Chapter 8 Gravitation

CHAPTER NO.8 GRAVITATION PART-2

 

8.1 INTRODUCTION

Early in our lives, we become 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. Historically ft was the Itallan Physicist Galfleo (1564-1642) 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 planea andarrived at a value of the acceleration due to gravity which is

close to the more accurate value obtained later.

 

Aseemingly 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 tn the sky with posttions unchanged year after year. The more tnteresting objects are the planets which seem to have regular motions against the background of stars. The earliest recorded model for planetary motions proposed by Ptolemy about 2000 years ago was 4 ‘geocentric’ model in which all celestial objects,

atars, the sun and the planets, all revolved around the earth.The only motion that was thought to be possible for celestial objects was motion ina circk. Complicated schemes of motion

were put forward by Ptolemy in order to describe the observed motion of the planets. The planets were described as moving in circles with the centre of the circles themselves moving in larger circles. Stmilar theories were also advanced by Indian astronomers some 400 years later. However a more elegant

model in which the Sun was the centre around which the planets revolved — the ‘heliocentric’ model - was already

mentioned by Aryabhatta (5" century A.D.) in his treatise. A thousand years later, a Polish monk named Nicolas

Copernicus (1473-1543) 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 (1546-1601)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 (1571-

1640}. He could extract from the data three elegant laws that now go by the name of Kepler’s laws. These laws were known to Newton and enabled him to make a great scientific lap inproposing his universal law of gravitation.

 

8.2 EEPLER’'S LAWS

The three laws of Kepler can be stated as follows:

1. Law of orbits : All planets move in elliptical orbits with the Sun sttuated at one of the foci of the ellipse (Fig. 8.18). This law was a deviation from the Copernican model which allowed only

circular orbits. The ellipse, of which the circle is @ special case, is a closed curve which can be drawn very simply aa follows.

 

Select two points F, and F,. Take a length ofa string and fix its enda at F, and F, 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. 8.1(b))

The closed curve you get is called an ellipse.Clearly for any point T on the ellipse, the sum of the distances from F, and F, is a constant. F,,F, are called the focii. Join the points F, and F,and extend the line to intersect the ellipse at points P and A as shown in Fig. 8.1(b). The midpoint of the Ime PA is the centre of the ellipse O and the length PO = AO is called the semi-

major axis of the ellipse. For a circle, the two focli merge into one and the semi-major axis becomes the radius of the circle.

 

2. Law of areas : The line that joins any planet to the sun sweeps equal areas in equal intervals of time (Fig. 8.2). This law comes from the observations that planets appear to move slower

when they are farther from the sun than when they are nearer.

 

3. Law of periods : The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced

out by the planet.

 

Table 8.1 gives the approximate time periods of revolution of eight* planets around the sun along with values of their semi-major axes.

Table 8.1 Data from measurement of

planetary motions given below confirm Kepler’s Law of Periods (a = Semf-major azis in unite of 10° m.T = Time period of revolution of the planet im yearafy).

9 = The quotient (T°/a’ ) in untts of

10 * y® mr?)

 

The law of areas can be understood as a

consequence of conservation of angular

momentum whch is valid for any central force .A central force is auch 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 posttion 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 At is (Fig. 8.2) AA given by ‘AA = % @&xwAd (8.1)

 

Hence

‘AA /At = (rx p)/m, (since v= p/m)

= L/@m (8.2)where v is the velocity, L is the angular

 

Momentum equal to {r x p ). Fora central

force, which is directed along r, Lis a constant (1571-1830) was a scientist of German origin. He formulated the three lawa of planetary motion based on the painstaking observations of Tycho

Brahe and coworkers. Kepler himself waa an assistant to Brahe and it took him sixteen long years to arrive at the three planetary laws. He is alao known as the founder of geometrical optics, being the first to describe what happens

to ight after it entera a telescope.

as the planet goes around. Hence, 4A /Atisa constant according to the last equation. This is the law of areas. Gravitation is a central force

and hence the law of areas follows.

Example 8.1 Let the speed of the planet

at the perihelion Pin Fig. 8.1 (a) be up and the Sun-planet distance SP be rp. Relate {vp, Up to the corresponding quantities at the aphelion {7, u,}. Will the planet take equal times to traverse BAC and CPB?

Answer The magnitude of the angular

momentum at Pis L,= m,Fp,v, since inspection tella us that r, and v, are mutually perpendicular. Similarly, L, = m,7,v, From angular momentum conservation My Tp Up = MyTaVa

fp ota or Uy rh <

Since ry, > 7, V, > Ug.

 

The area SBAC bounded by the ellipse and

the radius vectors SBand SC ts 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.

  

8.3 UNIVERSAL LAW OF GRAVITATION

Legend has it that observing an apple falling from a tree, Newton waa inspired to arrive at an universal law of gravitation that led to an explanation of terrestrial gravitation as well as

of Kepler's laws. Newton's reasoning was that the moon revolving in an orbit of radius RK, was subject to a centripetal acceleration due to earth's gravity of magnitude

Vv? 4 R,,

an = RR, FF (8.3)

 

where V is the speed of the moon related to the time period T by the relation V = 27 R,, /T. The time period T is about 27.3 days and R, was already known then to be about 3.84 x 10°m. If we substitute these numbers in Eq. (8.3), we get a value of a, much smaller than the value of acceleration due to gravity g on the suriace of the earth, arising also 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

axe

dt

 

The angular momentum of the particle is conserved, if the torque 7 = r x F due to the force F on it vanishes. This happens either when F ie zero or when F is along z. 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, ic., along the position vector of the point of application of the force with respect to the fixed point. (See Figure below.)

Furtha, the magnitude of a central force Fdepende on r, the distance of the point of application of the force from the fixed point; F = FU).

 

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 confined to a plane.

 

(2) 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=% rvein a.

 

An immediate application of the above discussion can be made to the motion of a planct 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 js directed towards the sun.This force also satisfics the requirement F = Pfr), eince F = G m,m,/r where m, and m, ure respectively the masecs of the planet and the sun and Gis 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 secand law of Kepler.

 

Tr (s the trejectary of the particle under the central force. At a position P, the force ts directed along OP, O ts the centre of the force taken as the orfgin. In time At, the particle moves from P to P’,arc PP’= As = v At. The fangent PQ at Pio te trajectory gtves the direction of the vetoctty at P. The

area swept in At ts the area of sector POP’ = (r sin &} PP’/2 = {rv stn a} At/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,, 0 R23 ga RZ and we get

oR

a, R; = 3600 (8.4)in agreement with a value of g = 9.8 ms? and the value of a, from Eq. (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.

 

The quotation is essentially from Newton’a famous treatise called ‘Mathematical Principlea of Natural Philosophy’ (Principia for short).

 

Stated Mathematically, Newton’'s gravitation law reads : The force F on a point masa m, due to another point mass m, has the magnitude

mM,

IFl= +

Gt (6.5)

Equation (8.5) can be expreased in vector form aa

F-g Ma (-1)=-G m, Ms»

r r

_-g ™ My ;

|

where G is the universal gravitational constant,; is the unit vector from m, to m, and r =r, —-r,as shown in Fig. 8.3.

 

The gravitational force is attractive, 1.e., the force F is along — r. The force on point mass m,due to m, is of course - F by Newton’a third law.

Thus, the gravitational force F,, on the body 1 due to 2 and F,, on the body 2 due to 1 are related as F,, = - F,,.

 

Before we can apply Eq. (8.5) 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 the other point Taases as shown in Fig 8.4.

 

The total force on m, is

F,= Gin, my mM re + Ging my mm fou + Ging my m™ ru mh 3 Ta

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)

Answer (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 Gm(2m) ;Fun = mr) j

Foy = cavtin (-i cos 30° — j sin 30° )

Foc = Gm(emn) (Hi cos 30° —j sin 30°)

 

From the principle of superposition and the law of vector addition, the resultant gravitational force F,, on (2m) is Fa = Faa + Fas + Foc Fy = 2G? j+ 2G? Ci cos 30°-j sin 30°)+ 2cm(i cos 30° = j sin 30°)= oO

 

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

(b) By symmetry the x-component of the

force cancels out. The y-component survives.Fp =4Gr?j- 2Gn?j =2Gm?j <

For the gravitational force between an

extended object (like the earth) and a point maas,Eq. (8.5) 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 in the same direction. We have to add up these forces

vectorially for all the point massea 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 :

! (1) The force of attraction between a follow spherical shell of uniform density and a point mass situated outelde 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 prependicular to this line. The components prependicular to this line ; cancel out when summing over all regions

of the shell leaving only a resultant force along the Hine joining the point to the centre.The magnitude of this force works out to be aa stated above.

Nowton’'s Principia Kepler had formulated his third law by 1619. The announcement of the underlying untversal law of gravitation came about seventy years later with the publication in 1687 of Newton's masterpiece Philosophiac 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 aaked him about the nature of the trajectory of a body moving under the

influence of an inverse aquare 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 hia work in book form and agreed to bear the coat of

publication. Newton accomplished this feat in eighteen months of superhuman effort. The Principia ia 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 apent 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.

 

(2) The force of attraction due to a hollow spherical ahell of uniform density, on a point mase situated inside it ia 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.

 

 

8.4 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 apparatue used by him is schematically ahown in figure.8.6

The bar AB has two small lead spheres

attached at fts ends. The bar is suspended from a rigid support by a fine wire. Two large lead spheres are brought close to the amall 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 aphere and

its neighbouring 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 @ is the angle of twist of the suspended wire, the restoring torque is proportional to @, equal to 76. Where 7 is the restoring couple per untt angle of twist. tcan 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

{ts neighbouring amall ball, M and m their masses, the gravitational force between the big sphere and its neighouring small ball is.

F= ot (8.6)

 

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 GGrL- 76 6.7)Observation of @ thus enables one to calculate G from this equation.

 

Since Cavendish’s experiment, the

measurement of G has been refined and the currently accepted value is

G=6.67x10" N m?/kg? (8.8)

 

8.8 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 amallest one at the centre and the largest one at fis 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 section 8.3. The total mass

of all the shells combined is fust the mass of the earth. Hence, at a point outside the earth, the gravitational force is just as if tts entire mass of

the earth is concentrated at its centre.

For a point inside the earth, the sttuation is different. Thia ia illustrated in Fig. 8.7.

 

Again conaider 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 aphere of radius r. For

the shells of radius greater than r, the point P lies inside. Hence according to result stated in the last section, they exert no gravitational force on mass mkept at P. The shells with radius <7

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 mat P as if ita mass M_is concentrated at the centre. Thus

the force on the mass mat P has a magnitude

Gm (M

reo) J (8.9)

 

We assume that the entire earth 1s of untform density and hence its mass is My = Ry p where M is the masa of the earth R, is ita radius and p is the density. On the other hand the ‘4g mass of the sphere M of radius ris 3?" and

hence / ; -

F=Gm| =p] fF = Gm Ie

. 30 R, }r

GmM,

“Re (8.10)

 

If the maas m is situated on the surface of earth, then r= R, and the gravitational force on it is, from Eq. (8.10)

, M,m

F=G (8.11)

 

The acceleration experienced by the mass m,which is usually dencted by the symbol g is related to F by Newton's 2™ law by relation F = mg. Thus

gh aM

~ mR; (6.12)

 

Acceleration g ia readily measurable. R,is a known quantity. The measurement of G by Cavendish's experiment (or otherwise), combined with knowledge of g and R, enables one to estimate M, from Eq. (6.12). This is the reason

why there ia a popular statement regarding Cavendish : “Cavendish weighed the 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 Fig. 8.8(@)}. The radius of the earth ts denoted by R,. Since this point is outside the earth,its distance from the centre of the earth is (R,+ h). If F(t) dencted the magnitude of the force

on the point mass m , we get from Eq. (8.5) :, GM,m F(R =e

 

The acceleration experienced by the point maas fs F(h)/ msg(h) and we get

Thus the force on the point mass is

Fi@= GM,m/(R-d)* 17 Substituting for M, from above , we get F@ =GM,m(R,-@2)/ R2 (6.18)and hence the acceleration due to gravity at

a depth d,

Fld)

KA=—, 8

ny Fld) _GMe pp |

gld)= —— Re (R, - a)

Re -d

=~ 9 ld / Re) (8.19)

 

Thus, a8 we go down below earth’s surface,the acceleration due gravity decreasea by a factor (l-d/R,). The remarkable thing about acceleration due to earth’a gravity is that it ia

Maximum on its surface decreasing whether you go up or down.

  

8.7 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 tts potential energy is just the amount of work done on the body by the force.

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 carth, 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 h, from the surface of the

earth and another point vertically above it at a height h, from the surface, the work done in lifting the particle of mass m from the first to the second posttion is denoted by W,,W,, = Force x displacement = mg (h,-h) (8.20)

 

If we associate a potential energy WIh) at a point at a height h above the surface such that

Wit) = mgh + W, (6.21)(where W, = constant) ;then it is clear that

W,, = Wh) - Wh) (8.22)

 

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, cancels out in Eq. (8.22). Setting

h= 0 in the last equation, we gett W(h=0)= W,. h=0 means points on the surface of the earth. Thus, W, 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 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 pect (8.23)where M, = 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, to r= Fr, (r, > r,) along a vertical path, we get instead of Eq. (6.20)2 GMm

W..= J adr 11)=-GM,;m (=-=| (6.24)

 

In place of Eq. (8.21), we can ths associate a potential energy W[7) at a distance r, such that Wire ew, (6.25)

valid forr>R,so that once again W,, = Wir.) - Wir).Setting r= infinity in the last equation, we get W (r= infinity) = W,. Thus, W, is the potential energy at infinity. One should note that only the difference of potential energy between

two points has a definite meaning from Eqs.(8.22) and (8.24). One  conventionally sets W,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 eariier discussion, we learn that the gravitational potential energy associated with two particles ofmasses m, and m, separated by distance by a

distance r is given by V=- ane (if we choose V = 0 aa r > ~)It should be noted that an isolated system of particles will have the total 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.

 

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.

Answer Cousider four masses each of mass m at the corners ofa square of side [; See Fig. 8.9.We have four mass pairs at distance ! and two diagonal pairs at distance J2 ;Hence,

_ 2Gm? 9 1) 5.4) c =——_ "kr =z— oO, =

The gravitational potential at the centre of the square (r=¥21/2) is

Ue-4y2 ty <

 

8.8 ESCAPE SPEED

Ifa atone is thrown by hand, we see it falla back to the earth. Of course using machines we can shoot an object with much greater apeeds and with greater and greater inftial apeed, the object

scalea 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 doea not fall back to the earth?

The principle of conservation of energy helps us to answer this queation. Suppose the object did reach infinity and that fta speed there was V, The energy of an object is the sum of potential and kinetic energy. Aa before W, denotes that gravitational potential energy of the object at infinity. The total energy of the projectile at

infinity then ts , mv;E (~)=W, TF (8.26)

If the object was thrown initially with a speed V, from a point at a distance (h+R,) from the centre of the earth (R, = radius of the earth), its energy initially was (ne Ry =dmye — SMe yy

+ Re =3 2 “h+R,) 1 (8.27)By the principle of energy conservation

Eqs. (8.26) and (8.27) must be equal. Hence mv? _ GMM, | mv;

 

2. (h+R,) 2 (8.28)The R.HLS. 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,is such that mV; GmM,

2 (he Ry) (8.29) The minimum value of V, 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 1 2 GmM,a)... "tar, (8.30)If the object is thrown from the surface of the earth, h = 0, and we get 2GM,,Yous - Poe (6.31)

Ee Using the relation g =GM,, / RZ, we get (Myon =V29R. (6.32)

 

Using the value of g and R,, numerically

(V)..%11.2 km/s. This is called the eacape speed, sometimes loosely called the escape veloctty.

 

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, replaced by the radiua of the moon. Both are smaller than their values on earth and the escape speed for the

moon turns out to be 2.3 km/s, about five times smalier. 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 eacape the gravitational pull of the moon.

 

Example 8.4 Two untform sold spheres

of equal radii R, but mass Mand 4 M have

a centre to centre separation 6 R, as shown in Fig. 8.10. The two apheres are held fixed.A projectile of mass mis projected from the surface of the sphere of mass Mdirectly 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.

Answer 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 ‘GMm_AGMm re (@R-ry (6R-7* =4F

6R-r=22r r=2R or-6R.

 

The neutral point r= -GRdoes 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 ts

 

: 1» GMm 4GMm E, == mv“ -—— - ——_.,

2 R SR

 

At the neutral point N, the speed approaches zero. The mechanical energy at N is purely potential.

, GMm_ AGMm

Ey =-—--—

2R 4R

From the principle of conservation of

mechanical energy ‘1 OM _4GM __GM_ OM

2 * R SR 2R R

or ee 2GM(4 1

OR (s 2)

samy?

“Cr) ‘

 

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.

 

8.8 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’a 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 a circular orbit of a distance (R, + h) from the centre of the earth,where R, = radius of the earth. If mis the mass

of the satellite and V its speed, the centripetal force required for this orbit is mv?

directed towards the centre. This centripetal force is provided by the gravitational force, which is

“GmM re Figravitation) = (R, +A)? (8.34)

where M, is the mass of the earth.

Equating R.H.S of Eqs. (8.33) and (8.34) and cancelling out m, we get

ye GM,

y= E

(Ry +h) (6.35)

 

Thus V decreases as h increases. From

equation (8.35),the speed V for h=0 is

V? (h=0) = GM/R, = gR; (8.36)

where we have used the relation

€ =GM/R,” . In every orbit, the satellite traverses a distance 27(K,,+ h) with speed V. Its time period T therefore is 2a(R, +h) 22(R, +h)?

T = BS

V IGM, (8.37)

 

on substitution of value of V from Eq. (8.35).Squaring both sides of Eq. (8.37), we get T7=k (R,+hP (wherek=47°/GM) (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 aurface of earth h can be neglected in comparison to R, in Eq. (8.38).Hence, for auch aatellites, Tis T,, where Ty = 2nJR,, /g (8.39)If we aubatitute the numerical values g = 9.8m 8? and R, = 6400 km., we get T=2n fetx10" s  9.8 Which is approximately 85 minutes.

 

Example 8.5 The planct Mars has two

moons, phobos and delmos. (f) phobos has

a period 7 hours, 39 minutes and an orbttal radius of 9.4 x105 km. Calculate the mass of mars. (i) Assime that earth and mars move in circular orbits around the sun,with the martian orbit being 1.52 times the orbital radtus of the carth. What is the length of the martian year in days ?

Answer (if) We employ Eq, (8.38) with the sun's mass replaced by the martian mass M,,: an?

T? =— R*

. GM,,, .

Ag? R*

Mian = a

GT

_ 4x(3.14)? x(9.4)" x10!"

6.67x107! x(459x 60)?

Mi 4%(3.14)* «(9.4)? x10!

" 6.67x(4.59x6)" x10”

= 6.48 x 10° kg,

(i) Once again Kepler's third law comes to our

aid,

Ta _ Riss

Tz Ros

 

where R,, is the mars -sun distance and R,, is the earth-sun distance.

o Ty = (1.52)5? x 365 = 684 days

We note that the orbits of all planets except Mercury, Mars and Phito* 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. 4 Example 8.6 Weighing the Earth : You are given the following data: g=9.81 ms,R, = 6.37x10° m, the distance to the moon R=3.84x10* m and the time period of the moon’s revchition is 27.3 days. Obtain the mass of the Earth M, in two different ways.

Answer From Eq. (8.12) we have

Ri Mps= oe

9.81x{6.37x10°)

~  6.67x10"

=5.97x 10% kg,

 

The moon is a satellite of the Earth. From the derivation of Kepler's third law [see Eq.

638]

72 AWRY

 GMy

. 47R8

Mp =—

GP

~ 667x107! x(27.3% 24x 60x60)?

= 6.02x10**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 s? m°. The moon ts at a distance

of 3.84 x 10°km from the earth. Obtain its time-period of revohition in days.

Answer Given k= 10% 6? m* : I 5 1

eof of (24x60x 60)? . (1 71000)" kan”

= 1.33 x10-"*d? km

 

Using Eg. (8.38) and the given value of k,the time period of the moon is T?= (1.33 x 10'4}(3.84 x 109° T =27.3d <

Note that Eq. (8.38) also holds for elliptical orbits if we replace (K,th) by the semi-major axis of the ellipse. The earth will then be at one of the foci of this ellipse.

 

8.10 ENERGY OF AN ORBITING SATELLITE

Using Eq. (8.35), the Kinetic energy of the satellite in a circular orbit with speed vu is KeE= 1a v 2 __GmM,

“OUR, #h)* (6.40)

 

Considering gravitational potential energy at infinity to be zero, the potential energy at distance (R+h) from the centre of the earth is

GmM,

PES ay R, +h) (8.41)

 

The K.E is positive whereas the P.E is

negative. However, in magnitude the KE. is half the P.E, so that the total E is

_ __GmM,EsK.E+P.E= aR, +h) (8.42)

 

The total energy of an 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 K.Z. and PE. vary from point to point. The total energy which remains constant is negative as in the 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 finite distance from the earth and hence their energies cannot be positive or zero.

 

Example 8.8 A 400 kg satellite is in a

circular orbit of radius 2K, about the

Earth. How much energy is required to

transfer it to a circular orbit of radius 4K, ?What are the changes in the kinetic and potential energies 7?

Answer Initially,‘E, ~_ GMym

4Rp While finally Ee GME m J BR:

The change in the total energy is

AE=5E,-E _GMgm_ G My mR  8Rp RE | 8

Ake 2 mR = 281 4006.57 910" =3.13%108d

The kinetic energy is reduced and it mimics AE, namely, AK = K,-K, = -3.13 x 10°J.The change in potential energy is twice the change in the total energy, namely AV=V, - V, =-6.25 x 10°J <

 

8.11 GEOSTATIONARY AND POLAR SATELLITES

An interesting phenomenon arises if in we arrange the value of (R+ } such that T in Eq. (8.37) becomes equal to 24 hours. If the circular orbit is in the equatorial plane of the earth, auch a satellite, having the same period

Example 8.8 A 400 kg satelitte is in a as the period of rotation of the earth about its circular orbit of radtua 2R, about the own axis would appear stationery viewed from Earth. How much energy is required to @ point on earth. The (R + 19 for this purpose transfer it to a circular orbit of radius 4K, 7 works out to be large as compared to R, :What are the changes in the kinetic and Tom," potential energies 7 R, +h (Ee) (8.43)e

Answer Initially, and for T= 24 hours, h works out to be 35800 Ian.“p  GMym which is much larger than R,. Satellites in a Ev=- 4Rr circular orbits around the earth in the India’s Leap into Space

India entered the space age with the launching of the low orbit satelitte Aryabhatta in 1975. In the frat

few yeara of its programme the launch vehicles were provided by the erstwhile Soviet Union. Indigenous launch vehicles were employed in the carly 1980's to send the Rohini scrice of satellites into space.‘The programune to send polar satellites into space began in late 1980's. A series of satellites labelled

TRS (indian Remote Sensing Satelittes} have been launched and this programme is expected to continue in future. The satellites have been employed for surveying, weather prediction and for carrying out expertmenta in apace. The INSAT {indian National Satellite} series of satellites were deaigned and made operational for communications and weather prediction purposes beginning in 1982. European launch vehicles have been employed in the INSAT sericea. India teated 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} ie the umbrella orgarisation that runs a number of centre. Ita main lauch centre at Sriharikota (SHAR) is 100 km north of Chennai. The National Remote Senatng Agency (NRSA) is near Hyderabad. Its national centre for reasearch in space and allied aciences is the Physical Research Laboratory (PRU at Ahmedabad.

equatorial plane with T= 24 hours are called Geostationery Satellites. Clearly, since the earth rotates with the same period, the satellite would

appear fixed from any point on earth. It takea 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.

 

It is known that electromagnetic waves above a certain frequency are not reflected from jonosphere. 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

broadcast or other forms of  commumication have much higher frequencies and thus cannot be

received beyond the line of sight. A

Geostationery satelite, 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 aateliites sent up by India are one such group of Geostationary satellites widely used for telecommunications in India.

 

Another class of satellites are called the Polar satellites (Fig. 8.11). These are low altitude (h»500 to 800 kin) satellites, but they go around

the poles of the earth in a north-south direction.whereas the earth rotatea around fis axis in an east-west direction. Since ita time period is

around 100 minutes it crosses any altitude many times a day. However, since its height h above the earth is about 500-800 lon, a camera fixed

on it can view only small strips of the earth in ene orbtt. Adjacent strips are viewed tn the next orbit, so that in effect the whole carth 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,

meterology as well aa for environmental studies of the earth.

 

6.12 WEIGHTLESSNESS

Weight of an object is the force with which the earth attracts ft. 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 ts pulled down a fittle 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 cefling of the room. Both ends of the spring as well as 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 recorded in the spring balance is zero since the

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, ft is weightless

and this phenomenon is usually called the phenomenon of weightlessness.

 

In a satellite around the carth, 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 toside 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 arethe same. Pictures of astronauts floating in a satellite show this fact.

 

SUMMARY

1. Newton's law of universal gravitation states that the gravitational force of attraction between any two particles of masses m, and m, separated by a distance r has the magnitude mm F=G—* r where Gis the universal gravitational constant, which hae the value 6.672 x10" N m* kg*.

 

2. If we have to find the resultant gravitational force acting on the particle m due to a number of masses M,, M,, ....M,etc. we use the principle of superposition. Let F;, F;. ....F,

be the individual forces due to M;,, M,, ....M,, each given by the law of gravitation. From the principle of superposition each force acts independently and uninfluenced by the

other bodies. The resultant force F, is then found by vector addition Fy, = F,+F,+....+F, = SF i=l where the symbol 'I stands for summation.

 

3. Kepler’s laws of planetary motion state that

(a) AD planets move in elliptical orbits with the Sun at one of the focal pointe

 

(b) The radtus vector drawn from the sun to a planet sweeps out equal areas in equal time intervals. This follows from the fact that the force of gravitation on the planet is central and hence angular momentum is conserved.

 

(c} The square of the orbital period of a planet is proportional to the cube of the semi-major axis of the elliptical orbit of the planet The period T and radius R of the circular orbit of a planet about the Sun are related

by 72-{ 25 GM,where M, is the mass of the Sun. Most planets have nearly circular orbits about the Sun. For elliptical orbits, the above equation is valid if R is replaced by the scami-major axis, a.

 

4. The acceleration due to gravity.

(@) at a height h above the Earth’s surface (R,; + h)7 cM, ( - 2h)

"RE Ry. for h << R,. ini 2h 1 GM;

h) = g(0 1 = 22) where g(Q) = —~£

gh) = gf l R, g(0) R ) at depth d below the Earth's surface ia _GM{, d)_.. | _ a) gia) = Pl - S) = wo [1 - S|

 

6. The gravitational force is a conservative force, and therefore a potential energy function can be defined. The gravitational potential energy associated with two particles separated by a distance ris given by

Vee G my Mg

_

where Via taken to be zero at r—> +. The total potential energy for a system of particlea ia the sum of energies for all pairs of particles, with each pair represented by a term of the form given by above equation. This prescription follows from the principle of

superposition.

 

6. If an taclated system consists of a particle of masa m moving with a apeed vu in the vicinity of a massive body of mass M, the total mechanical energy of the particle is gtven by pelt ee Gum

2 r That is, the total mechanical energy is the sum of the kinetic and potential encrgica.The total energy ie a constant of motion.

 

7. If mmoves in a circular orbit of radius a about M, where M >> im, the total energy of the system ia

‘Eee GMm 2a with the choice of the arbitrary constant in the potential energy given in the paint 5.,above. The total energy is negative for any bound system, that is, one in which the orhit

ia closed, such as an elliptical orbit. The kinetic and potential energies are

Kein 2a ye Gm a

 

8. The escape speed from the surface of the Earth ts 2G Me = Re” A2gR,,

and has a value of 11.2 km s*.

 

9. If a particle is outalde a uniform epherical shell or solid sphere with a spherically symunctric internal masa distribution, the sphere attracts the particle as though the mase of the sphere or shell were concentrated at the centre of the sphere.

 

10.If a particle is ineide a uniform spherical shell, the gravitational force on the particle is zero. If a particle is inside a homogencous solid sphere, the force on the particle acts toward the centre of the sphere. This force is exerted by the spherical mase interior to the particle.1LA geostationary (geosynchronous communication) satellite moves in a circular orbit in the equatorial plane at a approximate distance of 4.22 x 10* km from the Earth's centre.

 

POINTS TO PONDER

1. In considering motion of an object under the gravitational influence of another objectthe following quantities are conserved:

(a) Angular momentum

(b) Total mechanical energy

Linear momentum is not conserved

 

2. Angular momentum conservation leada to Kepler's second law. However, it ia not special to the inverse square law of gravitation. It holds for any central force.

 

3. In Kepler's third law (see Eq. (8.1) and T* = K,R°. The constant K; is the aame for all planets in circular orbits. This applies to satellites orbiting the Earth [(Eq. (8.38)].

 

4. An astronaut experiences weightleasness in a space aatellite. This is not because the gravitational force is small at that location in apace. It fa because both the astronaut

and the satellite are in “free fall" towarda the Earth.

 

5. The gravitational potential energy associated with two particles separated by a distance ris given by V= Sma + constant The constant can be given any value. The simplest choice is to take it to be zero. With thie choice y--Gmm r This choice implies that V— 0 as r—>» «. Choosing location of zero of the gravitational energy is the same as choosing the arbitrary constant in the potential energy. Note that the gravitational force is not altered by the choice of this constant.

 

6. The total mechanical energy of an object ia the sum of ite kinetic energy (which is always positive) and the potential energy. Relative to infinity [1.c. if we presume that the potential

energy of the object at infinity ie zero), the gravitational potential energy of an object is negative. The total energy of a satellite is negative.

 

7. The commonly encountered expression mgh for the potential energy is actually an approximation to the difference in the gravitational potential energy discussed in the point 6, above.

 

8. Although the gravitational force between two particles is central, the force between two finite rigid bodies fa not necessarily along the line joining their centre of masa. For a spherically symmetric body however the force on a particle external to the body is as if

the mass ia concentrated at the centre and this force is therefore central.

 

9. The gravitational force on a particle inside a spherical ahell ts zero. However, (unlike a metallic shell which ahields electrical forces) the shell does not ahield other bodies outside

it from exerting gravitational forces on a particle inside. Gravitational shielding is not possible.

 

EXERCISES

8.1 Answer the following :

(a) You can shield a charge from electrical forces by putting it inaide a hollow conductor.Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?

 

(b) An astronaut inaide a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ?

 

(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun's pull is greater than the moon's pull.(you can check this yourself using the data available in the succeeding exercises).However, the tidal effect of the moon's pull is greater than the tidal effect of sun.Why ?

 

8.2 Choose the correct alternative :

(a) Acceleration due to gravity increases/decreases with increasing altitude.

 

(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).

 

(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.

 

(d) The formula -G Mm{1/r, - 1/7,) is more/less accurate than the formula

mag(r,— 7,) for the difference of potential energy between two points r, and r, distance away from the centre of the earth.

 

8.3 Suppose there existed a planet that went around the sun twice as fast as the earth.What would be its orbital size as compared to that of the earth ?

 

8.4 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 x 10* m. Show that the mass of Jupiter is about one-thousandth that of the sun.

 

8.6 Let us aseume that our galaxy consists of 2.6 x 10° stars each of one solar mase. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be 10° ly.

 

8.6 Choose the correct alternative:

(a) If the zero of potential energy is at infintty, the total energy of an orbiting satellite is negative of its kinetic/potential energy.

 

(b) The energy required to launch an orhiting satellite out of earth’s gravitational influence is more/leas than the energy required to project a stationary object at the same height (as the satellite} out of earth’s influence.

 

8.7 Does the escape speed of a body from the earth depend on (a) the mass of the bady, (b)

the location from where ft is projected, (c) the direction of projection, (d) the height of the location from where the body is launched?

 

8.8 Acomet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a)lear speed, (b) angular speed, (c} angular momentum, (d) kinetic energy, {e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

 

8.9 Which of the following symptoms is likely te afflict an astronant in space (a) swollen feet, (b) swollen face, (c) headache, (d} orientational problem.

 

8.10 In the following two exercises, choose the correct answer from among the given ones:The gravitational intensity at the centre of a hemispherical shell of uniform mass denaity has the direction indicated by the arrow (see Fig 8.12) (i) a, (if) b, fit) c, fiv) 0.

 

8.11 For the above problem, the direction of the gravitational intensity at an arbitrary point P ie indicated by the arrow (i) d, Gi) ¢, Git) f, tiv) g.

 

8.12 Arocket is fired from the earth towards the aun. At what distance from the earth's centre ie the gravitational force on the rocket zero 7? Mase of the sun = 2x10” kg,mass of the earth = 6x10™ kg. Neglect the effect of other planeta etc. (orbital radius = 1.5 x 16" mj.

 

8.13 How will you ‘weigh the sun’, that is eatimate ite mass? The mean orbital radius o:the earth around the sun is 1.5 x 10° km.

 

8.14 Asaturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 x 10° km away from the sun?

8.15 Abody weighs 63 N on the surface of the earth. What is the gravitational force on i due to the earth at a height equal te half the radius of the earth ?

 

8.18 Assuming the earth to be a sphere of uniform masse denaity, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface 7

 

8.17 Arocket ia fired vertically with a speed of 6 km s" from the earth's surface. How fat from the earth does the rocket go before returning to the earth ? Mase of the earth = 6,0 x 10* kg: mean radius of the earth = 6.4 x 10° m; G = 6.67 x 10 N m*kg*.

 

8.18 The escape speed of a projectile on the earth's surface is 11.2 km s". A body kk projected out with thrice this speed. What is the speed of the bedy far away from the earth? Ignore the presence of the sun and other planeta.

 

8.19 A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite = 200 kg; masa of the carth = 6.0x10™ kg; radius of the carth = 6.4 x 10° m; G = 6.67 x 10" N m’kg*,

 

8.20 Two stars each of one solar mass (= 2x10” kg) are approaching each other for a heac on colliaion. When they are a distance 10° km, their speeds are negligible. What ts the speed with which they collide ? The radiua of each atar is 10° km. Assume the stare to remain undistorted until they collide. (Use the known value of Gj.

 

8.21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Ia an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ?

 

Additional Exercises

8.22 As you have learnt in the text, a geostationary satellite orbita the earth at a height o:nearly 36,000 km from the surface of the earth. What is the potential due to earth's gravity at the alte of this satellite ? (Take the potential energy at infinity to be zero).Mass of the earth = 6.0x10™ kg, radius = 6400 km.

 

8.23 Aastar 2.6 times the mase of the sun and collapsed to a aize of 12 km rotates with apeed of 1,2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objecta called pulsara belong to this category). Will an object placed on ita equator remain stuck to ita surface due to gravity 7 (masse of the

sun = 2x10” kg).

 

8.24 A spaceship is stationed on Mara. How much energy must be expended on the

spaceship to launch it out of the solar system ? Massa of the space ship = 1000 kg: mage of the sun = 2x10” kg; mase of mars = 6.4x10™ kg; radius of mara = 3395 km:radius of the orbit of mara = 2.28 x10° km; G = 6.67x10'™" N m* kg*.

 

6.25 Arocket is fired ‘vertically from the surface of mars with a speed of 2 km 9”. If 20% of ita initial energy is lost due to martian atmospheric resistance, how far will the