rotation of the earth, i.e. 24 hours.

When observed from the earth’s surface, this satellite appears stationary. Therefore, it is called a geostationary satellite. As its motion is synchronous with the rotational motion of the earth, it is called a geosynchronous satellite. The height of communication satellite above the surface of the earth is about 36,000 km.

e.g. INSAT series satellites, APPLE.

Notes:

• The orbit of the geosynchronous satellite is called geosynchronous orbit.
• It lies in an equatorial plane. i.e. the angle between geosynchronous orbit and equatorial plane is 0°.
• The radius of the geosynchronous orbit is about 42400 km.
• The satellite parking strip is an area over the equator is becoming congested with several hundreds of communication, weather, military and transmission satellites.

Uses of Communication Satellites:

• The communication satellites are used for sending microwave and TV signals from one place to another.

• The communication satellites are used for weather forecasting.

• The communication satellites are used for detecting water resource -locations and areas rich in ores.

• The communication satellites are used for spying In enemy countries i.e. It can be used for military purposes

Polar or Sun-synchronous Satellite:

A polar satellite is a low altitude satellite orbit around the earth in north-south orbit passing over the north pole and south pole. The orbit of the polar satellite is called the polar orbit. The polar orbit makes an angle of inclination of 90° with the equatorial plane. Polar satellites cross the equatorial plane at the same time daily.

The height of the polar satellite above the earth is about 500-800 km. Its time period is about 100 minute.

Advantages of Polar Satellite:

Geostationary satellites are fixed at one position w.r.t. the earth at height 36000 km above the Earth. Its long-range helps meteorologist to understand and analyze the weather. But to understand Earth’s atmosphere and changes in the atmosphere, the whole planet must be scanned periodically and most effectively. To this polar satellites are used.

Since its time period is about 100 minutes it crosses any altitude many times a day and its height h above the earth is about 500-800 km, a camera fixed on it can view only small strips of the earth in one orbit. Adjacent strips are viewed in the next orbit so that in effect the whole earth can be viewed strip by strip during the entire day. From the path shown in the figure, we can see that it covers almost all geographical area.

These satellites can view polar and equatorial regions at close distances with good resolution.

Information gathered from such satellites is extremely useful for remote sensing, meteorology as well as for environmental studies of the earth.

Uses of Polar Satellites:

• Information gathered from polar satellites is extremely useful for remote sensing, meteorology as well as for environmental studies of the earth.
• They are used for spying and surveillance.
• They are used for monitoring troop movements i.e. for military purpose.
• They are used to note land and sea temperature variation.
• They are also used to monitor the growth of crops.

Weightlessness in Satellite:

By the definition, the weight of a body is equal to the gravitational force with which the body is attracted towards the centre of the earth. When the astronaut is on the surface of the earth, his weight acts vertically downwards.  At the same time, the earth’s surface exerts an equal and opposite force of reaction on the astronaut.  Due to this force of reaction, the astronaut feels his weight.

When the astronaut is in an orbiting satellite, a gravitational force still acts upon him.  However, in this case, both the astronaut as well as the satellite are now attracted towards the earth and have the same centripetal acceleration due to gravity at that place.

As both astronaut and the surface of the satellite are attracted towards the earth centre with the same acceleration, and hence the astronaut can’t produce any action on the floor of the satellite.  So the floor does not give any reaction on the astronaut.  Hence the astronaut has a feeling of weightlessness. When we are moving down in accelerating lift, we can have the feeling of partial weightlessness.

Actual Weightlessness Condition:

As we go away from the earth’s surface, the acceleration due to gravity decreases. At some point in space, it becomes negligible. At such points in space, weightlessness can be experienced. The acceleration due to gravity is zero at the centre, hence at the centre of the earth, weightlessness can be experienced.

Under free fall of the body, weightlessness can be experienced. Weightlessness can be experienced at some point between the Earth and the Moon where the gravitational field created by them cancel each other.

Effect of Weightlessness Condition:

• One can lift a load of 1000 kg easily.
• One can overturn coffee mug without spilling the coffee in it on the ground.
• It leads to a headache and puffy faces.
• The height can grow by 1 mm to 2 mm.
• The cardiovascular system does less work.
• The muscle can get out of condition, and consequently one may have difficulty in doing routine work after returning back to the earth.

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Science > Physics > Gravitation > Satellites

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In this article we shall study numerical problems on Keppler’s third law of orbital motion, to calculate period of revolution of a planet around the sun.

Example – 01:

What would be the length of the year if the earth were at half its present distance from the sun?

Given: r₂ = 1/2 r₁, old period T₁ = 365 days

To Find:  New period T₂ =?

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: Thus length of the year would be 129 days

Example – 02:

What would be the duration of the year if the distance of the Earth from the sun gets doubled?

Given: r₂ = 2 r₁, old period T₁ = 365 days

To Find:  New period T₂ =?

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: Thus the length of the year would be 1032 days

Example – 03:

Calculate the period of revolution of the planet Jupiter around the Sun. The ratio of the radius of Jupiter’s orbit to that of earth’s orbit around the Sun is 5.2.

Given: r_J : r_e  = 5.2 , Time period of the Earth  T₁ = 1 year.

To Find:  Period of Jupiter T_J =?

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: The period of revolution of planet Jupiter is 11.86 years.

Example – 04:

A geostationary satellite is orbiting the earth at a height of 6R above its surface. What is the time period of a satellite orbiting at a height of 2.5 R above the earth’s surface? Where R is the radius of the Earth.

Solution:

Given: r₁ = R + 6R = 7R, r₂ = R + 2.5 R = 3.5 R, Time period of geostationary satellite T₁ = 24 hours.

To Find:  Period of second satellite T₂ =?

By Keppler’s law, we have T² ∝ r³

Ans: The period of revolution of a satellite orbiting at a height of 2.5 R above the earth’s surface is 8.458 hr.

Example – 05:

A satellite orbiting around the earth has a period of 8 hrs. If the distance of another satellite from the centre of the earth is four times that of above, what is its period?

Given: Ratio of radii of orbits r₁ = r, r₂ = 4r, Time period of first satellite  T₁ = 8 hours.

To Find:  Period of second satellite T₂ =?

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: The period of the second satellite is 64 hours.

Example – 06:

Calculate the period of revolution of a planet around the sun if the diameter of its orbit is 60 times that of Earth’s orbit around the Sun. Assume both orbits to be circular.

Given: d_P = 60 d_E,  r_P =  60 r_E, Time period of Earth  T_E = 1 year.

To Find:  Period of the planet T_P =?

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: The period of the planet is 464.8 years

Example – 07:

The planet Neptune travels around the sun with a period of 165 years. Show that the radius of its orbit is approximately thirty times that of the Earth.

Given: Period of NeptuneT_N = 165 years, Time period of Earth  T_E = 1 year.

To Show:  Radius of Neptune r_N =30 r_E

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: Thus the radius of its orbit is approximately thirty times that of the Earth.

Example – 08:

The radius of earth’s orbit is 1.5 ×10⁸ km and that of mars is 2.5 × 10¹¹ m. in how many years the mars completes its one revolution.

Given: Radius of the orbit of the Earth r_E = 1.5 ×10⁸ km = 1.5 ×10¹¹ m, Radius of the orbit of the Mars r_M = 2.5 ×10¹¹ m, Time period of Earth T_E = 1 year.

To Find:  Time period of Mars T_M =?

Solution:

By Keppler’s law, we have T² ∝ r³

T_M = 2.15 years

Ans: In 2.15 years Mars completes its one revolution.

Example – 09:

The mean distance of the earth from the Sun is 1.496 ×10⁸ km and its period of revolution around the Sun is 365.3 days. The time periods of revolutions of planets Venus and Mars are 224.7 days and 687.0 days respectively, where day means a terrestrial day. Calculate the distances of Venus and Mars from the Sun.

Given: Radius of the orbit of the Earth r_E = 1.496 ×10⁸ km, Time period of Earth T_E = 365.3 days, Time period of Venus  T_E = 224.7 days, Time period of Mars  T_M = 687.0 days.

To Show:  Radius of the orbit of the Venus r_V =? Radius of the orbit of the Mars r_M = ?,

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: Distance of Venus from the sun is 1.08 ×10⁸ km and distance of Mars from the sun is 2.279 ×10⁸ km

Example – 10

Suppose Earth’s orbital motion around the Sun is suddenly stopped. What time the Earth shall take to fall into the Sun.

Given: Radius of the orbit of the Earth = r, Time period of Earth T = 365 days,

To Show:  Time taken by the Earth to fall into the Sun T₂ = ?,

Solution:

In this problem, we are assuming there is no effect of temperature on the Earth. If the Earth suddenly stops rotating around the sun and falls into the sun and let us assume it comes back to the point on the orbit. Thus it starts orbiting along a highly flattened ellipse with major axis  = r. Thus semimajor axis = a = r₂ = r/2

By Keppler’s law, we have T² ∝ r³

This is the total period of revolution. It should take half the time period to fall into the Sun

Time required = 130/2 = 65 days

Ans: Earth shall take 65 days to fall into the Sun.

Note: Thus, in general, the time taken by a planet to fall into the sun = Period x 0.1768 (This relation can be used for MCQ)

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In this article, we shall study Keppler’s laws of orbital motion and their use to derive Newton’s law of gravitation.

In the second century A.D., Ptolemy proposed a geocentric theory of the universe. According to this theory, the Earth is the centre of the universe and all other planets, sun and other stars revolve around the Earth. Indian astronomer, mathematician and philosopher Aryabhatta in 498 A.D. proposed that not only the Earth revolves around the sun but it rotates around its own axis. Polish astronomer Nicolaus Copernicus (1473 -1543) proposed the heliocentric theory of the universe. According to this theory, the Sun is the centre of the galaxy and all other planets revolve around the Sun. The Danish astronomer Tyco Brahe made the most accurate observations possible before the invention of the telescope. These observations showed discrepancies within Ptolemy’s geocentric theory of the universe.

Johannes Kepler (1571 – 1630) studied data of motion of planets by Danish astronomer Tycho Brahe and put forward his laws of planetary motion. These laws are known as Keppler’s law of orbital motion. He also proposed that the motion of the planet around the sun is not in circular orbit but it is in an elliptical orbit with the Sun at one of the foci of the elliptical orbit. A Netherland’s spectacle maker Hans Lippershey assembled first reflecting telescope. Using this concept Galileo developed his own telescope. Using the telescope he observed skies and substantiated Copernicus’s heliocentric theory of the universe. Further, he observed there are many stars in the universe and the Sun is one of them. He observed that only planets move around the sun. The moon moves around the Earth and Some planets like Jupiter has more than one moon.

There was a curiosity among astronomers about the motion of planets around the sun. For the study of the motion of planets around the sun or satellites around the Earth, we consider gravitational attraction between the sun and the planet or earth and the satellite only. Here we ignore the disturbing effect of the gravitational forces due to other bodies. Another major assumption is that the centre of mass of the sun (earth) and the planet (satellite) system is at their geometrical centre.

Keppler’s Laws:

First Law (Keppler’s Law of Elliptical Orbits – 1609):

Every planet moves around the sun in a closed elliptical orbit with the sun at one of its foci.

The following figure shows the path of the planet around the Earth.

Due to elliptical motion and the Sun at one of the foci, the distance of a planet from the sun changes continuously. The closest point on the orbit of the planet from the sun is B. This closest point on the orbit of the planet to the sun is called perihelion and this minimum distance is called perigee. Thus Perigee = SB = a – ae = a (1 – e). Where ‘a’ is semi-major axis and e is the eccentricity of the ellipse. The farthest point on the orbit of the planet from the sun is A. This farthest point on the orbit of the planet to the sun is called aphelion and this maximum distance is called apogee. Thus Apogee = SA = a + ae = a (1 + e). Where ‘a’ is semi-major axis and e is the eccentricity of the ellipse.

Second Law (Keppler’s Law of Equal Areas – 1609):

The radius vector drawn from the Sun to the planet sweeps out equal areas in equal time. i.e. aerial velocity of the radius vector is constant. Thus the Aerial velocity of the satellite is always constant. i.e.  A/t = Constant.

In this case, the planet covers unequal distances in equal time. Thus the planet has variable speed. When the planet is very close to the Sun, the maximum distance is covered in given time. Thus the planet has maximum velocity when it is at perihelion. Hence it has maximum kinetic energy at perihelion. When the planet is very far to the Sun, the minimum distance is covered in given time. Thus the planet has minimum velocity when it is at aphelion. Hence it has minimum kinetic energy at perihelion.

Keppler’s Third Law (Keppler’s Law of Period – 1618)

The square of the period of a satellite is directly proportional to the cube of the semimajor axis of its elliptical orbit.

If T is the period of satellite and r is the semimajor axis or the radius of the circular orbit of the satellite.

A similar result is obtained for the elliptical orbit and then the radius r of the circular orbit is replaced by semi-major axis ‘a’.

Note: 

Keppler’s laws are empirical laws because they are based on observations and not on theory. Kepler’s laws are applicable whenever an inverse square law is involved. They are applicable to the solar system and artificial satellite systems.

Proof of Keppler’s Law of Equal Areas:

Consider a planet moving in an elliptical orbit from point P₁ to point P₂, Let in a small interval of time Δt, it traces small area ΔA at the focus S. Let the angle traced by radius vector be Δθ at the Focus S.

The area of the sector is given by

Where ω is the angular velocity of the planet.

Now, the instantaneous angular momentum is given by

L = m r ω²

Thus r ω² = L/m

Substituting in equation (1) we have

As net torque acting on the planet is zero, the angular momentum is constant. Thus R.H.S. of equation (2) is constant.

Thus dA / dt = constant

Thus the areal velocity of the planet is constant. This proves Keppler’s second law.

Proof of Newton’s Law of Gravitation Using Keppler’s Law:

Let, m = Mass of planet
M = Mass of sun
R = Radius of the orbit of the planet
ωw= Angular velocity of the planet.
F = Centripetal force exerted on planet by the Sun

We know that

By Keppler’s law, we have T² ∝ r³

T² =  k r³ 

Substituting in equation(1)

The quantities in the bracket are constant.

Thus force exerted by the sun on the planet is a) directly proportional to the mass ‘m’ of the planet. b) Inversely proportional to the square of the distance ‘R’ between the planet and the sun. Now, the planet will exert equal and opposite force on the sun such that

Now, the planet will exert equal and opposite force on the sun such that

Thus force exerted by the planet on the Sun is a) directly proportional to mass ‘M’ of the Sun and b) inversely proportional to the square of the distance ‘R’ between the planet and the sun.

From (2) and (3) we have

This relation is known as Newton’s law of gravitation.

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