Course Content

Force of Gravity  Lesson

Force of Gravity  Quiz 1 and 2
Gravitational Force
Introduction
We are aware of the tendency of all material objects to be attracted towards the earth. Anything thrown up falls down towards the earth, going uphill is lot more tiring than going downhill, raindrops from the clouds above fall towards the earth and there are many other such phenomena. The Italian Physicist Galileo who recognised the fact that all bodies, irrespective of their masses, are accelerated towards the earth with a constant acceleration. It is said that he made a public demonstration of this fact. To find the truth, he certainly did experiments with bodies rolling down inclined planes and arrived at a value of the acceleration due to gravity which is close to the more accurate value obtained later.
An unrelated phenomenon, observation of stars, planets and their motion has been the subject of attention in many countries since the earliest of times. Observations since early times recognised stars which appeared in the sky with positions unchanged year after year. The more interesting objects are the planets which appear to have regular motions against the background of stars. The earliest recorded model for planetary motions proposed by Ptolemy about 2000 years ago was a ‘geometric’ model in which all celestial objects, stars, the sun and the planets, all revolved around the earth. The only motion that was thought to be possible for celestial objects was motion in a circle. Complicated schemes of motion were put forward by Ptolemy in order to describe the observed motion of planets. The planets were described as moving in circles with the centre of the circles themselves moving in larger circles. Similar theories were also advanced by Indian astronomers some 400 years later. However a more refined model in which Sun was the centre around which the planets revolved the ‘heliocentric’ model was already mentioned by Aryabhatta (5th century A.D.) in his treatise. A thousand years later, Polish monk named Nicolas Copernicus (14731543) proposed a definitive model in which the planets moved in circles around a fixed central sun. His theory was discredited by the church, but notable amongst its supporters was Galileo who had to face prosecution from the state for his beliefs.
It was around the same time as Galileo, a nobleman called Tycho Brahe (15461601) hailing from Denmark, spent his entire lifetime recording observations of the planets with the naked eye. His compiled data were analysed later by his assistant Johannes Kepler (15711640). He could extract from the data three superb laws that now go by the name of Kepler’s laws. These laws were known to Newton and enabled him to make a scientific surge in proposing his universal law of gravitation.
Kepler’s Laws
The three laws of Kepler can be stated as follows:

 Law of orbits: All planets move in elliptical orbits with the sun situated at one of the foci of the ellipse (Fig 1a). This law was a deviation from the Copernican model which allowed only circular orbits. The ellipse, of which the circle is a special case, is closed curve which can be drawn very simply as follows:
Fig 1(a) An ellipse traced out by a planet around the sun. The closest point P and the farthest point A, P is called the perihelion and A the aphelion. The semimajor axis is half the distance AP.
Fig. 1(b) Drawing an ellipse, A string has its ends fixed at F_{1} and F_{2}. The tip of a pencil holds the string taut and is moved around.Select two points F_{1} and F_{2}. Take a length of a string and fix its ends at F_{1} and F_{2}by pins. With the tip of a pencil stretch the string taut and then draw a curve by moving the pencil keeping the string taut throughout[Fig 1(b)]. The closed curve you get is called is an ellipse. Clearly for any pint T on the ellipse, the sum of the distances from F_{1} and F_{2} is a constant. F_{1} and F_{2} are called the focii. Join the points F_{1} and F_{2} and extend the line to intersect the ellipse at points P and A as shown in Fig 1(b). The midpoint of the line PA is the centre of the ellipse O and the length PO = AO is called the semimajor axis of the ellipse. For a circle, the two focii merge in to one and the semimajor axis becomes the radius of the circle.
 Law of areas: The line that joins any planet to the sun sweeps equal areas in equal intervals of time (Fig. 2). This law comes from the observations that planets appear to move slower when they are farther from the sun than when they are nearer.
Fig. 2 The P moves around the sun in an elliptical orbit. The shaded area is the area ΔA swept out in a small interval of time Δt.  Law of periods: The square of the time period of revolution of a planet is proportional to the cube of the semimajor axis of the ellipse traced out by the planet.
The table below gives the approximate time periods of revolution of nine planets around the sun along with values of their semimajor axes.
 Law of orbits: All planets move in elliptical orbits with the sun situated at one of the foci of the ellipse (Fig 1a). This law was a deviation from the Copernican model which allowed only circular orbits. The ellipse, of which the circle is a special case, is closed curve which can be drawn very simply as follows:
Table 1 – Data from measurement of planetary motions given below confirm Kepler’s Law of Periods
a = Semimajor axis in units at 10^{10}m.
T = Time period of revolution of the planet in years(y).
Q = The quotient (T^{2}/a^{2}) in units of 10^{34}y^{2}m^{2}
Planet  a  T  Q 

Mercury  5.79  0.24  2.95 
Venus  10.8  0.615  3.00 
Earth  15.0  1  2.96 
Mars  22.8  1.88  2.98 
Jupiter  77.8  11.9  3.01 
Saturn  143  29.5  2.98 
Uranus  287  84  2.98 
Neptune  450  165  2.99 
Pluto  590  248  2.99 
The law of areas can be understood as a consequence of conservation of angular momentum which is valid for any central force. A central force is such that the force on the planet is along the vector joining the sun and the planet. Let the sun be at the origin and let the position and momentum of the planet be swept out by the planet of mass m in time interval Δt is (Fig. 2) ΔA given by
ΔA = 1/2 (r × vΔt) …………………………………………………..(1)
Hence ΔA/Δt = 1/2 (r × p)/m, (since v = p/m)
= L/(2m) …………………………………………….(2)
where v is the velocity, L is the angular momentum equal to (r × p). For a central force, which is directed along r, L is a constant as the planet goes around. Hence ΔA/Δt is a constant according to the last equation. This is thee law of areas. Gravitation is a central force and hence the law of areas follows.
Example 1
Let the speed of the planet at the perihelion P in Fig. 1(a) be v_{p} and the Sunplanet distance SP be r_{p}. Relate {r_{p}, v_{p}} to the corresponding quantities at the aphelion {r_{A}, v_{A}}. Will the planet take equal times to traverse BAC and CPB?
Answer
The magnitude of the angular momentum at P is L_{p} = m_{p}r_{p}v_{p}, since inspection tells us that r_{p} and vp are mutually perpendicular. Similarly, L_{A} = m_{p}r_{A}v_{A}. From angular momentum conservation,
m_{p}r_{p}v_{p }= m_{p}r_{A}v_{A}
or v_{p}/v_{A }= r_{A}/r_{p}
Since r_{A }>r_{p}, v_{p }>v_{A}
The area SBAC bounded by the ellipse and the radius vectors SB and SC is larger than SBPC in Fig. 1. From Kepler’s second law, equal law, equal areas are swept in equal times. Hence the planet will take times. Hence the planet will take a longer time to traverse BAC than CPB.
Universal Law of Gravitation
Legend has it that observing an apple falling from a tree, Newton was inspired to arrive at a universal law of gravitation that led to an explanation of terrestrial gravitation as well as of Kepler’s laws. Newton’s reasoning in an orbit of radius R_{m }was subject to a centripetal acceleration due to earth’s gravity of magnitude
a_{m }= V^{2 }/R_{m }= 4π²R_{m }/T² ……………………………………………………………(3)
where V is the speed of the moon related to the time period T by the relation V = 2πR_{m }/T. The time period T is about 27.3 days and R_{m }was already known then to be about 3.84 ×10^{8}m. If we substitute these numbers in equation (3), we get a value of a_{m }much smaller than the value of acceleration due to gravity g on the surface of the earth, arising also due to earth’s gravitational attraction.
Central Forces
We know the rate of the time rate of change of angular momentum of a single particle about the origin is
dl/dt = r × F
The angular momentum of the particle is conserved, if the torque τ = r × F due to the force F on it vanishes. This happens either when F is zero or when F is along r. We are interested in forces which satisfy the latter condition. Central forces satisfy this condition. A ‘central’ force is always directed towards or away from a fixed point. i.e., along the posit vector of the point of application of the force with respect to the fixed point. (See Figure below). Further, the magnitude of a centre central force F depends on r, the distance of the point of application of the force from the fixed point; F = F(r).
In the motion under a central force the angular momentum is always conserved. Two important results follow from this:
 The motion of a particle under the central force is always confined to a plane.
 The position vector of the particle with respect to the centre of the force (i.e., the fixed point has a constant areal velocity. In other words the position vector sweeps out equal areas in equal times as the particle moves under the influence of the central force.
Try to prove both these results. You may need to know that the areal velocity is given by: dA/dt = 1/2 rv sin α .
An immediate application of the above discussion can be made to the motion of a plane under the gravitational force of the sun. For convenience the sun may be taken to be heavy that it is at rest. The gravitational force of the sun on the planet is directed towards the sun. This force also satisfies the requirement F = F(r), since F = G m_{1}m_{2}/r² where m_{1 }and m_{2 }respectively by the masses of the planet and the sun and G is the universal constant of gravitation. The two results (1) and (2) described above, therefore, apply to the motion of the planet. In fact the result (2) is the well known second law of Kepler.
At a position P, the force is directed along OP, O is the centre of the force taken as the origin . In time Δt, the particle moves from P to arc PP’ = Δs = vΔt. The tangent PQ at P to the trajectory gives the direction of the velocity at P. The area swept in Δt is the area of sector PQP’ ≈ (r v sin α)/2
This clearly shows that the force due to earth’s gravity decreases with distance. If one assumes that the gravitational force due to the earth decreases in proportion to the inverse square of the distance from the centre of the earth, we will have a_{m }α(R_{m})^{2}; g_{ }α(R_{E})^{2 }and we get
g/a_{m }= (R_{m})^{2 }/(R_{E})^{2}; 3600 ……………………………………………..(4)
in agreement with a value of g ; 9.8 m s^{2 }and the value of a_{m }from Eq. (3). These observations led Newton to propose the following the Universal Law of Gravitation:
Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
The quotation is essentially from Newton’s famous treatise called ‘Mathematical Principles of Natural Philosophy’ (Principia for short).
Stated Mathematically, Newton’s gravitation law reads: The force F on a point mass m_{2 }due to another point mass m_{1 }has the magnitude
Fig. 3 Gravitational force on m_{1 }due to m_{2 }is along r where the vector r is (r_{2} – r_{1}).
The gravitational force is attractive, i.e., the force F is along –r. The force on point a mass m_{1 }due to m_{2 }is of course –F by Newton’s third law. Thus, the gravitational force F_{12 } on the body 1 due to 2 and due to 1 are related as F_{12 } = – F_{21}.
Before we can apply Eq. (5) to objects under consideration, we have to be careful since the law refers to point masses whereas we deal with extended objects which have finite size. If we have a collection of point masses, the force on any one of them is the vector sum of the gravitational forces exerted by other point masses as shown in Fig.4.
The total force on m_{1 }is
Example 2
Three equal masses of m kr each are fixed at the vertices of an equilateral triangle ABC.
(a) What is the force acting on a mass 2m placed at the centroid G of the triangle?
(b) What is the force if the mass at the vertex A is doubled?
Take AG = BG = CG =1m (see Fig. 5)
Answer
(a) The angle between GC and the positive xaxis is 30° and so is the angle between GB and the negative axis. The individual forces in vector rotation are
For the gravitational force between an extended object (like earth) and a point mass. Eq. (5) is not directly applicable. Each point mass in the extended object will exert a force on the given point mass and these force will not all be able in the same direction. We have to add up these forces vectorially for all the point masses in the extended object to get the total force. This is easily done using calculus. For two special cases, a simple law results when you do that:
 The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is concentrated at the centre of the shell. Qualitatively this can be understood as follows: Gravitational forces caused by the various regions of the shell have components along the line joining the point mass to the centre as well as along a direction perpendicular to this line cancel out when summing over all regions of the shell leaving only a resultant force along the line joining the point to the centre. The magnitude of this force woks out to be as stated above.
 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.
Newton’s Principia
Kepler had formulated his third law by 1619. The announcement of the underlying universal law of gravitation came about seventy years later with the publication in 1687 of Newton’s masterpiece Philosophiae Naturalis Principia Mathematica, often simply called the Principia.
Around 1685, Edmund Halley (after whom the famous Halley’s comet is named), came to visit Newton at Cambridge and asked him about the nature the trajectory of a body moving under the influence of an inverse square law. Without hesitation Newton replied that it had to be an ellipse, and further that he had worked it out long ago around 1665 when he was forced to retire to his farm house from Cambridge on account of a plague outbreak. Unfortunately, Newton had lost his papers. Halley prevailed upon Newton to produce his book form and agreed to bear the cost of publication. Newton accomplished this feat in eighteen months of superhuman effort. The Principia is a singular scientific masterpiece and in the words of Lagrange it is “the greatest production of the human mind”. The Indian born astrophysicist and Nobel laureate S. Chandrasekhar spent ten years writing a treatise on the Principia. His book, Newton’s Principle for the Common Reader brings in to sharp focus the beauty, clarity and breath taking economy on Newton’s methods.
The Gravitational Constant
The value of the gravitational constant G entering the Universal law of gravitation can be determined experimentally and this was first done by English scientist Henry Cavendish in 1798. The apparatus used by him is schematically shown in Fig. 6.
The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire. Two large lead spheres are brought close to the small force to the small ones but on opposite sides as shown. The big spheres attract the nearby small ones by equal and opposite force as shown. There is no net force on the bar but only a torque which is clearly equal to F times the length of the bar, where F is the force of attraction between a big sphere and its neighboring small sphere. Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque. If θ is the angle of twist of the suspended wire, the restoring torque is proportional to θ, equal to τ θ, where τ is the restoring couple per unit angle of twist. τ can be measured independently e.g., by applying a known torque and measuring the angle of twist. The gravitational force between the spherical balls is the same as if their masses are concentrated at their centres. Thus if d is the separation between the centres of the big and its neighboring small ball, M and m their masses, the gravitational forces between the big sphere and its neighboring small ball is
F = G(Mm/d^{2}) …………………………………………..(6)
If L is the length of the bar AB, then the torque arising out of F is F multiplied by L. At equilibrium, this is equal to the restoring torque and hence
G(Mm/d^{2})L = τθ …………………………………………..(7)
Observation of θ thus enables one to calculate G from this equation.
Since Cavendeish’s experiment, the measurement of G has been refined and the currently accepted value is
G = 6.67 × 10^{–}^{11} N m^{2}/kg^{2} …………………………………………..(8)
Acceleration due to Gravity of Earth
The earth can be imagined to be a sphere made of a large number of concentric spherical shells with the smallest one at the centre and the largest one at its surface. A point outside the earth is obviously outside all the shells. Thus, all the shells exert a gravitational force at the point outside just as if their masses are concentrated at their common centre according to the result stated in the last section. The total mass of all the shells combined is just the mass of the earth. Hence, at a point outside the earth, the gravitational force is just as if its entire mass of the earth is concentrated at its centre.
For a point inside the earth, the situation is different. This is illustrated in Fig. 7.
Again consider the earth to be made up of concentric shells as before and a point mass m situated at a distance r from the centre. The point P lies outside the sphere of radius r, the point P lies inside. Hence according to result stated in the last section, they exert no gravitational force on mass m kept at P. The shells with radius ≤ r make up a sphere of radius r for which the point P lies on the surface. This smaller sphere therefore exerts a force on a mass m_{r }is concentrated at the centre. Thus the force on the mass m at P has a magnitude
F = Gm(m_{r}/r^{2}) …………………………………………..(9)
We assume that the entire earth is of uniform density and hence its mass is M_{E} = 4π/3 (R_{z})^{3}ρ where M_{E }is the mass of earth R_{E }is its radius and ρ is the density. On the other hand the mass of the sphere M_{r }of radius r is 4π/3 ρr^{3 }and hence
F = Gm(4π/3 ρ) r^{3}/r^{2 }= Gm[M_{E}/(R_{E})^{3}] r^{3}/r^{2}
=Gm[M_{E}/(R_{E})^{3}] r …………………………………………..(10)
If the mass m is situated on the surface of earth, then r =R_{E }and the gravitational force on it is, from Eq. (10)
F = G[M_{E}m/(R_{E})^{2}] ………………………………………….(11)
The acceleration experienced by the mass m, which is usually denoted by the symbol g is related F by Newton’s 2nd law by relation F = mg. Thus
g = F/m = GM_{E}/(R_{E})^{2} …………………………………………(12)
Acceleration g is readily measurable. R_{E }is a known quantity. The measurement of G by Cavendish’s experiment (or otherwise), combined with knowledge of g and R_{E }enables one to estimate M_{E }from Eq. (12). This is the reason why there is a popular statement regarding Cavendish: “Cavendish weighed the earth”
Acceleration due to Gravity below and above the surface of Earth
Consider a point mass m at a height h above the surface of the earth as shown in Fig 8(a). The radius of the earth is denoted by R_{E }. Since this point is outside the earth.
g at a height h above the surface of the earth.
Its distance from the centre of the earth is (R_{E }+ h). If F(h) denoted the magnitude of the force on the point mass m, we get from Eq. (5):
F(h) = GM_{E }m/(R_{E }+h)² …………………………………………….(13)
The acceleration experienced by the point mass is F(h)/m = g(h) and we get
g(h) = F(h)/m = GM_{E }/(R_{E }+h)² ………………………………..(14)
This is clearly less than the value of g on the surface of earth: g = GM_{E }/(R_{E })². For h << R_{E }, we can expand the RHS of Eq. (14):
g(h) = GM/(R_{E })²(1+h/R_{E })² = g(1+h/R_{E })^{2 }
For h/R_{E }<<1, using binomial expression,
g(h) ≅ g(12h/R_{E }) …………………………………………………….(15)
Equation (15) thus tells us that for small heights h above the value of g decreases by a factor (12h/R_{E }).
Now, consider a point mass m at a depth d below the surface of the earth (Fig. 8(b)), so that its distance from the centre of the earth is (R_{E }– d) as shown in the figure. The earth can be thought of as being composed of a smaller sphere of radius (R_{E }– d) and a spherical shell of thickness d. The force on m due to the outer shell of thickness d is zero because the result quoted in the previous section. As far as the smaller sphere of radius (R_{E }– d) is concerned, the point mass is outside it and hence according to the result quoted earlier, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre. If M_{s }is the mass of the smaller sphere, then,
M_{s }/M_{E }= (R_{E }– d)³/(R_{E })³ …………………………………………….(16)
Since the mass of a sphere is proportional to be cube of its radius.
In this case only the smaller sphere of radius (R_{E }– d) contributes to g.
Thus the force on the point mass is
F(d) = G M_{s }m/(R_{E }– d)² ………………………………………….(17)
Substituting for M_{s }from above, we get
F(d) = G M_{E }m(R_{E }– d)/(R_{E })³_{ } ……………………………………(18)
and hence the acceleration due to gravity at a depth d,
g(d) = F(d)/m is
g(d) = F(d)/m = G M_{E }/(R_{E })³(R_{E }– d)
= g (R_{E }– d)/R_{E }= g(1d/R_{E }) ……………………………………..(19)
Thus, as we go down below earth’s surface, the acceleration due to gravity decreases by a factor (1d/R_{E }). The remarkable thing about acceleration due to earth’s gravity is that it is maximum on its surface decreasing whether you go up or down.
Gravitational Potential Energy
We had discussed earlier the notion of potential energy as being the energy stored in the body at its given position. If the position of the particle changes on account of forces acting on it, then the change in its potential energy is just the amount of work done is independent of the path are the conservative forces.
The force of gravity is a conservative force and we can calculate the potential energy of a body arising out of this force, called the gravitational potential energy. Consider points close to the surface of earth, at distances from the surface much smaller than the radius if the earth. In such cases, the force of gravity is practically constant equal to mg. directed towards the centre of the earth. If we consider a point at a height h_{1 }from the surface of the earth and another point vertically above it at a height h_{2 }from the surface, the work done in lifting the particle of mass m from the first to the second position is denoted by W_{12}
W_{12 }= Force × displacement
= mg (h_{2 }h_{1 }) …………………………………………………(20)
If we associate a potential energy W(h) at a point at a height h above the surface such that
W(h) =mgh + W_{0 }…………………………………………..(21)
(where W_{0 }= constant):
then it is clear that
W_{12 }= W (h_{2 }) W (h_{1 }) ………………………………………(22)
The work done in moving the particle is just the difference of potential energy between its final and initial positions. Observe that the constant W_{0 }cancels out in Eq. (22). Setting h = 0 in the last equation, we get W(h=0) = W_{0 }. h = 0 means points on the surface of the earth. Thus, W_{0 }is the potential energy on the surface of the earth. If we consider points at arbitrary distance from the surface of the earth, the result just derived is not valid since the assumption that gravitational force mg is a constant is no longer valid. However from our discussion we know that a point outside the earth, the force of gravitation on a particle directed towards the centre of the earth is
F_{ }= GM_{E }m/r² …………………………………………………..(23)
where M_{E }= mass of earth. m = mass of the particle and r its distance from the centre of the earth. If we now calculate the work done in lifting a particle from r = r_{1 }to r = r_{2 }t(r_{2 }> r_{1 }) along a vertical path, we get instead of Eq. (20),
In place of Eq. (21), we can thus associate a potential energy W(r) at a distance r, such that
W(r) = GM_{E }m/r + W_{1} ………………………………………(25)
valid for r>R,
so that once again W_{12 }= W(r_{2 }) – W(r_{1 }).
Setting r =infinity in the last equation, we get W (r = infinity) = W_{1}. Thus, W_{1} is the potential energy between two has different meaning from Eqs. (22) and (24). One conventionally sets W_{1 }equal to zero, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point. We have calculated the potential energy at a point of a particle due to gravitational forces on it due to the earth and it is proportional to the mass of the particle. The gravitational potential due to the gravitational force of the earth is defined as the potential energy of a particle of unit mass at that point. From the earlier discussion, we learn that the gravitational potential energy associated with two particles of masses m_{1 }and m_{2 }separated by distance by a distance r is given by
V = – G m_{1 }m_{2}/r (if we choose V = 0 as r →∞)
It should be noted that an isolated system of particles will have the potential energy that equals the sum of energies (given by the above equation) for all possible pairs of its constituent particles. This is an example of the application of the superposition principle.
Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square?
Answer
Consider four masses each of mass m at the corners of a square of side l; See Fig. 9. We have four mass pairs at distance l and two diagonal pairs at a distance √2,
Hence,
W(r) = 4Gm^{2}/l – 2Gm^{2}/√2l
= 2Gm^{2}/l(2+1/√2) = –5.41 Gm^{2}/l
The gravitational potential at the centre of the square (r = √2l/2) is
U(r) = 4√2Gm/l
Escape Speed
If a stone is thrown by hand, we see it falls back to the the earth. Of course using machines we can shoot an object with much greater speeds and with greater and greater initial speed, the object scales higher and higher heights. A natural query that arises in our mind is the following:
‘can we throw an object with such high initial speeds that it does not fall back to the earth?’
The principle of conservation of energy helps us to answer this question. Suppose the object did reach infinity and that its speed there was V_{f}. The energy of an object is the sum of potential and kinetic energy. As before W_{1 }denotes that gravitational potential energy of the object at infinity. The total energy of the projectile at infinity then is
E (∞) = W_{1 }+ m(V_{f})²/2 ……………………………………………….(26)
If the object was initially with a speed V_{i }from a point at a distance (h+ R_{E }) from the centre of the earth (R_{E }= radius of the earth), its energy initially was
E (h +R_{E} ) = m(V_{i})²/2 – GmM_{E}/(h + R_{E}) + W_{1 } ……………….(27)
By the principle of energy conservation Eqs. (26) and (27) must be equal. Hence
m(V_{i})²/2 – GmM_{E}/(h + R_{E}) ≥ 0 …………………………………… (29)
The minimum value of V_{i }corresponds to the case when the LHS of Eq.(29) equals zero.
Thus, the minimum speed required for an object to reach infinity (i.e., escape from the earth) corresponds to
m{(V_{i})²}_{min} = GmM_{E}/(h + R_{E}) ≥ 0 …………………………………… (30)
If the object is thrown from the surface of the earth, h = 0, and we get
(V_{i})_{min }= √(2GM_{E})/R_{E} ………………………………………(31)
Using the relation g = GM_{E}/(R_{E})^{2} ………………………………..(32)
Using the value of g and R_{E }, numerically (V_{i})_{min}_{≅ }11.2 km/s. This is called the escape speed, sometimes loosely called the escape velocity.
Equation (32) applies equally well to an object thrown from the surface of the moon with g replaced by the acceleration due to Moon’s gravity on its surface and r_{E }replaced by the radius of the moon. Both are smaller their values on earth and the escape speed for the speed for the moon turns out to be 2.3 km/s, about five times smaller. This is the reason that 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.
Example 4
Two uniform solid spheres of equal radii R, but mass M and 4 M have a centre separation 6 R, as shown in Fig. 10. The two spheres are held fixed. A projectile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the the minimum speed v of the projectile so that it reaches the surface of the second sphere.
Answer
The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see Fig.10) is defined as the position where the two forces cancel each other exactly. If ON = r, we have
GMm/r² = 4GMm/(6R – r)²
(6Rr)² = 4r²
6Rr = + 2r
r = 2R or 6R
The neutral point r = 6R does not concern us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a speed which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface of M is
E_{i}= 1/2mv² GMm/R –4GMm/5R
At the neutral point N, the speed approaches zero. The mechanical energy at N is purely potential.
E_{N }= 1/2GMm/2R – 4GMm/4R
From the principle of conservation of mechanical energy
1/2v² GM/R –4GMm/5R = GM/2R – GM/R
or
v² = –2GM/R (4/5 1/2)
v = √(3GM/5R)
A point to note is that the speed of the projectile is zero at N, but is nonzero when it strikes the heavier space 4M. The calculation of this speed is left as an exercise to the students.
Earth Satellites
Earth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun and hence Kepler’s laws of planetary motion are equally applicable to them. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis. Since, 1957, advances in technology have enabled many countries including India to launch artificial earth satellites for practical use in fields like telecommunication, geophysics and meteorology.
We will consider a satellite in circular orbit of a distance (R_{E}+ h) the centre of the earth, where R_{E }= radius of the earth. If m is the mass of the satellite and V its speed, the centripetal force required for this orbit is
F(centripetal) = mV²/ (R_{E}+ h) ……………………………………….(33)
directed towards the centre. This centripetal force is provided by the gravitational force, which is
F = GmM_{E}/ (R_{E}+ h)² ……………………………………………………..(34)
where M_{E }is the mass of the earth.
Equating R.H.S of Eqs. (33) and (34) and cancelling out m, we get
V²= GM_{E}/ (R_{E}+ h) ……………………………………………………..(35)
Thus V decreases as h increases. From equation (35), the speed V for h = 0 is
V² (h=0) = GM/ R_{E }= gR_{E} ……………………………………….(36)
where we have used the relation g = GM/ (R_{E })². In every orbit, the satellite traverses a distance 2π(R_{E}+ h)/V with speed V. Its time period T therefore is
T=2π(R_{E}+ h)/V= 2π(R_{E}+ h)^{3/2}/√(G M_{E}) …………………..(37)
on substitution of value of V from Eq. (37), we get
T²= k(R_{E}+ h)^{3}/(where k = 4π²/GM_{E}) …..…………………..(38)
which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a satellite very close to the surface of earth h can be neglected in comparison to R_{E }in Eq. (38).
Hence, for such satellites, T is T_{0}, where
T_{0 }= 2π√(R_{E}/g) ……………………………………………………..(39)
If we substitute the numerical values g: 9.8 m s^{2 }and R_{E }= 6400 km., we get
T_{0 }= 2π√(6.4 × 10^{6 }/9.8) s which is approximately 85 minutes.
Example 5
The planet Mars has two moons, phobos and demos. (i) phobos has a period of 7 hours, 39 minutes and an orbital radius of 6.4 × 10^{3 }km. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martin orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days?
Answer
(i) We employ Eq. (38) with the sun’s mass replaced by the martian mass M_{m}
T ² = (4π²/GM_{m})R³
M_{m }= 4π²R³/GT²
= 4 × (3.14)² × (9.4)³ × (10)^{18}/6.67 × 10^{11}×(459 × 60)²
M_{m }= 4 × (3.14)² × (9.4)³ × (10)^{18}/6.67 × (459 × 6)² × 10^{5}
= 6.48 × 10^{23 }kg
(ii) Once again Kepler’s third law comes to our aid,
(T_{M})²/(T_{E})² = (R_{MS})³/ (R_{ES})³ where R_{MS }is the marssun distance and R_{ES }is the earth – sun distance.
∴ T_{M }= (1.52)^{3/2} × 365
= 684 days
We note that the orbits of all planets except Mercury, Mars and Pluto are very close to being circular. For example, the ratio of the semiminor to semimajor axis for our Earth is b/a = 0.99986.
Example 6
Weighing the Earth: You are given the following data: g = 9.81 ms^{2}. R_{E }= 6.37 × 10^{6 }m, the distance to the moon R_{ }= 3.84 × 10^{8 }m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth M_{E }in two different ways.
Answer
From Eq. (12) we have
M_{E}= g(R_{E})^{2} /G
=9.81 × (6.37 × 10^{6 })²/(6.67 × 10^{11}
= 5.97 × 10^{24}kg.
The moon is a satellite of the Earth. From the derivation of Kepler’s third law (see Eq. (38))
T²= 4π²R³/GM_{E}
M_{E }= 4π²R³/GT²
= 4 × 3.14× 3.14 × 3.84³× 10^{24}/(6.67 × 10^{11}× (27.3×24 × 60 ×60)²
= 6.02 × 10^{24 }kg
Both methods yield almost the same answer, the difference between them being less than 1%.
Example 7
Express the constant k of Eq. (38) in days and kilometers. Given k = 10^{13 }s^{2 }m^{3 }.The moon is at a distance of 3.84× 10^{5 }km from the earth. Obtain its timeperiod of revolution in days.
Answer
Given k = 10^{13 }s^{2 }m^{3 }
= 10^{13 }[{1/(24 × 60 ×60)²}d²][1/(1000)³ km³]
= (1.33 × 10^{14 })(3.84 × 10^{5})³
Using Eq. (38) and the given value of k, the time period of the moon is
T² = (1.33 × 10^{14})(3.84× 10^{5)})³
T = 27.3 d
Note that Eq. (38) also holds for elliptical orbits if we replace (R_{E }+ h) by the semimajor axis of the ellipse. The earth will then be at one of the foci of this ellipse.
Energy of an Orbiting Satellite
Using Eq. (35), the kinetic energy of the satellite in a circular orbit with speed v is
KgE = 1/2 mv²
= GmM_{E }/2(R_{E }+ h) ………………………………………………..(40)
Considering gravitational potential energy at infinity to be zero, the potential energy at distance (R_{0}+ h) from the centre of the earth is
P.E =GmM_{E }/2(R_{E }+ h) …………………………………………….(41)
The K.E is positive whereas the P.E is negative. However, in magnitude the K. E is half the PE so that total E is
E = K.E +P.E =GmM_{E }/2(R_{E }+ h) ……………………………….(42)
The total energy of a circularly orbiting satellite is thus negative, with the potential energy being negative but twice is magnitude of the positive kinetic energy.
When the orbit of a satellite becomes elliptic, both the KE and PE vary from point to point, the total energy which remains constant as in the case of circular orbit case. This is what we expect, since as we have discussed before if the total energy is positive or zero, the object escapes to infinity. Satellites are always at infinite distance from the earth and hence their energies cannot bee positive or zero.
Example 8
A 400 kg satellite is in a circular orbit of radius 2R_{E }about the earth. How much energy is required to transfer it to a circular orbit of radius 2R_{E }? What are the changes in kinetic and potential energies?
Answer
Initially,
E_{i} =GmM_{E }/4R_{E }
while finally,
E_{f} =GmM_{E }/8R_{E }
The change in total energy is
ΔE = E_{f} – E_{i}
= GmM_{E }/8R_{E } = [(GmM_{E }/(R_{E })²]mR_{E }/8
ΔE = gmR_{E }/8 = 9.81 × 400 × 6.37 × 10^{6}/8 = 3.13 × 10^{9}J
The kinetic energy is reduced and its mimics ΔE, namely, ΔK = K_{f} – K_{i}
= 3.13 × 10^{9}J
The change in potential energy is twice the change in the total energy, namely
ΔV = V_{f} – V_{i}= 6.25 × 10^{9}J
Geostationary and Polar Satellites
An interesting phenomenon arises if in we arrange the value of (R_{E} + h) such that T in Eq. (37) becomes equal to 24 hours. If the circular orbit is in the equatorial plane of the earth, such a satellite, having the same period as the period of rotation of the earth about its own axis would appear stationary viewed from a point on earth. The (R_{E} + h) for this purpose is works out to be large as compared to R_{E} :
R_{E} + h = (T²GM_{E} /4π ²)^{1/3} …………………………(43)
and for T = 24 hours. h works out to be 35800 km., which is much larger than R_{E}. Satellites in a circular orbit around the earth in the equatorial plane with T=24 hours are called Geostationary Satellites. Clearly, since the earth rotates with the same period, the satellite would appear fixed from any point on earth. It takes very powerful rockets to throw up a satellite to such large heights above the earth but this has been done in view of the several benefits of many practical applications.
India’s Leap into Space
India started its space programme in 1962 when Indian National Committee for Space Research was set up by the Government of India which was superseded by the Indian Space Research Organisation (ISRO) in 1969. ISRO identified the role and importance of space technology in nation’s development and bringing space to the service of the common man . India launched its low orbit satellite Aryabhata in 1975., for which the launch vehicle was provided by the erstwhile Soviet Union. ISRO started employing its indigenous launching vehicle in 1979 by sending Rohini series of satellites into space from its main launch site at Satish Dhawan Space Center, Sriharikota, Andhra Pradesh.
The tremendous progress in India’s space programme has made ISRO one of the six largest space agencies in the world. ISRO develops and delivers application specific satellite products and tools for broadcasts, communication, weather forecasts, disaster management tools, Geographic Information System, cartography, navigation, telemedicine, dedicated distance education satellite etc. In order to achieve complete selfreliance in these applications, cost effective and reliable Polar Satellite Launch Vehicle (PSLV) was developed in early 1990s. PSLV has thus become a favored carrier for satellites of various countries, promoting unprecedented international collaboration. In 2001, the Geosynchronous Satellite Launch Vehicle (GSLV) was developed for launching heavier and more demanding Geosynchronous communication satellites. Various research centers and autonomous institutions for remote sensing, astronomy and astrophysics, atmospheric sciences and space research are functioning under the aegis of the Department of Space, Government of India. Success of lunar (Chandrayaan) and inter planetary (Mangalyaan) missions along with other scientific projects has been landmark achievements of ISRO. Future endeavors of ISRO include human space flight projects, the development of heavy lift launchers, reusable launch vehicles, semicryogenic engines, single and two stage to orbit (SSTO and TSTO) vehicles, development and use of composite materials for space application etc. In 1984 Rakesh Sharma became the first Indian to go into outer space aboard in a USSR spaceship.
It is known that electromagnetic waves above a certain frequency are not reflected from ionosphere. Radio waves used for radio broadcast which are in the frequency range 2 MHz to 10 MHz, are below the critical frequency. They are therefore reflected by the ionosphere. Thus radio waves broadcast from an antenna can be received at points far away where the direct wave fail to reach on account of the curvature of the earth. Waves used in television broad cast or other forms of communication have much higher frequencies 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.
A strip on earth’s surface (shown shaded) is visible from the satellite during one cycle. For the next revolution of the satellite, the earth has rotated a little on its axis so that an adjacent strip becomes visible.
Another class of satellites are called Polar satellites (Fig. 11). These are low altitude (h ≈500 to 800 km) satellites, but they go around the poles of the earth in a northsouth direction whereas the earth rotates around its axis in an eastwest direction. Since its time period is around 100 minutes it crosses any altitude many times a day. However, since its height h above the earth is about 500800 km, a camera fixed on it can view only small strips of of the earth in one orbit, so that in effect the whole earth can be viewed strip by strip during the entire day. 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.
Weightlessness
Weight of a object is the force with which the earth attracts it. We are conscious of our own weight when we stand on a surface, since the surface exerts a force opposite to our weight to keep us at rest. The same principle holds good when we measure the weight of an object by a spring balance hung from a fixed point e.g., the ceiling. The object would fall down unless it is subject to a force opposite to gravity. This is exactly what the spring exerts on the object. This is because the spring is pulled down a little by the gravitational pull of the object and in turn the spring exerts a force on the object vertically upwards.
Now, imagine that the top end of the balance is no longer held fixed to the top ceiling of the room. Both ends of the spring as well as the object move with identical acceleration g. Th spring spring is not stretched and does not exert any upward force on the object which is moving down with acceleration g due to gravity. The reading recorded in the spring balance is zero since spring is not stretched at all. If the object were a human being, he or she will not feel his weight since there is no upward force on him. Thus when an object is in free fall, it is weightless and this phenomenon is usually called the phenomenon of weightlessness.
In a satellite around the earth, every part and parcel of the satellite has an acceleration towards the centre of the earth which is exactly the value of 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 reflect show this fact.