Chapter 8: Gravitation

Introduction

In our daily life, we all have noticed that everything that is thrown up will fall back to earth. Going uphill is a lot more tiring than going downhills and raindrops from clouds move towards the earth. There are a lot more such phenomena. An Italian scientist Galileo recognized every object experiences a constant acceleration toward the earth, irrespective of their masses.

The stars, moon and planets have been observed since ancient times. The motion of the moon around the earth and the motion of the earth around the sun were some phenomena that needed to be explained.

In early times it was believed that the earth is the center of the universe and everything revolves around the earth. This was known as the geocentric theory. This prevails for a very long time and at that time there was not much advancement in this subject. Later Galileo and other astronomers found that the Sun is the center and not the earth, and all the planets including the earth move around the Sun. This theory was known as the heliocentric theory. After the establishment of the heliocentric theory, the rapid advancement of various subjects of Science happens. Therefore Galileo is known as the Father of Modern Science.

In this chapter, we will discuss the laws that will explain all the phenomena discussed above.

Kepler’s Law

A nobleman Tycho Brahe from Denmark spent his entire life recording observations for the planets with naked eyes. His compiled data was analyzed later by his assistant Johannes Kepler. He could extract from the data three law’s called Kepler’s Laws.

Three Laws of Kepler

1. Law of the orbit: All Planets move around the sun in an elliptical orbit with the sun at one of its foci.

2. Law of Area: The line that joins any planet to the sun sweeps equal areas in equal intervals of time. This law comes from the observations that planets appear to move slower when they are farther from the sun than when they are nearer.

3. Law of period:  The Square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.  Mathematically,   T²  α a³.

The graph between the Time period and their semi-major axis is drawn below

This law is also consistent with the conservation of angular momentum.

Since the angular momentum of the planet revolving around the sun at any point of time is conserved.

m v1 r1= m v2 r2    so we have  v1/v2= r2/r1

The universal law of Gravitation

Kepler’s laws were known to Newton and enabled him to make a great scientific leap in proposing his universal law of gravitation.

Statement of Universal law of gravitation:

Every object in the Universe attracts every other object with a force directed along the line of the center for the two objects that is proportional to the product of their masses and inversely proportional to the square of the separation between the two objects.

Where G= universal gravitational constant

The Universal Gravitational Law can explain almost anything, right from how an apple falls from a tree to why the moon revolves around the earth.

• The gravitational force is always attractive. Also the magnitude of gravitational force on mass 1 due to mass 2 is equal to the magnitude of the force on mass 2 due to mass 1.   | F₁₂|= | F₂₁| But their directions are opposite. This is consistent with the third law of the motion.
• This law refers to point masses whereas we deal with extended objects which have finite size, so we use the concept of center of mass here. For example, The force of attraction between a hollow spherical shell of uniform density and a point mass of the shell is just as if the entire mass is concentrated at the center of the shell.
• If we have a collection of point masses the force on any one of them is the vector sum of the gravitational force exerted by the other point masses. This is actually the principle of superposition.

Net force on m1 = force on m1 due to m2 + force on m1 due to m3

• The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it is zero. Qualitatively, we can again understand this result. Various regions of the spherical shell attract the point mass inside it in various directions. These forces cancel each other completely.

Acceleration Due to Gravity

When you throw something in the air, you notice that it goes up and then it comes down toward the earth and finally lands on the ground. The same thing will happen if you drop an object from some height. It will also land up on the ground. This happens because of the gravitational pull of the earth.

Now let's take two objects of mass m1 and m2 (where m1>m2). When we drop these masses from some height you will notice that both the masses hit the ground at the same time irrespective of their masses. So we can conclude from here that the gravitational pull of the earth on various objects near it is independent of the masses of the objects.

The conclusion from the above discussion is that earth pulls every object towards it with a force that is independent of the mass of the object. This pull is due to the gravitational force between the earth and the object.

Let the acceleration due to gravity on the object be ‘g’ , so gravitational force due to earth on an object of mass ‘m’ would be  ‘mg’ which is actually equal to the weight of the object.

Let the mass of the earth is me and the radius of the earth is R, gravitational force between the earth and the object near the earth would be

The value of g near the earth is approximately 9.8 m/s²

From the above-derived formula of acceleration due to gravity ‘g’ is it clear that it is independent of the mass of the object and only depends on the mass of the earth and the distance of the object and the center of the earth.

When the objects are near the earth when (h<<R), then we take the radius of earth ‘R’ as the distance between the earth and the object.

But when we move to a greater height from the earth or at a certain depth inside the earth then the value of g would be different. In the next section, we would see the variation of ‘g’ with height and depth.

Variation of g with height and depth

The value of g varies with altitude and depth from the surface of the earth. We will discuss the derivation of the variation of g with depth and height. After that, we will see from the results that the value of g is maximum at the surface of the earth and decreases with both altitude and depth.

Variation of acceleration due to gravity with depth.

Suppose we have to calculate ‘g’ at the depth‘d’ from the surface of the earth. Suppose M be the mass of the whole earth and M1 be the mass of the inner sphere of the earth below the depth d as shown in the figure.

While calculating the value of g at the depth‘d’ we have to consider the gravitational force due to inner mass M1 and not the mass of the whole earth.

The acceleration due to gravity at depth d is g_d, the mass of the inner sphere is M1 and the distance of it from the center of the earth is (R_e-d) so we can write the value of acceleration due to gravity at depth d as

   ….1

Now simplify the value of M1.

     

Now put the value of M1 in equation 1.

So we have finally the result.

The acceleration due to gravity varies with depth as )

Variation of acceleration due to gravity with height ‘h’

When we want to calculate the value of acceleration due to gravity at an altitude ‘h’ where h≈R Then when we calculate ‘g’ we have to consider the distance (R+h) instead of just R. As now ‘h’ is of the same order as R and hence cannot be neglected.

Acceleration due to gravity at height h g_o is given as

So now we have results,

Acceleration due to gravity varies with height as  

The graph given below concludes the above discussion. It is clear that the value of g is maximum at the surface of the earth and decreases with both altitude and depth.

A fun thing to do: Virtual lab

Below is the link to understand the universal law of gravitation.

Force of gravity lab

What can we do in this virtual lab simulation?

• We can change the masses of the objects
• We can change the distance between them

What can we observe?

• We can see the direction of the gravitational force on each body
• We can also see the magnitude of the gravitational force both in decimal and scientific notations.

Gravitational potential Energy near the surface of the earth

We have already discussed the notion of potential energy as being the energy stored in the body due to its position and configuration. We attach the concept of potential with only conservative forces like electrostatic, gravitational force etc.

If we change the position or configuration of the system its potential energy would change. An external force has to do some work on the system to change the position of the configuration of the system. Actually, this external work done by the external force gets converted into the potential energy of the system.

The force of gravity is a conservative force and we can calculate the potential energy of the body arising due to this force, called gravitational potential energy.

Potential energy of the object of mass ‘m’ at height ‘h’ near the earth's surface is given by,  Potential energy at height h= mgh+ P.E(0)

Where P.E (0) is the potential energy of the object at the surface of the earth which is a constant.

If we are only interested in knowing the change in potential of the object when we lift the object from the surface to height ‘h’ then we actually drop the constant term and simply write P.E=mgh , near the earth

Suppose we have an object of mass ‘m’ kept at height ‘h1’ from the surface of the earth. If we try to lift that object to a height ‘h2’ from the surface of the earth, we need to do some work against gravity.

Work done in lifting the particle of mass m from the first to the second position is denoted by W12. 

W12= mg(h2-h1)

The work done in moving the particle is just the difference of potential energy between its final and initial positions (change in potential energy)

Gravitational potential energy at an arbitrary distance

If we wish for the gravitational potential energy at an arbitrary distance from the surface of the earth (not near the earth), then we cannot use the above-discussed formula P.E= mgh, as this is not valid here since the assumption that the gravitational force mg is a constant is no longer valid.

However, from our discussion, we know that at a point outside the earth, the force of gravitation on a particle directed towards the center of the earth is 

Where r is the distance of the object from the center of the earth.

Me= mass of earth and m= mass of the object.

If we wish to move the object from a distance r1 to a distance r2 from the center, an external force has to do some work against the force of gravitation

W12=-∫_r1^r2  F dr= -GMe m  ∫_r1^r2   dr/r^2 =-GMe m [ 1/r2-1/r1  ]

The negative sign indicated a bound state.

When we take an object of mass m from infinity (r1) into the gravitational field of massive object M at a distance ‘r’ (r2) from the center of the object of mass M.

When the work is done in doing this will be, W=-GMm/r.

This work is stored in the form of the potential energy of the system.

Gravitational Potential energy due to a system of masses

When we wish to calculate the potential energy of a system of masses, for example, three masses at the corner of an equilateral triangle etc. We need to find the work done by the external force in assembling the system of mass. As the work done by the external force in assembling the masses at their respective positions. For example

Escape velocity

All of us have experienced that when we throw anything from earth, it eventually falls back on earth. Greater the speed of throw would be, the higher the object can go and then return back to earth. This happens because the gravitational pull of the earth pulls the object back on earth.

Now the question is with what speed the object should be thrown so it would never come back to earth again and just escape the earth's gravitational field. That velocity would be the escape velocity of the earth.

While calculating the escape velocity, we make use of the concept of mechanical energy conservation.

Let the initial mechanical energy of the rocket at the surface of the earth would be  

    ... 1

We assume that when the object just escapes the gravitational field of the earth its final gravitational potential energy would be zero. And to calculate the minimum escape velocity we assume that the object just stops after escaping the gravitational field of the earth so its final kinetic energy would also be zero.

T.Ef= P.Ef+K.Ef=0+0=0

According to the principle of conservation of mechanical energy T.Ei=T.Ef

So,  

Therefore we have escape velocity     from the surface of the earth.  The numerical value of the escape velocity from the surface of the earth =11.2 km/hr.

If we want to calculate escape velocity from the surface of the moon then M would be the mass of the moon and R= radius of the moon.

Since the gravitational pull of the moon is very less compared to the gravitational pull of the earth, escape velocity from the surface of the moon is about five times smaller than the escape velocity from the surface of the earth.  Acceleration due to gravity on moon

So escape velocity at surface of moon =  = 2.3 km/s

The smaller value of escape velocity is the reason that the moon has no atmosphere. Gas molecules if formed on the surface of the moon having velocities larger than this will escape the gravitational pull of the moon.

In general for any planet of radius ‘r’ and mass ‘M’ we have the value of escape velocity.

If the object was to be thrown out of the gravitational pull of the earth from a height ‘h’ above the surface of the earth, then the initial distance of the object from the center of the planet will be  r= R+h for this case. Therefore escape velocity would be    from a height ‘h’ from the surface of the earth.

You will notice that as the distance from the center increases, escape velocity decreases.

Orbital velocity

Earth is surrounded by various satellites hovering miles above our heads. Our own moon also remains above the planet at all hours. But why don’t these objects come crashing down onto the planet’s surface? After all, other items in the sky, like an airplane or a hot air balloon, will eventually crash down if they run out of power. The reason that man-made satellites and the moon do not come crashing down is because they have achieved orbital velocity.

Did you know that space missions like GPS satellites, geosynchronous satellites, etc. are possible only because of the correct calculations of the orbital velocity?

The word "orbit" is defined as the path a body follows when being acted upon by the force of gravity. For example, a communications satellite follows an orbital path around the Earth and the Earth follows an orbital path around the Sun.

Orbital velocity, a velocity sufficient to cause a natural or artificial satellite to remain in orbit. The inertia of the moving body tends to make it move on in a straight line, while gravitational force tends to pull it down. The orbital path, elliptical or circular, thus represents a balance between gravity and inertia.

For a satellite of mass m to move around a planet of mass M around a circular path at a distance ‘r’ from the center of the planet with a constant speed ‘v’, there is a requirement of centripetal force. This centripetal force is provided by the gravitational pull of the planet on the satellite.

So we have

So we have an orbital velocity  

If the satellite is at altitude ‘h’ above the earth’s surface then r= R+h

So  )

It is clear that orbital velocity is inversely proportional to the square root of the radius of the orbit. As we go far from the planet, orbital velocity decreases.

The time period of revolution of the satellite in orbital radius at height ‘h’ above the earth's surface is given by

For a satellite very close to the surface of earth h can be neglected in comparison to R, so we can approximate  = 

After putting the value of   and R= 6400 km, we get

Numerical value of the Time period of a satellite near the surface of the earth is equal to 85 minutes.

Energy of the orbiting satellite.

The total energy of the orbiting satellite of mass ‘m’ would be the sum of kinetic energy and potential energy of the satellite.

The kinetic energy of the satellite in the orbit would be =  , where v_o be the orbital velocity of the satellite in that orbit.

    …1

The potential energy of the satellite in an orbit at a height ‘h’ above the surface of the earth would be given as,

P.E. = -GMm/(R+h)               … 2

If we add P.E. and K.E. we will get the total energy of the orbiting satellite

Points to be noted:

• The Total energy of the orbiting satellite is negative which tells us that the satellite is in a bound state with the planet.

T.E. < 0

• The total energy of an orbiting satellite is half the value of its potential energy.

• The absolute value of the total energy of the orbiting satellite is equal to its kinetic energy.

T.E= - K.E.   ; |T.E.|= K.E.

Geostationary Satellite and polar satellites

Geostationary satellites

As the name suggests geostationary, is a satellite that appears to be located at a fixed point in space when viewed from the earth’s surface. This is possible when the time period of the satellite in the orbit matches the rotation time of the earth about its own axis which is 24 hours.

 = 24 hours if we put the value of G, M and R and try to calculate altitude h, you will get something close to 36000 km.

A geostationary satellite is in an orbit that can only be achieved at an altitude of 35,786 km (22,236 miles) precisely and which keeps the satellite fixed over one longitude at the equator.

It is known that electromagnetic waves above a certain frequency are not reflected from the ionosphere. Waves used in television broadcasts or other forms of communication have much higher frequencies than critical frequency and thus cannot be received beyond the line of sight. A Geostationary satellite, appearing fixed above the broadcasting station can however receive these signals and broadcast them back to a wide area on earth. The INSAT group of satellites sent up by India are one such group of geostationary satellites widely used for telecommunications in India.

Advantages of Geostationary satellites

• A single geostationary satellite is on a line of sight with about 40 percent of the earth's surface. Three such satellites, each separated by 120 degrees of longitude, can provide coverage of the entire planet
• The geostationary orbit has the advantage that the satellite remains in the same position throughout the day, and antennas can be directed towards the satellite and remain on track.
• The geostationary orbit is used by many applications including direct broadcast as well as communications or relay systems.
• Satellites are always in the same positions relative to earth so antennas do not need re-orientation.

Disadvantages of geostationary satellites.

• Satellites are more costly to install in geostationary orbits in view of its greater altitude.
• Geostationary satellite's orbits can only be above the equator and therefore the Polar Regions cannot be covered.
• Long path length introduces delays and also some losses in the signal.
• The signal travel delay is about 120ms in one direction. The distance of 35786 Km gives 120 ms latency with 3x10⁸ m/sec speed of the signal. Hence it is not suitable for point-to-point applications requiring time-critical applications such as real-time voice, video etc.

Polar satellites

Polar Satellite - A polar orbit is one in which a satellite passes above or nearly above both poles of the body being orbited on each revolution. It, therefore, has an inclination of (or very close to) 90 degrees to the equator.

Polar satellites are also known as sun-synchronous satellites orbits adobe the earth at about 700-1700 km and its orbital period is about 100 minutes.

Advantages of polar satellite

• Polar orbits are often used for earth-mapping, earth observation, capturing the earth as time passes from one point, reconnaissance satellites, as well as for some weather satellites.
• Since the orbit is lower for the geostationary satellites, the data resolution is higher and losses and delays are minimal.
• They provide global coverage of climate studies.

Disadvantages of solar orbits

Cannot provide continuous viewing of one location and it is not geostationary.

Weightlessness

Weightlessness is a term used to describe the sensation of a complete or near-complete absence of weight. Astronauts orbiting the Earth often experience the sensation of weightlessness. These sensations experienced by the orbiting astronauts are the same sensations experienced by anyone who has been temporarily suspended above the seat on an amusement park ride. The causes of the sensation of weightlessness in both these cases are the same.

Weightlessness is a sensation experienced by an individual where there are no external objects touching one’s body. In other words, the sensation of weightlessness exists when all contact forces are removed. These sensations are common to the state of free fall.

During free fall, the only force acting on the body is the force of gravity. As gravity is a non-contact force, it cannot be felt without any opposing force. This is the reason why you feel weightless when in a state of free fall.

In a satellite around the earth, every part of the satellite has an acceleration towards the center of the earth which is exactly the value of the earth’s acceleration due to gravity at that position. Thus in the satellite, everything inside it is in a state of free fall. This is just as if we were falling towards the earth from a height. Thus, in a manned satellite, people inside experience no gravity. Gravity for us defines the vertical direction and thus for them, there are no horizontal or vertical directions, all directions are the same. Pictures of astronauts floating in a satellite show this fact.

A fun thing to do: Virtual lab

Below is the link of the simulation of gravity and orbits

Gravity and orbit

In this simulation, we can see the orbit of a celestial body around another massive celestial body (Sun, earth, moon and artificial satellite here)

What can we do following things in this simulation?

• We can choose our system between the given options ( Sun, earth, moon), ( Sun, earth), ( Earth, moon), ( Earth and artificial satellite)
• We can change the masses of the body of the chosen system.
• We can switch on or off the gravity and see how the object will move.
• We can select options like grid, Force vector, velocity vector etc.,
• We can also zoom in and out of the screen.

What will we observe in this simulation?

• After selecting the system and enabling the option like force vector, velocity we can see the direction of force and velocity of the body.
• After clicking on the Play button, the lesser massive body will orbit around the massive body and we can see the path traced by the body.
• When the body is moving, we can see how the direction of velocity is changing.
• Also, we can see the time taken by the body to move one complete revolution.