15 pius augustine gravitation

Gravitation
Newton did not discover
gravity?
Dr. Pius Augustine, SH College, Kochi

Gravitation
The discovery of gravitation dates back to
earlier times, when earth dwellers
experienced the consequence of tripping
and falling.
Newton discovered that gravity is
universal – that is not unique to earth, as
others of his time assumed.

Gravitation
Newton compared fall of apple with
fall of moon.
Moon falls away from the straight
line it would follow if there were no
forces acting on it.
Because of tangential velocity it
falls around earth.

• Newton’s idea was unaccepted for 20
years, during this time he developed
geometric optics and became
famous.
• With the prodding of astronomer
friend Edmund Halley, Newton did
some corrections in his experimental
data on gravitation and published his
data which had far reaching
generalizations of the human mind.

Universal law of gravitation
Every body in the universe attract
every other body with a force which
is directly proportional to product of
their masses and inversely
proportional to the square of the
distance

Note the different role of mass here. Thus far
mass was treated as measure of inertia and
was called inertial mass.
Now mass is a measure of gravitational force,
and is called gravitational mass. But both are
proved to be equal which is the foundation of
Einstein’s general theory of relativity.

Planets – wanderer
1543 – Nicholas Copernicus – heliocentric
theory without proof.
Danish Tycho did not accept it.
After Tycho’s death, his student Kepler
analysed Tycho’s data and presented
his study results known as Kepler’s
laws of planetary motion.

Scientist before Newton believed that
things fell on the earth because it was
the inherent property of matter, and
need no further explanation.

Newton was not satisfied and his
thought process, brought universal
gravitation (at the age of 23 in 1666
but published his book Principia
Mathematica in 1687)

First systematic study of falling
bodies by Galileo who died in the
year Newton was born.

Galeleo did experiments on bodies rolling
down inclined plane, falling from leaning
tower etc.
All bodies fall with same constant
acceleration in the absence of air
resistance.

Assuming moon’s orbit as circular
speed of moon in orbit
v = 2πR/T =1.02 x 103m/s.
R = 3.84 x 108m
T = 27.3 days
ac = v2/R = 2.72 x 10-3m/s2.
Value of centripetal accn of moon

Moon and apple are accelerated
towards earth.
Difference arise due to tangential
velocity of moon, which apple does
not have.

Newton found
ac(= 2.72 x 10-3m/s2)was about 1/3600
of the value of g.
Also Re/R = 1/60 = dapple/dmoon.
Re is radius of earth (distance to apple)
and R distance from centre of earth to
moon.

ac/g = (Re/R)2
ie. acceleration and hence
force is inversely
proportional to square of
distance.

Apple must exert equal and
opposite force on the earth which
must depend on mass of earth.

Combining the two, force of
attraction between bodies should
depend on their masses and vary
inversely as square of distance
between them

Another approach :
using Kepler’s laws

Kepler’s laws of planetary motion
i. All planets move in elliptical orbits
with the sun at a focus.
ii. The radius vector from the sun to
the planet sweeps equal area in
equal time
iii.The square of the time period of a
planet is proportional to the cube of
the semi major axis of the ellipse.

Kerpler’s 1st law : reqmt of force to
provide centripetal force. Idea
gravitational force.
2nd law: Angular momentum (L) is a
constant. ie. Force has no perpendicular
component or force is fully radial.

3rd law :
T2 = KR3.
F = mv2/R = m /R x (2πR/T)2.
= (4π2/K) m/R2.
ie. F α m F α 1/R2.
Since force is mutual,
F α Msun. F = G Ms mp/R2.Dr. Pius Augustine, SH College, Kochi

Gravitational constant G
Defined to be numerically
equal to the force of
attraction between two
unit masses kept at unit
distance apart.

Determination of G

Two small lead balls, each of mass m,
are attached to the opposite ends of a
light rod suspended by fine quartz fibre
with its axis horizontal.
Two big masses are attached near to
them on another rod.

Small balls are attracted by big balls,
and fibre twists.
Gravitational couple F l = Cθ
θ – by lamp and scale arrangement,
C – known, r – distance between the
centres of small and big spheres , G – can
be calculated. Dr. Pius Augustine, SH College, Kochi

Equivalence of gravitational and inertial
mass.
In the derivation of equation
T = 2π (l/g)1/2, mi is taken equal
to mg.
If the two masses were taken to be
different T = 2π (mil/mgg)1/2
Slide change
Dr. Pius Augustine, SH College, KochiDr. Pius Augustine, SH College, Kochi

Equivalence of gravitational and inertial mass .
Newton used a pendulum with
hollow bob made of tin. Filled
with different substances and
given oscillations of same
amplitude.
Slide change

Equivalence of gravitational and inertial mass .
Difference in acceleration could be due
to difference in inertial mass , and
time period should be different.
But time period was found to be same
in all cases.
So two masses should be equal.

Gravitational force is
conservative.
It is weakest and longest range
force.

Value of G on earth is 6.7
x 10 -11 Nm2kg-2. what is
its value on the moon?

Moon falls around earth rather
than straight into it. If the
Moon’s tangential velocity were
zero, how would it move?
It would move straight down to
earth and crash into earth!!

What happens to the force between
two bodies if the mass of one of the
bodies is doubled? If both masses are
doubled?
One mass is doubled, force b/w it and
the other one doubles.
Both doubled, F becomes four times

Gravitational force acts on all bodies in
proportion to their masses. Why, then
, doesn’t a heavy body fall faster than
a light body ?
Because both of them have same
ratio of weight to mass (g)

How much would a
kilogram of matter would
weigh at the centre of
the earth?

The moon’s mass is only 1/80 of the
earth’s mass. Is the gravitational force
exerted by the moon on the earth less
than, equal to or greater than the
gravitational force exerted on the
moon?
Equal -
Newton’s third lawDr. Pius Augustine, SH College, Kochi

If a planet existed whose mass and
radius were both half of the earth,
what would be the value of
acceleration due to gravity on its
surface compared to what it is on the
earth?
g = GM/R2.

Two artificial satellites, one close
to surface and the other away, are
revolving round the earth. Which
has longer time period of
revolution ?

For the moon to have the same orbit as it
has now, how would it period of revolution
change if i) the mass of moon were double
of its present mass ii) the earth’s mass
were double of its present mass?
i)no change
ii) T/ √2

Does the escape
velocity depend on the
direction of projection
No Dr. Pius Augustine, SH College, Kochi

Why is Newton’s law of gravitation
called a universal law ?
Applicable irrespective of the nature of two
bodies
ie. Big or small, temperature, chemical
composition, at all times, at all locations
and for all distances in the universe.

Violation of universal law of gravitation
In the vicinity of sun, sun’s attraction is very
great and law does not give satisfactory
result.
Eg. Perihelion of the planet mercury
undergoes slight rotation in every hundred
years which is not explained by the law of
gravitation.

Violation of universal law of gravitation
Law of gravitation is found
invalid in a very small distance
of the order of 10-7cm or still
smaller intermolecular
distance. Dr. Pius Augustine, SH College, Kochi

Gravitational force for extended bodies
If distance is very large – point mass
concept
If not, force between wo particles
to be found out and add
vectorially. - very difficult.

Gravitational force for extended bodies
Solid or hollow sphere – centre
of mass concept can be used.
Heavenly bodies can be taken
as spherical.

Shell theorm
Newton solved apple – earth problem by
stating the shell theorm.
A uniform spherical shell of matter attracts
a particle lying outside the shell as if the
whole mass of shell be concentrated at
the centre of the shell.
No gravitational force acts on a particle
due to spherical shell, if the particle is
present inside spherical shell.

Difference between gravitation and gravity
1. Gavitation is the force of attraction
between any two bodies in the
universe.
Gravity is the earth’s gravitational pull
on the body lying on or near surface
of earth. Dr. Pius Augustine, SH College, Kochi

2.The direction of gravitational force
on body A due to body B is along AB
Gravity is acting along the line joining
the body and the centre of the earth
and is directed to the centre of earth.

3.Gravitational force between two
bodies can be zero, if the
separation is infinity.
Force of gravity on a body is zero at
the centre of earth.

Acceleration due to gravity
g = GM/R2

Variation in acceleration due to gravity (g)
Due to following reasons
i. Non spherical shape of earth
ii. Altitude of the place
iii.Depth inside the earth
iv.Diurnal (latitude effect) motion of
the earth.
v. Non uniform density of the earth.

i. Non spherical shape of earth
Equatorial radius is about 21
km more than polar radius.
g at the pole at sea level is
1.80 cm/s2 greater than g at
the equator sea level.

Variation of g with height (altitude)
g‘ = g [1 – 2h/R]
g - g’ = 2hg/ R
where h is the altitude
in air.

Variation of g with depth
g = G 4 π R ρ
3
gd = G 4 π (R-d) ρ
3
Ratio gives g’ = g [ 1 – d/R]
g - gd = dg/R
ρ - density of earth

Taking mass of inner sphere M’
= density x vol of inner sphere
= (R-d)3 M/R3
F on a mass m by inner sphere
= GM’m/(R-d)2
gd = GM ( R-d) = (GM) r
R3 R3

Note: i) if a body is taken above the
earth, the value of g varies inversely
as the square of the distance from
the centre of the earth.
If a body is taken inside the earth, g
decreases linearly with distance
from the centre of earth.Dr. Pius Augustine, SH College, Kochi

Note: ii) The rate of decrease of the g with
height is twice as compared to that with
depth.
iii) If the rate of rotation of the earth
increases, the value of acceleration due
to gravity decreases at all places on the
surface earth except at poles.Dr. Pius Augustine, SH College, Kochi

Note: iv) g is min for the planet
Mercury and max for Jupiter.
V)If earth stops rotating, g at
equator increases by Rω2 = 0.034
m/s2, but no change at poles.

g at a point above the surface of the
earth is less than the acceleration
due to gravity at the
corresponding point inside the
earth. g at a height is same as g
at a depth if d = 2h

Earth is not a homogenous sphere.
Density of earth at upper layer is
much less while lower layers is very
dense. So g decreases rapidly with
the depth initially, but later on rate of
decreases becomes less .

Plot graph showing the variation of g with
distance from the centre of earth
surface
gh = g [R2/r 2] ie.
Inversely
proportional to the
square of the
distance from the
centre. r = ( R+h)
gd = g [R/r] ie
falls linearly
with distance.
r = ( R –d)

Variation of g with latitude

Variation of g with latitude
Let φ be the latitude,
Radius of circular path = Rcos φ.
Fcf = mRcos φ ω2
Fcf will have a component away from centre
of earth = Fcf cos φ
Effective wt mg’ = mg - mRcos φ ω2 cos φ
g’ = g - Rcos φ ω2 cos φ
At equator φ = 0 poles φ = 90o.

How will the value of g be affected
if the earth begins to rotate at a
speed greater than its present
speed ?
Acceleration due to gravity
decreases

Effective force of gravity is not to the
centre of earth. Comment .
Plumb line is not vertically to the
centre of earth
Wall of a building is not to the centre
of earth.

When a body falls towards earth, earth
moves towards the body. Why is
earth’s motion not noticed ?
Acceleration produced in earth is
infinitesimally small due to larger
mass of the earth.

Gravitational field
Region around a mass where its
attraction can be felt.
Intensity of gravitational field is
defined as the force /unit mass
placed at that point.
Vector Unit – N/kg.
Intensity I = g = F/m = GM/R2.
Zero at centre of earth and infinity.Dr. Pius Augustine, SH College, Kochi

Gravitational fields and potential
Solid sphere, spherical shell and ring – REF: ADC

Intensity of gravitational field
I = F /m = g.
Units SI and CGS

Gravitational potential
Work done in bringing a unit mass
from infinity to that point.
At a distance x,
F on unit mass = GM/x2.
dW = F.dx = Fdx
W = ∫[GM/x2] dx limit ∞ to r.
= -GM / r. = Gravitational potential.

Gravitational potential
As r increases , gravitational
potential increases and attains
max value = 0 at infinite distance.
Zero at infinity (-ve sign shows that
gravitational forces is attractive)

Gravitational field intensity I at each
point is opposite to the direction in
which force exerted by the external
agent .
dw = - I . dl = - I dl cos0 = -Idl = Idr
dl increases dr decreases

W = integrate (Idr) with limit infinity
to r
W = ∫[GM/r2] dr
limit ∞ to r.
= - GM / r. with limit
= - GM/r Dr. Pius Augustine, SH College, Kochi

Gravitational potential energy
[-GM/r ] * m = mgh
For a body at a height h
GPE = PE at h – PE at surface .
= - GMm/(R+h) - {-GMm/R}
= GMmh/R2. neglect h/R
= mgh

What is the value of
gravitational potential
energy at infinity ?
Zero

In a certain region of space gravitational
field is given by I = (-K/r). Taking the
reference point to be at r = ro, with
gravitational potential v = vo, find the
gravitational potential at distance r.
I = -dV/dr
dV/dr = K/r
dV = K/r dr
Integrating with limits vo to v and ro to r
V = Vo + K log r/ro.

Where is the gravitational field
zero and where is the
gravitational potential zero, in
case of earth ?
Field is zero at centre of earth
and at infinity.
Potential zero at infinity.

Can you shield a body from the
gravitational influence of nearby
matter by putting it inside a hollow
sphere or by some other means ?
No.
Gravitational forces are
independent of the intervening
medium. Dr. Pius Augustine, SH College, Kochi

Give the dimension of gravitational
potential ?

The magnitude of gravitational field at
distance r1 and r2 from the centre of a
uniform sphere of radius R and mass M are
I1 and I2 respectively . Find the ratio of I1/I2
if a) r1 > R and r2>R b) r1 and r2 less than R
Hint: a unit mass body lying at P inside the sphere will
experience gravitational pull due to radius x(centre
to P)
M’ = ρ V’
I1 = Gρ(4πr1
3/3 ) / r1
2.
I1/I2 = r1/r2. Dr. Pius Augustine, SH College, Kochi

If the radius of the earth is increased or
decreases by n%, keeping mass constant, find
the % increase or decrease in the value of g.
g2/ g1 = ( R1 / R2) 2.
R1 = R R2 = ( R ± n%R) = R ( 1 ± n%)
g2 / g1 = [ 1 ± n% ]-2
% Change in g = [ (g2 – g1 )/g1 ] x 100
= [ (1 ± n% )-2 - 1]x100

Two identical copper spheres of radius R
are in contact with each other. The
gravitational attraction between them is F .
Find the relation between F and R.
F = G M M
( 2R ) 2
= G ( 4πR3 ρ/3)2
4R2
F α R4.

How does the weight of a spacecraft
going from the earth to the moon vary?
i. Its weight start decreasing
ii. Zero at the point where the force of
attraction on the spacecraft due to
the earth and the moon will just
become equal and opposite.
iii. It will again start increasing as the
spacecraft further moves towards
the moon. Dr. Pius Augustine, SH College, Kochi

Taking radius of moon’s orbit around the
earth to be ‘r’ and mass of the earth be 81
times the mass of the moon, find the
position of the point from the earth ,
where the net gravitational field is zero.
Let ‘x’ be the distance to the point
from centre of earth.
GMe = GMm
X2 ( r-x)2
Solving, x = 0.9 r

Distiniguish b/w mass and weight
i. Defn
ii. Mass cannot be zero, weight zero
at centre of earth and free fall
iii.Scalar , vector
iv.Kg, N
v. Spring balance and physical
balance Dr. Pius Augustine, SH College, Kochi

Orbital velocity
Velocity that must be imparted
to a body to keep it in an orbit
round the earth is known as its
orbital velocity.
Vo = √gR

Escape velocity
The velocity with which a body must be
projected in order that it may escape
from the earth’s atmosphere.
For earth - 11.2km/s
Moon - 2.4 km/s Sun - 620km/s
Mercury = 4.2km/s,
Jupiter – 61km/s

Note:
For O2, N2, CO2 and water vapour
average velocity at moderate
temperatures is of the order of 0.5 to 1
km/s and for lighter gases (H and He) 2
to 3 km/s.
ie. Lighter gases average molecular
velocities is of the order of escape
velocity of moon and hence escapes.

Kinetic energy of projection = work
done in taking the body from
surface to infinity.
½ mve
2 = ∫ GMm/ r2 dr limit R to ∞
Ve = √ 2gR
Escape velocity is independent of
mass of the body

Escape velocity - dependence
i. Depends upon the location
ii. Depends on gravitational potential at the
point from which it is projected which in turn
depends on the height of the location.
Note: independent of mass of the body
independent of direction of projection.
Body can easily attain the escape velocity, if it is
projected in the direction the launch site is
moving, as the planet rotates about the axis .

Why does hydrogen escape faster from
earth’s atmosphere than oxygen ?
Vrms = (3kT /m) ½
ie vrms α 1 / m

If the kinetic energy of a satellite revolving
in an orbit close to the earth happens to be
doubled, will the satellite escape?
Yes
Velocity becomes equal
to escape velocity

An elephant and an ant are to be
projected out of earth into space.
Do we need different velocities to
do so ?
Need same velocity to
project.

Does a rocket really need the escape
velocity of 11.2km/s initially to escape
from the earth ?
No.
Rocket can have any initial velocity
lower than escape velocity.
But its velocity should continue to
increase.
Will escape only if velocity becomes
11.2 km/s Dr. Pius Augustine, SH College, Kochi

Absence of atmosphere around moon
Because of very small value of
escape velocity from surface of
moon.
Even rms velocity of gas
molecules is greater than Ve.

Vrms = [ 3RT / M] ½.
T – temp, M – relative molecular
mass (min for H2)
Hence H2 escape from almost
all planets.
H2 must have been present long
ago.

If a body is projected with velocity v, greater
than escape velocity ve from the surface of
earth, find its velocity in interstellar space.
Let v’ be the velocity of projected body
in interstellar space .
Apply conservation of energy
½ mv2 + ( -GMm/R) = ½ mv’2 +0
Using ve = ( 2GM/R)½
V’ = ( v2 – ve
2)½ .

Maximum height of a projectile
Total energy at bottom = TE at top
½ mv2 = U2 – U1.
= -(GMm/R+h) – (-GMm/R)
= GM/R { mh/R+h}
= GM/R2 { mh/1+(h/ R)}
= g{ mh/1+(h/ R)}
V2 = 2gh/1+(h/ R)
h = v2R/2gR-v2.
i. For small velocities , h = v2/2g
ii. When v2 = 2gR , h = ∞, V = √2gR which is
escape velocity

• A geosynchronous satellite is
a satellite in geosynchronous orbit, with an orbital
period the same as the Earth's rotation period. Such
a satellite returns to the same position in the sky after
each sidereal day, and over the course of a day traces
out a path in the sky that is typically some form of
analemma. Dr. Pius Augustine, SH College, Kochi
Synchronous satellite

Parking orbit
• A parking orbit is a temporary orbit used
during the launch of a satellite or other space
probe. A launch vehicle boosts into
the parking orbit, then coasts for a while, then
fires again to enter the final desired trajectory.

Is it possible to put an artificial satellite on
an orbit in such a way that it always
remains visible directly over Chandigarh?
No, possible only in
equitorial plane.
Chandigarh does not lie on
equitorial plane.

Weightlessness in a satellite
Weightlessness – sense of apparent
loss in weight in situations like when
falling freely under gravity when
bodies are in zero gravity region etc is
called weightlessness.
In satellite weight (gravitational
attraction) is used for providing
centripetal force.

An astronaut, while revolving in a
circular orbit happens to throw a ball
outside. Will the ball reach the surface
of the earth ?
NO.
it will continue to move in
the same circular orbit and
chase the astronaut.

The acceleration due to gravity at the
surface of the moon is 1.67m/s2. if
the radius of the moon is 1.74 x 106
m, find the mass of the moon .
G =6.67 x 10-11 Nm2kg-2.
ans. 7.58 x 1022 kg.

An astronaut on the moon measures g
to be 1.67m/s2. He knows that the
earth is about 80 times more massive
than the moon . What is his estimate of
the ratio of the radius of the earth to
that of the moon ?
Re/Rm = [Me gm / Mm ge ]1/2
use eqn for g = GM/ R2. Dr. Pius Augustine, SH College, Kochi

A rocket is fired from the earth towards the
moon . At what distance from the moon is
the gravitational force on the rocket is zero
. Me = 6 x 1024 kg, Mmoon = 7.4 x 1022kg and
orbital radius = 3.8 x 108m. Neglect effect
of sun and other planets.
GmMe/(r-x)2 = GmMm/x2.
x = r/10 Dr. Pius Augustine, SH College, Kochi

A body weigh 63N on the
surface of earth, How
much will it weigh at a
height equal to half the
radius of the earth ?
Ans. 28N

What is the value of the
acceleration due to gravity
at an altitude of 500km?
Assume the earth to be a
sphere of radius 6400km .
Ans. 8.27m/s2.

Assuming that the earth is a
sphere of radius 6400km,
at what altitude will the
value of the acceleration
due to gravity, be half its
value at the surface of the
earth ?
Ans : 2649.6km Dr. Pius Augustine, SH College, Kochi

Assuming earth is a sphere of uniform
mass density , find the % decrease in the
weight of a body when taken to the end
of a tunnel, 32km below the surface of
the earth. Radius of earth = 6400km.
Hint : (mg – mgd ) x 100 =0.5%
mg

A hole is drilled half way to the
centre of the earth . A body is
dropped into the hole. How much
will it weigh at the bottom of the
hole, if it weights 250N on the
surface of the earth ? Assume the
earth to be a sphere of uniform
mass density .
Ans : 125N Dr. Pius Augustine, SH College, Kochi

A star 2.5 times the mass of the sun
collapses to a size of just 12km and rotates
with a speed of 1.5rev/s (such extremely
small and dense stars are called neutron
stars). If an object is placed at its equator,
will it remain stuck to it due to its gravity
or will it fly off ? Mass of sun = 2 x 1030kg.

F = GMm/R2 = 9.26 x 1012 m N
Centrifugal force
= mRω2 = 5.33 x 105 m N
(ω = 2π x 1.5 rad/s)
Since gravitational force is
greater , will not fly off.

Calculate the accn produced in the
earth when a stone of mass 6kg
falls on it. Mass of earth = 624kg and
g = 9.8 N/kg.
Hint: mg = M a
a = 9.8-24m/s2.

Estimate the mass of the sun, assuming
the orbit of the earth around the sun
to be a circle. The distance between
the sun and the earth is 1.49x1011 m
and G=6.67-11.
Hint : GMeMs/r2 = mv2/r
v = 2πr/T, T = 365 days Ans : 1.9730kg

A rocket is launched vertically from the
surface of the earth with initial speed
of 10km/s. how far above the surface
of the earth would it go? Radius of
earth = 6400km. Ignore the
atmospheric resistance.

Law of conservation of energy
TE at point of projection = TE at
highest point.
½ mv2 - GMm/R = - GMm/(R+h)
at highest point , v = 0
v2/2 = GM/R2{hR/(R+h)}
=g {hR/(R+h)}
h= 3.93R

A body is released from above at a
distance r from the centre of the
earth . Compute the velocity of the
body when it strikes the surface of
the earth .

- GMm/r = ½ mv2 - GMm/R
Initial velocity = 0 and
final velocity = v
v2 = 2GM [ 1/R – 1/r ]

Satellites
A body which is
constantly revolving in
an orbit around a
planet.

Artificial Satellites
Man made object placed at a height above
the earth and given sufficient velocity so
as to revolve round the earth in closed
orbit.
Sputnik -1 Russia 0ct 4 , 1957 first.
India’s first – Aryabhatta April 19
1975

Geostationary or synchronous
satellites
Used for communication .
Corotate with the earth.
T = 24 h.
Orbital height = 35,870 km.
Orbit is called parking orbit.

Height of geostationary satellite.
Gravitational force on satellite = centripetal force
GmM/r2 = mac.
ac = GM/r2 = gR2/r2.
Orbital velocity V =2πr/T
Centripetal acceleration ac = V2 /r = 4π2r / T2.
Radius of orbit r = ac T2 /4π2.
= (gR2/r2 ) T2 /4π2.
r3 = gR2 T2 /4π2.
r = [gR2 T2 /4π2]1/3.
h = r – R = 42250 - 6380 = 35,870 km.

Polar satellites or remote sensing
satellites.
Revolves in polar orbit.
Used to record the land and sea
temperature, take pictures of
clouds, make forecasting of
climatic changes and monitor the
threat of green house gases.
Used for remote sensing .

Period of Saturn is 29.5 years.
Calculate the average distance of
Saturn from the sun. ( The radius of
the earth’s orbit is 1.58km)
(T1/T2)2 = (d1/d2)3.
Ans : 14.38km

1st law : law of orbits ( in
1609)
The orbit of a planet is an
ellipse with sun at one of
its foci.

2nd law : law of areas( 1609)
The line joining a ploanet to
the sun sweeps out equal
areas in equal intervals of
times, ie. Areal velocity dA/dt
of the planet is a constant .

Geometrical meaning of angular momentum
L = 2m (dA/dt) = twice mass times areal
velocity .
If planet moves through a very small distance dr,
dA = ½ (r x dr)
Dividing by dt,
dA/dt = ½ ( r x v) = ½ ( r x p ) /m
= L/2m .
Since no external torque, dA/dt = constant.

3rd law : law of periods or
harmonic law (in 1619)
The square of the time period T
of revolution of a planet around
the sun is directly proportional
to the cube of its semimajor axis
a. T2 α a3.

If earth be at one half its present
distance from the sun , how many days
will there be in a year ?
T2 = T1 (R2/R1)3/2 =
129 days
T1 = 365 days

In an imaginary planetary system , the central star has
the same mass as our sun, but is much brighter so
that on ly a planet twice the distance between the
earth and the sun can support life. Assuming
biological evolution including ageing etc) on that
planet similar to ours , what would be the average life
span of human on that planet in terms of its natural
year ? The average life span of human on the earth
may be taken to be 70 years.

T1 = 1 year R2 = 2R1
T2 = T1 (R2/R1)3/2
= 23/2 = 2.828 years
Life span = 70 / 2.828
= 24.75 years

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135
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Dr. Pius Augustine, Dept of Physics, Sacred Heart College, Thevara
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Dr. Pius Augustine, Asst. Professor, Sacred Heart College, Thevara, Kochi.

15 pius augustine gravitation

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