Chapter 8 Gravitation
CHAPTER NO.8 GRAVITATION PART2
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 (15641642) 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 (14731543) proposed a definitive model
in which the planets moved in circles around a fixed central sun. His theory
was discredited by the church, but notable amongst its supporters was Galileo
who had to face
prosecution from the state for his beliefs.
It was around the same time as Galileo, a nobleman
called Tycho Brahe (15461601)hailing from Denmark, spent his entire lifetime
recording observations of the planets with the naked eye. His compiled data
were analysed later by his assistant Johannes Kepler (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 semimajor 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 semimajor 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
semimajor axes.
Table 8.1 Data from measurement of
planetary motions given below confirm Kepler’s Law
of Periods (a = Semfmajor 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 (15711830)
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
Sunplanet 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
wellknown 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
Fg 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 xaxis is 30° and so is the angle between
GB and the negative xaxis. 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 xcomponent of the
force cancels out. The ycomponent 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/(Rd)* 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 (ld/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
Ue4y2 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 (@Rry (6R7* =4F
6Rr=22r r=2R or6R.
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
earthsun 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 semiminor to semimajor 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
timeperiod 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 semimajor 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 northsouth direction.whereas
the earth rotatea around fis axis in an eastwest 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 500800 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 semimajor 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 scamimajor 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 yGmm 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 onethousandth 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