8.0 Preview:

• This chapter revolves around Gravitation and various laws that govern this force of attraction.
• We begin by stating the three laws of Kepler, which describe planetary motion. First law—Law of orbits: All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse. Second law—Law of areas: The line that joins any planet to the sun sweeps equal areas in equal intervals of time as it revolves. Third law—Law of periods: 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.
• We will then study Newton’s law of gravitation in some depth called, Universal Law of Gravitation and is stated as: “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”.
• Henry Cavendish’s experiment will be illustrated to explain the planetary attraction and to arrive at the value of Gravitational constant.
• We shall obtain equations for acceleration of gravity on an object at a certain height, “h” above the ground and at a depth, “d” below the ground.
• Next, we define the work done in raising an object from one height to another from the difference in potential energies of the object corresponding to those heights. We will again talk of gravitational potential at a point due to the gravitational force of the earth as the potential energy of a particle of unit mass at that point.
• Further, we shall learn about an interesting concept called, Escape velocity as the least velocity which a body must have in order to escape the gravitational attraction of a particular planet or of any other object is called escape velocity.
• We will cover some of the interesting aspects of a satellite like time period of revolution and the energy of an orbiting satellite; types of satellites: Geostationary Satellites and Polar Satellites. Satellites in circular orbits around the earth in the equatorial plane with T = 24 hours are called Geostationary Satellites while satellites located at close radii of the earth which record images of landforms in strips are called Polar Satellites. We shall also look at their practical applications.
• The phenomenon of “weightlessness” will be dealt with. Weightlessness occurs when there is no force of support (or reaction force) on our body. When we say, a body is in “free fall”, it means that it is accelerating downward under the acceleration of gravity. The knowledge of weightlessness will be used to answer questions such as, why astronauts in space feel weightless!

8.1 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 so on.

Physicist, Galileo conducted 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.

The earliest recorded model for planetary motions proposed by Ptolemy was a ‘geocentric’ model in which all celestial objects, stars, the sun and the planets, all revolved around the earth. However, a more elegant model in which the Sun was the centre around which the planets revolved – the ‘heliocentric’ model – was mentioned by Aryabhatta. Nicholas Copernicus then proposed a definitive model in which planets moved in circles and was supported by Galileo. Tycho Brahe recorded observations of planets and his assistant, Johannes Kepler analysed the compiled data to propose three laws, which is now known as Kepler’s laws. These laws enabled Newton to make a great leap in proposing the universal law of gravitation.

8.2 Kepler’s Laws

The three laws of Kepler can be stated as follows:

First law—Law of orbits: All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse (Refer Figure: 8.1a).

• The ellipse can be drawn very simply as follows:
• Refer Fig. 8.1(b), select two points: $F_1$ and $F_2$. Take a certain 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 until it is tightly stretched and then draw a curve by moving the pencil keeping the string taut (or tightly stretched) throughout.
• The closed curve we get is called an ellipse. Clearly, for any point T on the ellipse, the sum of the distances from $F_1$ and $F_2$ is a constant. $F_1, F_2$ are called the foci. Join the points F1 and F2 and extend the line to intersect the ellipse at points P and A as shown in Fig. 8.1(b). The midpoint of the line PA is the centre of the ellipse O and the length, PO = AO is called the semi major axis of the ellipse.

Note box: 1. For a circle, the two foci merge into one and the semi-major axis becomes the radius of the circle. 2. Copernic’s model allowed only circular orbits.

Second law—Law of areas: The line that joins any planet to the sun sweeps equal areas in equal intervals of time as it revolves (refer Figure 8.2 above). This law comes from the observations that planets appear to move slower when they are farther from the sun than when they are nearer.

Third law—Law of periods: 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.
The table below gives the approximate time periods of revolution of nine planets around the sun along with values of their semi-major axes.

Table 1: Data from measurement of planetary motions given below confirm Kepler’s Law of Periods

a = Semi-major axis in units of $10^{10}$ m.

T = Time period of revolution of the planet in years(y).

Q = The quotient $( T^2/a^3 )$ in units of $10^{-34} y^2 m^{-3}$.

• The law of areas can be considered as a consequence of conservation of angular momentum which is valid for any central force.
• Again, refer figure 8.2b, 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 denoted by r and p, respectively. Then the area swept out by the planet of mass, m in time interval, Δt is ΔA given by:
  ΔA = ½ (r × vΔt)…………….. (8.1)
  Hence,
  ΔA /Δt =½ (r × p)/m ; (since, v = p/m)
  = L / (2 m); (since, angular momentum= radius x linear momentum)……….. (8.2)
  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 the second law, the law of areas. Gravitation is a central force and hence the law of areas follows.

Note box: Kepler was also known as the founder of geometrical optics, being the first to describe what happens to light after it enters a telescope.

Example 8.1: Let the speed of the planet at the perihelion P in Fig. 8.1(a) be $v_P$ and the Sun-planet 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?

Solution: 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 $v_p$ are mutually perpendicular. Similarly, $L_A = m_p r_A v_A$. From angular momentum conservation
$$ m_p r_p v_p = m_p r_A v_A $$
$$ or \space \frac {v_p}{v_A} = \frac {r_A}{r_p} $$
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. 8.1. From Kepler’s second law, equal areas are swept in equal times. Hence the planet will take a longer time to traverse BAC than CPB.

Questions for section 8.2:

1. State Kepler’s first law. Explain the construction of ellipse with a diagram.
2. State Kepller’s second law. On the basis of what observation was it arrived at?
3. State Kepler’s third law.

8.3 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. The law helped explain terrestrial gravitation as well as Kepler’s laws.
• Newton reasoned that, the moon revolving in an orbit of radius Rm, was subject to a centripetal acceleration, am due to earth’s gravity having a magnitude;
  $$ a_m = \frac {V^2}{R_m}= \frac {4 \pi^2 R_m}{T^2} ……(8.3)$$
  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 × 108m. By substituting these values in equation (8.3), we get a value of “$a_m$” much smaller than the value of acceleration due to gravity, “g” on the surface of the earth, although “g” also arises due to earth’s gravitational attraction.

Central Forces

• We know the time rate of change of the angular momentum of a single particle about the origin is
  dl/dt = r x F
• The angular momentum of the particle is conserved, if the torque τ = r × F because 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 condition, F along r.

Tr is the trejectory of the particle under the central force. 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 P′, arc PP′ = Δs = v Δt. 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 POP′ ≈ (r sinα ) PP′ /2 = (r v sin a) Δt/2.)

• Central forces always lie along r. A ‘central’ force is always directed towards or away from a fixed point, that is, along the position vector of the point of application of the force with respect to the fixed point (See Figure above). Further, the magnitude of a central force F depends on r, the distance of the point of application of the force from the fixed point (or), F = F(r).
• In the motion under a central force the angular momentum is always conserved. Two important results follow from this:
  (1) The motion of a particle under the central force is always in a plane.
  (2) The position vector of the particle with respect to the centre of the force (that is, the fixed point) has a constant areal velocity. In other words, the position vector sweeps out equal areas in equal time intervals as the particle moves under the influence of the central force.
  The areal velocity is given by :
  dA/dt = ½ r v sin α.
• An immediate application of the above discussion can be made to the motion of a planet under the gravitational force of the sun. For convenience, the sun may be taken to be so 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 it is given by:
  $$ F = G m_1m_2/r^2;$$
  Where, $m_1$ and $m_2$ are respectively the masses of the planet and the sun; “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 itself!
• 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: $a_m = aR_m^{-2}$;$ g = aR_E^{-2}$. Thus, we get,
  $$ \bigg( \frac {g}{a_m} \bigg) = \bigg( \frac {R^2_m}{R^2_E} \bigg) $$
  $$ \frac {g}{a_m} = \frac {R^2_m}{R^2_E};3600 …….(8.4) $$
• The above equation is found to be in agreement with the value of g = 9.8ms-2 and the value of am from Equation 8.3.
• These observations led Newton to propose the following 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.
• 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;

Let $ \vec F $ be the force on a point $m_2$ due to another point mass $m_1$ at a distance r.The magnitude of this force is given by $ F = \frac {Gm_1m_2}{r^2} ….(8.5)$

where G is the universal gravitational constant.In vector form,this equation can be expressed as,
$ \vec F = \frac {Gm_1m_2}{r^2} ( – \hat r)=- \frac {Gm_1m_2}{r^2} \hat r $

where $ \hat r $ is the unit vector from $m_1$ to $m_2$

In the figure $ \vec r = r_2-r_1$.Since $ \vec r = r \hat r $,

$$ \vec F = – \frac {Gm_1m_2}{r^3}\vec r $$

• The gravitational force is attractive, that is, the force F is along “– r”. The force on point 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 body 2 and $F_{21}$ on the body 2 due to 1 are related as $F_{12} = – F_{21}$.
• Before we can apply Equation-8.5 (or, Newton’s law of gravitation) 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.
• Thus, 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 the other point masses on it as shown in Fig 8.4 below:

Figure 8.4: Gravitational force on point mass $m_1$ is the vector sum of the gravitational forces exerted by $m_2, m_3$ and $m_4$.

• The total force on $m_1$ is;
  $$ F_1 = \frac {Gm_2m_1}{r^2_{21}}r_{21}+\frac {Gm_3m_1}{r^2_{31}}r_{31}+\frac {Gm_4m_1}{r^2_{41} } r_{41} $$

Example 8.2: Three equal masses of m kg 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. 8.5)

Solution:

(a) The angle between GC and the positive x-axis is 30° and so is the angle between GB and the negative x-axis. The individual forces in vector notation are
$$ F_{GA} = \frac {Gm(2m)}{1} \hat j $$
$$ F_{GB} = \frac { Gm(2m)}{1} ( – \hat i \space cos 30^o – \hat j \space sin 30^o ) $$
$$ F_{GC} = \frac {Gm(2m)}{1} ( + \hat i \space cos 30^o – \hat j \space sin 30^o ) $$

From the principle of superposition and the law of vector addition, the resultant gravitational force $F_{R}$ on (2m) is

$$ F_R = F_{GA} + F_{GB }+ F_{GC} $$

$$ F_R =2Gm^2 \hat j + 2Gm^2 ( – \hat i \space cos 30^o – \hat j \space sin 30^o) + 2Gm^2( \hat i \space cos 30^ο − \hat j \space sin 30^ο)= 0 $$

Alternatively, one expects on the basis of symmetry that the resultant force ought to be zero.

(b) By symmetry the x-component of the force cancels out. The y-component survives.
$$ F_R = 4Gm^2 \hat j − 2Gm^ 2 \hat j = 2Gm^2 \hat j $$

• For the gravitational force between an extended object (like the earth) and a point mass, Equation (8.5) is not directly applicable. This is because, each point mass in the extended object (earth) will exert a force on the given point mass and these forces will not all be in the same direction.
• Thus, we have to add up these forces (by vector addition), coming from all the point masses in the extended object to get the total force.
• This is easily done using calculus. Let us apply this to two special cases, to arrive at a simple law;

(1) 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. The components perpendicular to this line cancel out when summing over all the forces at different regions of the shell, leaving only a resultant force along the line joining the point to the centre.

(2) 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 (that is, the point masses that constitute the outer shell) attract the point mass inside it in various directions. These forces cancel each other completely.

Concept box:  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 of 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 work in 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 Principia for the Common Reader brings into sharp focus the beauty, clarity and breath-taking economy of Newton’s methods.

Questions for section 8.3:

1. Give the expression for centripetal acceleration caused due to acceleration due to gravity of the earth, on the moon.
2. State the Universal law of gravitation. Explain with necessary equations, its significance.

8.4 The Gravitational Constant

The value of the gravitational constant, G by the Universal law of gravitation can be determined experimentally and this was first done by English scientist Henry Cavendish. We will now describe Cavendish’s experiment;

• Refer figure, 8.6; the bar, AB has two small lead spheres attached at its ends (show attachment).
• The bar is suspended from a rigid support by a fine wire. Two large lead spheres are brought close 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 (no differentiation in sizes).
• 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 neighbouring small sphere.
• Due to this torque, the suspended wire gets twisted, until the restoring torque of the wire equals the gravitational torque (and the wire now begins to unwind).
• If θ is the angle of twist of the suspended wire, the restoring torque is proportional to θ. Thus, Restoring Torque= τθ.
  Where, “τ” is the restoring couple per unit angle of twist. τ can be measured independently by applying a known torque and measuring the angle of twist.
• The gravitational force between the spherical balls is as if their masses are concentrated at their centres.
• Thus, if “d” is the separation between the centres of the big and its neighbouring small ball, with masses M and m respectively, the gravitational force between the big sphere and its neighbouring small ball is;
  $$ F=G(mM/d^2) ……..(8.6)$$
  If “L” is the length of the bar AB, then the torque arising out of F is: F times L;
  At equilibrium, this is equal to the restoring torque and hence,
  $$ G \bigg( \frac {Mm}{d^2} \bigg) L= τθ ……(8.7) $$
  Observing θ thus enables one to calculate G from this equation.

Note box: The value of G has been refined and the currently accepted value is: $$ G = 6.67 × 10^{-11} \space  N  \space m^2/kg^2…………………….      (8.8)  $$

8.5 Acceleration due to Gravity of the Earth

• The earth can be imagined to be a sphere made of a large number of concentric spherical shells with the smallest circle at the centre and the largest circle being earth’s surface.
• A point outside earth is obviously outside all the shells.
• Thus, all the shells exert a gravitational force together at the point outside as if their masses are concentrated at the 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, we can generalize that for a point outside the earth, the gravitational force would be exerted as though the entire mass of the earth is concentrated at its centre.

• For a point inside the earth, the situation is different. This is illustrated in Figure- 8.7 above.
• Again, consider the earth to be made up of concentric shells as before and let us assume a point mass, m situated at a distance, r from the centre. Let this point be point P, which lies outside the sphere of radius, r.
• However, for the shells of radius greater than r, the point P lies inside. 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 (surface of the inner circle). This smaller sphere therefore exerts a force on a mass m at P as if its mass “mr” is concentrated at the centre. Thus, the force, F on the mass m, at P has a magnitude;

$$ F = \frac {Gm(m_r)}{r^2} …….(8.9) $$

We assume that the entire earth is of uniform density and hence its mass is $M_E = \frac {4 \pi}{3} R^3_E \rho $ where $M_E$ is the mass of the 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 $ \frac {4 \pi}{3} \rho r^3 $ and hence
$$ F = Gm \bigg( \frac {4 \pi}{3} \rho \bigg) \frac {r^3}{r^2}=Gm \bigg( \frac {M_E}{R^3_E} \bigg) \frac {r^3}{r^2} $$
$$=\frac {GmM_E}{R^3_E}r ….(8.10)$$
If the mass m is situated on the surface of earth, then $r = R_E$ and the gravitational force on it is, from Eq. (8.10)
$$F=G \frac {M_Em}{R^2_E} ….(8.11)$$
The acceleration experienced by the mass m, which is usually denoted by the symbol g is related to F by Newton’s 2nd law by relation F = mg. Thus
$$ g = \frac {F}{m}=\frac {GM_E}{R^2_E} …(8.12)$$
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 Equation-8.12. This is the reason why there is a popular statement regarding Cavendish: “Cavendish weighed the earth”.

Questions for the sections: 8.4 and 8.5:

1. Describe Henry Cavendish’s experiment to determine Gravitational constant, G.
2. Arrive at equation for acceleration due to gravity, “g” in the form of equation, 8.12; with the illustration of concentric model of earth.

8.6 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 Figure- 8.8(a) above. The radius of the earth is denoted by $R_E$.
• Since this point is outside 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 Equation 8.5;

$$ F(h)=(GM_Em)/(R_e+h)^2 ……….. (8.13) $$

• The acceleration experienced by the point mass is F(h)/m $\equiv $ g(h)and we get;

$$ g(h)= \frac {F(h)}{m} = \frac {GM_E}{(R_E+h)^2} …….(8.14) $$
This is clearly less than the value of g on the surface of earth: $g=\frac {GM_E}{R^2_E}$.For , E h<<R we can expand the RHS of Eq. (8.14) :
$$ g(h)= \frac {GM}{R^2_E(1+h/R_E)^2} = g(1+h/R_E)^{-2} $$
For $ \frac {h}{R_E} << 1 $
$$g(h) \cong \bigg( 1- \frac {2h}{R_E}\bigg) ….(8.15) $$
Equation (8.15) thus tells us that for small heights h above the value of g decreases by a
factor $(1- 2h / R_E)$.

• Equation (8.15) thus tells us that for small values of heights h above, the value of g decreases by a factor: $(1- 2h / R_E)$.

Figure 8.8 (b): g at a depth d. In this case only the smaller sphere of radius ($R_E$–d) contributes to g.

• Now, consider a point mass, m at a depth, d below the surface of the earth (Refer Figure 8.8b), such that, its distance from the centre of the earth is: ($R_E$ – d) as shown in the figure.
• Subsequently, we can think of the earth being composed of a smaller sphere of radius: ($R_E4 – d) and a spherical shell of thickness, d.
• The force on point mass, “m” due to the outer shell of thickness “d” is zero according to the result quoted in the previous section.
• As far as the smaller sphere of radius, ($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 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,
  $$ M_s / M_E = (R_E – d)^3/R_E^3 …………… (8.16) $$
  Since mass of a sphere is proportional to be cube of its radius.
  Thus the force on the point mass is
  $$ F(d) = GM_sm/(R_E-d)^2 ……(8.17) $$
  Substituting for $M_s$ from above , we get
  $$ F (d) = G M_E m ( R_E – d ) / R_E^3 ……… (8.18)$$
  and hence the acceleration due to gravity at a depth d,
  $$ g(d)=\frac {F(d)}{m} \space is $$
  $$ g(d) = \frac {F(d)}{m} = \frac {GM_E}{R^3_E}(R_E-d) $$
  $$ =g \frac {R_E-d}{R_E} =g(1-d/R_E) ……..(8.19) $$

Thus, as we go down below earth’s surface, the acceleration due gravity decreases by a factor (1-d)/$R_E$.

Note box: The remarkable thing about acceleration due to earth’s gravity is that it is maximum on the earth’s surface and decreasing whether we go up or down!

Questions for section 8.6:

1. Explain and arrive at the equations for acceleration of gravity on an object at a certain height, “h” above the ground
2. Explain and arrive at the equations for acceleration of gravity on an object at a depth, “d” below the ground.

8.7 Gravitational Potential Energy

• We have already studied potential energy as 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 the work done on the body by the force (PE= $F_s$).
• As we had discussed earlier, forces for which the 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 of the earth. In such cases, the force of gravity is practically a constant equal to mg, directed towards the centre of the earth. If we consider a point at a height h1 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 W12 and given by;
  $$ W_{12} = Force × displacement$$
  $$ = mg (h_2 – h_1)………….. (8.20)$$
• If we associate a potential energy W(h) at a point at a height, h above the surface such that
  $$ W (h) = mgh + W_o ……………. (8.21)$$
  (Where, $W_o$ = constant) ;
  then,
  $$ W_{12} = W(h_2) – W(h_1) …………… (8.22)$$
• The work done in moving the particle is just the difference of potential energy between its final and initial positions as discussed earlier.
• The constant $W_o$ cancels out in Eq. (8.22) since h = 0 and thus, we get W(at h = 0) as $W_o$. h = 0 means points on the surface of the earth. Thus, $W_o$ 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 (equations, 8.21 and 8.22) just derived are not valid since the assumption that the 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;

$$ F = \frac {GM_Em}{r^2} ….(8.23)$$
Where, $M_E$ = mass of earth, m = mass of the particle and r= distance of the particle from the centre of the earth.

• If we now calculate the work done in lifting a particle from r = $r_1$ to r = $r_2$ ($r_2 > r_1)$ along a vertical path, we get instead of Equation (8.20);

$$ W_{12} = \int_{r_1}^{r_2} \frac {GMm}{r^2}dr $$
$$ =-GM_Em \bigg( \frac {1}{r_2}- \frac {1}{r_1} \bigg) ……(8.24)$$

• In place of Equation- 8.21, we can thus associate a potential energy W(r) at a distance r, such that;

$$ W(r)=-\frac {GM_Em}{r}+W_1 ……(8.25) $$

The equation is 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 (at r = infinity) = $W_1$. Thus, $W_1$ is the potential energy at infinity. (why is first term in equation 8.25=0?)
• When we set $W_1$ equal to zero, we get potential energy at a point as the amount of work done in displacing the particle from infinity to that point.
• We now know that the potential energy on a point due to gravitational forces of earth is proportional to the mass of the particle.
• Thus, the gravitational potential at a point 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_24 separated by distance by a distance r is given by,

$$ V= – (Gm_1m_2)/r \hspace{10mm} (if we choose V = 0 as r →∞) $$

• It should be noted that an isolated system of particles will have total potential energy equals the sum of energies for all possible pairs of its constituent particles (in other words, the energy for each pair of particles is found using the above equation and summed up for total energy). This is an example of the application of the superposition principle (refer concept box that follows).

Concept box: The superposition principle, also known as superposition property, states that, for all linear systems, the net response at a given place and time caused by two or more stimuli is the sum of the responses that would have been caused by each stimulus individually.

Example 8.3: 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.

Solution: Consider four masses each of mass m at the corners of a square of side l ; See Fig. 8.9. We have four mass pairs at distance l and two diagonal pairs at distance $ \sqrt 2 l $ . Hence,
$$ W(r) = -4 \frac {Gm^2}{l} = -2 \frac {Gm^2}{ \sqrt 2l} $$
$$ = – \frac {2Gm^2}{l} \bigg( 2 + \frac {1}{\sqrt 2} \bigg) = -5.41 \frac {Gm^2}{l} $$
The gravitational potential at the centre of the square (r= $ \sqrt {2}$ l/2) is
$$ U(r) = – 4 \sqrt 2 \frac {Gm}{l} $$

Questions for section 8.7:

1. Give the equation for work done in raising a particle from r1 to r2.
2. Define Gravitational potential at a point due to gravitational force.

8.8 Escape Speed

Concept box: Escape velocity: The least velocity which a body must have in order to escape the gravitational attraction of a particular planet or of any other object is called escape velocity.

• If a stone is thrown by hand, we see that it falls comes down to land on the earth. Using machines, we can shoot an object with much greater initial speed; raising the object to a higher height.
• 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.
• Now, let us suppose that the object did reach infinity and its speed there was $V_f$. Let us find its energy at this point.
• The energy of an object as you know is the sum of potential and kinetic energy. As discussed earlier, $W_1$ denotes that gravitational potential energy of the object at infinity. The total energy of the projectile at infinity then is,

$$ E(∞ )=W_1+ \frac {mV^2_f}{2} …..(8.26) $$
If the object was thrown initially with a speed $V_i$ from a point at a distance $(h+R_E)$ from the center of the earth ($R_E$ = radius of the earth), its energy initially was
$$ E(h+R_E)= \frac {1}{2}mV^2_i – \frac {GmM_E}{(h+R_E)}+W_1 …..(8.27)$$
By the principle of energy conservation Eqs. (8.26) and (8.27) must be equal. Hence
$$ \frac {mV^2_i}{2}-\frac {GmM_E}{(h+R_E)}=\frac {mV^2_f}{2} ……(8.28) $$
The R.H.S. is a positive quantity with a minimum value zero hence so must be the L.H.S.
Thus, an object can reach infinity as long as $V_i$ is such that
$$ \frac {mV^2_i}{2}-\frac {GmM_E}{(h+R_E)} \geq 0 $$
The minimum value of $V_i$ corresponds to the case when the L.H.S. of Eq. (8.29) equals zero. Thus, the minimum speed required for an object to reach infinity (i.e. escape from the earth) corresponds to
$$ \frac {1}{2} m(V^2_i){min}= \frac {GmM_E}{h+R_E} ……(8.30)$$ If the object is thrown from the surface of the earth, h=0, and we get $$ (V_i){min} = \frac { \sqrt {2GM_E}}{R_E} ……(8.31) $$
Using the relation $ g = GM_E/ R^2_E$ , we get
$$ (V_i){min} =\sqrt {2gR_E} ….(8.32) $$ Inserting the values of g and RE, we get, $(V_i){min}$ ≈ 11.2 km/s.

• This is called the escape speed, sometimes loosely called the escape velocity.
• Thus, equation (8.32) 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 acceleration due to gravitation and radius of moon are smaller than that of the earth and the escape speed for the moon turns out to be 2.3 km/s, which is about five times smaller.
• This is the reason that moon has no atmosphere.

Note box: Since the escape velocity for moon is so low, Gas molecules even with low velocities but higher than 2.3 km/s, if formed on the surface of the moon will escape the gravitational pull of the moon.

Example 8.4: Two uniform solid spheres of equal radii R, but mass M and 4 M have a center to centre separation 6 R, as shown in Fig. 8.10. 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 minimum speed v of the projectile so that it reaches the surface of the second sphere.

Solution: The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see Fig. 8.10) is defined as the position where the two forces cancel each other exactly. If ON = r, we have
$$ \frac {GMm}{r^2} = \frac {4GMm}{ (6R-r)^2} $$
$$ (6R – r)^2 = 4r^2 $$
$$ 6R – r = ±2r $$
$$ r = 2R \space or \space – 6R $$.
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
$$ E_i = \frac {1}{2} mv^2 – \frac {GMm}{R} – \frac {4GMm}{5R} $$
At the neutral point N, the speed approaches zero. The mechanical energy at N is purely potential.
$$ E_N = – \frac {GMm}{2R} – \frac {4GMm}{4R} $$
From the principle of conservation of mechanical energy
$$ \frac {1}{2}v^2 – \frac {GM}{R} – \frac {4GM}{5R} = – \frac{GM}{2R}- \frac {GM}{R} $$
$$ or \space v^2 = \frac { 2GM}{R} \bigg( \frac {4}{5}- \frac {1}{2} \bigg) $$
$$ v = \bigg ( \frac {3GM}{5R} \bigg)^{1/2} $$
A point to note is that the speed of the projectile is zero at N, but is nonzero when it strikes the
heavier sphere 4 M. The calculation of this speed is left as an exercise to the students.

Q and A:

Why is it not possible for the moon at have an atmosphere?

Moon having a low acceleration due to gravity of its own, possesses a low escape velocity of 2.3 Km/s which is even smaller than the velocity of gas molecules. Thus, gas molecules escape from the surface of moon, leaving no room for creation of an atmosphere.

Questions for section 8.8:

1. What is escape velocity?
2. Arrive at the equation for escape velocity and thus obtain, its value.

8.9 Earth Satellites

• Earth’s satellites are the 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 motion of satellites.
• The satellites’ orbits around the earth are circular or elliptic. Hence, it is in accordance with Kepler’s law of orbits.
• Moon is the only natural satellite of the earth with an almost circular orbit. The moon has almost the same time period of one revolution as it does for one rotation, approximately 27.3 day.
• Artificial earth satellites have been launched for practical use in fields like telecommunication, geophysics and meteorology.
• We will consider a satellite in a circular orbit of a distance $(R_E + h)$ from 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;
  $$ F(centripetal) = \frac {mV^2}{(R_E+h)} ……(8.33)$$
  directed towards the center. This centripetal force is provided by the gravitational force, which is
  $$ F(gravitation) = \frac {GmM_E}{(R_E+h)^2} ……(8.34)$$
  Where $M_E$ is the mass of the earth.
  Equating R.H.S of Eqs. (8.33) and (8.34) and cancelling out m, we get
  $$ V^2 = \frac {GM_E}{(R_E+h)} ……(8.35)$$
  Thus V decreases as h increases. From equation (8.35),the speed V for h = 0 is
  $$ V^2 (h=0)=GM/R_E=gR_E ….(8.36) $$
  where we have used the relation $ g =GMR^2_E $.In every orbit, the satellite
  traverses a distance 2π($R_E$ + h) with speed V. It’s time period T therefore is
  $$ T = \frac {2 π(R_E+h)}{V} = \frac {2 π(R_E+h)^{3/2}}{\sqrt {GM_E}} ……..(8.37)$$
  on substitution of value of V from Eq. (8.35).Squaring both sides of Eq. (8.37), we get
  $$ T^2 =k(R_E+h)^3 \space (where \space k=4 π^2/GM_E) …..(8.38) $$
  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. (8.38).Hence, for such satellites, T is $T_o$, where
  $$ T_o =2 π\sqrt {R_E/g} ….(8.39) $$
  If we substitute the numerical values g ; 9.8 m $s^{-2}$ and $R_E$ = 6400 km., we get
  $$ T_o =2 π \sqrt {\frac {6.4 \times 10^6}{9.8} }s $$
  Which is approximately 85 minutes.

Example 8.5: The planet Mars has two moons, phobos and delmos. (i) phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 × 103 km. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days?

Solution: (i) We employ Eq. (8.38) with the sun’s mass replaced by the martian mass $M_m$
$$ T^2 = \frac {4π^2}{GM_m}R^3 $$
$$ M_m = \frac {4π^2}{G} \frac {R^3}{T^2} $$
$$ = \frac { 4×(3.14)^2×(9.4)^3×10^{18}}{6.67×10^{-11}×(459×60)^2} $$
$$ M_m= 6.48 × 10^{23} \space kg $$.
(ii) Once again Kepler’s third law comes to our aid,
$$ \frac {T^2_M}{T^2_E} = \frac { R^3_{MS}}{R^3_{ES}} $$
where $R_{MS}$ is the mars -sun distance and $R_{ES}$ is the earth-sun distance.
∴ $$T_M = (1.52)^{3/2} × 365 $$
$$ = 684 \space days $$
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 semi minor to semi-major axis for our Earth is, b/a
= 0.99986.

Example 8.6: 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 ME in two different ways.
Solution: From Eq. (8.12) we have
$$ M_E = \frac {gR^2_E}{G} $$
$$ = \frac { 9.81× ( 6.37 × 10^6)^2}{6.67×10^{-11}} $$
$$ = 5.97× 10^{24} \space kg $$.
The moon is a satellite of the Earth. From the derivation of Kepler’s third law [see Eq. (8.38)]
$$ T^2 = \frac {4π^2R^3}{GM_E} $$
$$ M_E = \frac {4π^2R^3}{GT^2} $$
$$ = \frac {4× 3.14 × 3.14×(3.84)^3×10^{24}}{6.67×10^{-11}×(27.3×24×60×60)^2} $$
$$=6.02×10^{24} \space kg $$
Both methods yield almost the same answer, the difference between them being less than 1%.

Example 8.7: Express the constant k of Eq. (8.38) in days and kilometres. 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 time-period of revolution in days.

Solution: Given $ k = 10^{–13} \space s^2 \space m^{–3} $
$$ = 10^{-13} \bigg[ \frac {1}{(24 ×60× 60)^2}d^2 \bigg] \bigg[ \frac {1}{(1/1000)^3km^3} \bigg] $$
$$ = 1.33 × 10^{–14} \space d^2 \space km^{–3} $$
Using Eq. (8.38) and the given value of k, the time period of the moon is
$$ T^ 2 = (1.33 × 10^{-14})(3.84 × 10^5)^3 $$
T = 27.3 d
Note that Eq. (8.38) also holds for elliptical orbits if we replace ($R_E$+h) by the semi-major axis of the ellipse. The earth will then be at one of the foci of this ellipse.

Questions for section, 8.9:

1. List the applications of an artificial satellite.
2. Arrive at the value of time period for an artificial satellite with the necessary equations.

8.10 Energy of an Orbiting Satellite

Using Equation (8.35), the kinetic energy of the satellite, $K_g$E in a circular orbit with speed v is,
$ KgE = \frac {1}{2}mv^2 $$
$$ = \frac {GmM_E}{2(R_E+h)} ……(8.40)$$
Considering gravitational potential energy at infinity to be zero, the potential energy at
distance ($R_e$+h) from the center of the earth is
$$ P.E = -\frac {GmM_E}{(R_E+h)} …..(8.41)$$
The K.E is positive whereas the P.E is negative. However, in magnitude the K.E. is half
the P.E, so that the total E is
$$ E = K.E+P.E = – \frac {GmM_E}{2(R_E+h) } ……(8.42) $$

• The total energy of a circularly orbiting satellite is thus negative, with the potential energy being negative but twice as the magnitude of the positive kinetic energy.
• When the orbit of a satellite becomes elliptic, both the Kinetic Energy and Potential Energy vary from point to point.
• The total energy which remains constant is negative as in the circular orbit case for planetary revolution. 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 finite distance from the earth and hence their energies cannot be positive or zero.

Example 8.8: A 400 kg satellite is in acircular orbit of radius 2$R_E$ about the Earth. How much energy is required to transfer it to a circular orbit of radius $4R_E$ ? What are the changes in the kinetic and potential energies?
Solution: Initially,
$$ E_i = – \frac {GM_Em}{4R_E} $$
While finally
$$ E_f = – \frac {GM_Em}{8R_E} $$
The change in the total energy is
$$ ΔE = E_f – E_i $$
$$ = \frac {GM_Em}{8R_E} = \bigg( \frac {GM_E}{R^2_E} \bigg) \frac { mR_E}{8} $$
$$ΔE = \frac {g \space mR_E}{8}= \frac {9.81×400×6.37×10^6}{8} =3.13×10^9 \space J $$
The kinetic energy is reduced and it mimics ΔE, namely, $ΔK = K_f – K_i = – 3.13 × 10^9 \space J $.
The change in potential energy is twice the change in the total energy, namely
$$ ΔV = V_f – V_i = – 6.25 × 10^9 \space J $$

Questions for section 8.10:

1. Arrive at the equation for total energy of an orbiting satellite.

8.11 Geostationary and Polar Satellites

• An interesting phenomenon occurs if in Equation-8.37 (check-page 210, lhs, data missing), we arrange the value of ($R_E$+ h) such that T becomes equal to 24 hours.
• If the circular orbit of a satellite is in the equatorial plane of the earth, having the same time period of revolution as the time period of rotation of the earth about its own axis, then it would appear stationery when viewed from any point on earth.
• When we try to make ($R_E$ + h) such that T becomes 24 hours as discussed, the value of $R_E+h$ becomes a much larger than $R_E$.

$$ R_E+h = \bigg( \frac {T^2GM_E}{4 \pi^2} \bigg) ^{1/3} ……(8.43) $$
Thus, for, T = 24 hours, h works out to be 35800 km, which is much larger than $R_E$.

• Such satellites in circular orbits around the earth in the equatorial plane with T = 24 hours are called Geostationary Satellites.
• 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.

Story:

India’s Leap into Space:

• India entered the space age with the launching of the low orbit satellite, Aryabhatta in 1975. In the first few years of its programme the launch vehicles were provided by the erstwhile Soviet Union.
• Indigenous launch vehicles were employed in the early 1980’s to send the Rohini series of satellites into space. The programme to send polar satellites into space began in late 1980’s. A series of satellites labelled IRS (Indian Remote Sensing Satellites) have been launched and this programme is expected to continue in future.
• The remote sensing satellites have been employed for surveying, weather prediction and for carrying out experiments in space.
• The INSAT (Indian National Satellite) series of satellites were designed and made operational for communications and weather prediction purposes, beginning in 1982. European launch vehicles have been employed in the INSAT series.
• India tested its geostationary launch capability in 2001 when it sent an experimental communications satellite (GSAT-1) into space.
• In 1984 Rakesh Sharma became the first Indian astronaut. The Indian Space Research Organisation (ISRO) is the umbrella organisation that runs a number of centres. Its main launch centre at Sriharikota (SHAR) is 100 km north of Chennai. The National Remote Sensing Agency (NRSA) is near Hyderabad. Its national centre for research in space and allied sciences is the Physical Research Laboratory (PRL) at Ahmedabad.

8.12 Weightlessness

Concept box:   The phenomenon of “weightlessness” occurs when there is no force of support (or reaction force) on our body. When we say, a body is in “free fall”, it means that it is accelerating downward under the acceleration of gravity.

• Weight of an object is the force with which the earth attracts it.
• Weights of objects are always balanced by a reaction which is equal in magnitude and opposite in direction.
• The principle can be extended to a spring balance, where we measure the weight of an object by attached to its fixed end and attached to the ceiling of a room. 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. Both, ends of the spring and the object move with identical acceleration g. The 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 now recorded in the spring balance is zero since the spring is not stretched at all.
• If the object attached to the spring were a human being, he or she will not feel his weight since there is no upward force on him (from the instant of the spring being suspended). Thus, when an object is in free fall, it is weightless and this phenomenon is usually called the phenomenon of weightlessness.

Example box: Astronauts and spacecraft in an orbit about the Earth do not escape Earth’s gravitational pull.  In fact, at the altitude of the International Space Station (ISS), the gravitational pull of the Earth is only about 12% less than it is at Earth’s surface. The reason astronauts experience weightlessness is that they are in free fall.  A spacecraft in orbit is falling towards the Earth, because of gravity, but it is moving forward in its orbit fast enough that the path it follows is a curve that is a closed ellipse.   When we stand on a scale, we have weight because gravity is pulling us down and the scale can’t get out of the way.  In space, both: our body and the scale are falling together, so the scale detects no force from our mass (weight) in this state.

Questions from sections, 8.11 and 8.12:

1. What are Geostationary and Polar satellites? State their applications.
2. What is weightlessness? Explain with the illustration of a spring balance.
3. Why do astronauts in space experience free fall?