Gravitational laws

In the wonderland of celestial objects

The tell of two G -- g

Johannes Kepler (1571-1630)• Tycho Brahe’s recorded

detailed observations of planetary positions and tried to make sense of them, using mathematics

• He could not succeed

• Kepler studied Tycho Brahe’s data & suggested three laws of planetary motion (Known as Keppler’s laws)

Johannes Kepler (Danish astronomer)

Kepler’s 1st Law:All planets move in elliptic orbits, with the Sun at one focus of the ellipse.

e

For understanding purposed ‘e’ is shown largeReal value = 0.017%

Distance from sun (All distance in million KM

Minimum : 146 million km

Maximum : 152 million km

Mean : 149 million km

Eccentricity (e) = (3/149) X 100 =0.02%

Exact eccentricity = 0.017%

152

146 Mean radius = 149 m km

e 29.8km/ sec

• For normal day to day calculations this small

eccentricity is insignificant & earth orbit can

be considered as round

• But for astronomical calculation & events this

eccentricity plays major roll

Planet Orbital Eccentricity (e)

Mercury 0.206Venus 0.007Earth 0.017Mars 0.093Jupiter 0.048Saturn 0.056Uranus 0.046Neptune 0.010

Eccentricity of other planets

Kepler’s 2nd Law:A line joining any planet to the Sun sweeps out equal areas in equal times.

Kepler laws

– Planet move in Elliptical orbits & not round

SUNPlanet

Mean distance ‘R’ Area ‘A’

in time ‘t’

Line joining Sun & planet

Area ‘B’ covered in same time interval‘t’

–A line joining planet & sun sweeps equal area in equal interval of time. Area ‘A’ = Area ‘B’–Square of time required by planet to complete one revolution around sun ( rotation period ‘T’) is proportional to cube of it’s mean distance from sun ---- T 2 R 3

Kepler’s 3rd Law:The square of the period of any planet is proportional to the cube of the semi-major axis of its elliptical orbit.

T2 is proportional to r3

1AU

1AU = 149million KmAstronomical unit of distance

Solar System

Let us study this data with graphs

This suggest explosion theory of planet formation

Orbital period

Mean orbital radius in AU

Note uneven distribution of planets around sun

Kepler’s 3rd Law

Straight line graph as per kepler lawT2 R3

T2

R3

Mean orbital radius in AU3

T2

Example: Jupiter’s Orbit Jupiter’s mean orbital radius is rJ = 5.20 AU (Earth’s orbit is 1 AU). What is the period TJ of Jupiter’s orbit around the Sun?

2 32 3 so J J

E E

T rT Cr

T r

3/2 3/2(5.20 AU)(1.0 yr) 11.9 yr(1.00 AU)

JJ E

E

rT T

r

Newton's theory of gravitation

Newton’s theory of Gravitation

• Experimentally acceleration & force on falling body are measured

• Acceleration & attractive forces on moon were calculated based on measurements of moon rotational period & moon to earth distance

With this Newton made a bold statement that the type force responsible for Moon’s rotation around Earth, Apple falling on earth & Planets rotating around sun is of same nature & governed by same laws & called them as laws of gravitation

Newton’s ingenuity

Newton proposed formula with which he could calculate the measured force on apple & moon

Newton’s ingenuity

Newton proposed that • Every object in the universe attracts every other • This force is inversely proportional to the square of

the distance between the objects.• The force is directly proportional to the product of

the masses of the two objects.

1 21 on 2 2 on 1 2

mmF F Gr

211 26.673 10 N m /kgG

Newton derived kepler’s law using law of gravity

(L3 M-1T-2)

Newton proposed that • Every object in the universe attracts every other • This force is inversely proportional to the square of

the distance between the objects.• The force is directly proportional to the product of

the masses of the two objects.

1 21 on 2 2 on 1 2

mmF F Gr

Newton’s Law of Gravity

1 21 on 2 2 on 1 2

mmF F Gr

Why r2 in the formula ??

1 21 on 2 2 on 1 2

mmF F Gr

If gravitational acceleration (g =9.8m/sec2 )is considered same at earth surface and for moon & then calculated value of moon’s orbital period will be 11 hr instead of 27.3 days as observed

Gravitational force

Alternatively if gmoon is calculated from for moon’s rotation period of 27.3 days, then we getgmoon= 2.72 x 10-3 m/s2

How to explain this difference in gmoon & gearth ?

Only possibility is that gravitational force due to earth is varying with distance from earth( hence the difference)

Then what is the relation ?

If we try different possible relation between g & distance ‘r’ for apple & moon, like g x r, g2 x r2 , g x r2 etc

We observe that

gM x Rm2 = (2.72x10-3 m/s2) (3.84x108 m)2

= 4.01x1014 m3/s2

gE x RE2 = (9.81 m/s2)(6.37x106 m)2

= 3.98x1014 m3/s2

Rm Re

gEgM

• Thus product of g x R2 ( 4.01x1014

m3/s2) is constant for moon and earth

• Hence g 1/R2

• This is in line with Newton's law of gravitation

Units of G

1 21 on 2 2 on 1 2

mmF F Gr

211 26.673 10 N m /kgG

Force in Newton = G x ( Kg x Kg) /m2 )

Unit of G = N m2 / kg2

= (L3 M-1T-2)

This results in formula

Thus addition of all forces due to mass at different points gives net force in direction of center line as if all mass is at center

For gravity calculations mass can be considered as point mass at center for deciding ‘ r ‘.

Net force along center line due to mass A & B

Force due to mass ‘A’

Mass B

r1

EARTH

Force due to mass ‘B’

r2

r

Mass A

These forces cancel.Hence net force perpendicular to center line is zero

r

m1 m2

F F

1 21 on 2 2 on 1 2

mmF F Gr

Q1: Which of these systems has the largest force of gravitational attraction ?

M1 = 4

.a)

.b)

.c)

.d)

.e)

Ans to q14/16=1/4

1/1=1

1/4

4/16=1/4

16/64=1/4

1 21 on 2 2 on 1 2

mmF F Gr

Ans to q1

4/16=1/4

1/1=1

1/4

4/16=1/4

16/64=1/4

• G is a very small number; this means that the force of gravity is negligible, unless there is a very large mass involved (such as the Earth).

• Gravitational forces are more predominant for celestial objects having huge mass

• If an object is being acted upon by several different gravitational forces, the net force on it is the vector sum of the individual forces.

• This is called the principle of superposition.

• Gravitational force between two object on earth is very small & can be neglected. Ex– Force between to 100 kg balls one meter

apart is 6.7 10 –7 N

(G ) ( m1 m2 )

r 2F =

(6.7 10 –11 ) ( 100 100)=

( 11 )

6.7 10 –11 10 4 =

6.7 10 –7 N =

Gravitational force between a point mass and a sphere is the same as if all the mass of the sphere is concentrated at its center.

Even for large objects like earth & moon mass can be considered as point mass at center for deciding ‘ r ‘.

Mass B

r1

EARTH

Mass AForce due to mass ‘A’

Force due to mass ‘B’

Net force along center line due to mass A & B

These forces cancel.Hence net force perpendicular to center line is zero

r2

Thus addition of all forces due to mass at different points gives net force in direction of center line as if all mass is at center

r

As G is constant, it’s value must be found out by experimentHenry Cavendish did experiment to measure ‘G’

Henry Cavendish experiment to measure ‘G’

211 26.673 10 N m /kgG

Measured value of G

Gravitational acceleration due to earth

Mass m1

Earth mass Me

F = m1 g

A big stone , small stone ,a piece of paper dropped from same height will reach ground in same time

g is independent of mass of body.

g

(G ) ( m1 Me)

r 2F = &

F m1 g =

(G ) ( m1 Me)

r 2

= 9.80 m / sec2

g =G Me

r 2

rValue of g can be measured experimentally

1 ½ (9.8 )12 = 4.9 m

2 ½ (9.8)22 = 19.6 m3 ½ (9.8)32 = 44.1 m4 ½ (9.8)42 = 78.4 m

Distance traveled

Speed of falling

Timesec

S = 1/ 2 g t2

Free fall upward motion

• Density of earth = Me / Vol = 5.5 gm cm 3

• This 5.5 times more than density of earth’s crust which consist water & soil (Density @ 1)

• From G we came to know how density is earth’s core

Density of earth’s core = app. 8.8 gm / cm 3!!!

Earth’s crustDensity @ 1

Earth’s coreDensity =8.8

Calculating mass of sun:

Ms = 2.0 10 30 kg .300,000 times more than earth

Now knowing Ms & rotation period of diff planets we can calculate their distance from sun

r 2

(G ) ( Ms Me )F = Fc =

• From period earth’s revolution can calculate centrifugal force on earth ( Fc= Me r 2

)

Earth

Sun

F

Fc

Ms

Me

• Now Fc = F, gravitational force due to sun

r

End

Gravitation and tides in sea

Reference slides

Limitation of gravitational: constant G• The measurement of nearest position of

mercury from sun shows different value than calculated by Newton law

• So if we calculate value of G from this measurement it shows different value

Contradiction to value of GPerihelion of Mercury

43 secs change per 100 years

Chang in position of nearest point of Mercury to ein in it’s orbit. Mercury

orbit• 430sec/ 100 years displacement is small amount• But such change should not occur as per

Newton’s laws• Hence Newton’s laws can not explain this

observed phenomenon of Perihelion of Mercury

Hypothesis:Value of G depends on average distribution of mass in UniverseIf we come very near to a massive object (Mercury near sun) that time value of g will change

Satellite motion

GeostationarySatellites

RotatingEarth

Satellite Rotating at same angular velocity

Geostationary Satellites

PE of satellite at rotating at

radius r(G ) ( Me m )

r= -

22

2E EGM m GMmv vr r r

r

v

Total energy = KE + PEMe

212

EGM mE K U mvr

12 2

E E EGM GM m GM mE mr r r

EGM mUr

12E U

Total energy of a satellite in a circular orbit around the Earth is half of its gravitational potential energy.

= gUGMvr m

2 2 2

2 2

4E Eg

M m GMmv rF Gr r r T

Angular speed V=r

R3 = GM2

Geostatic Satellite Orbits : earth = Satellite

The product GM is called as the geocentric gravitational constant = 398,588. km3 s−2:

The angular speed ω = 2π rad / Sederal day =0.00007288

Angular speed V=rR3 = GM

2

Orbital Period = 23 h 56 m 4. sec( 1 sidereal day)= 86164 Sec

Geostatic Satellite Orbits : earth = Satellite

Angular speed V=rR3 = GM

2

Geostatic Satellite Orbits : earth = Satellite

R3 = 6.67428 10 -11 x 5.972 x 1024

(7.288 10-5)2

6.67428 x 5.972 x 1024 x10 -11 x 10 10

(7.288 x 7.228)=

Angular speed V=rR3 = GM

2

Geostatic Satellite Orbits : earth = Satellite

R3 = 0.75034 x 1023

= 0.075034 x 1024 meters = 0.42178 x 108 meters

= 42178 Km R

R

Geostatic Satellite Orbits : earth = Satellite

R = 42178Km

6378km

r

.r = 42178 – 6378 = 35800km

Let us use this for finding the geostationary orbit of an object in relation to mars. The geocentric gravitational constant GM (which is μ) for mars has the value of 42,828 km3s-2, and the known rotational period (T) of mars is 88,642.66 seconds. Since ω = 2π/T, using the formula above, the value of ω is found to be approx .00007088218. Thus, r^3 = 8,524,300,617,153.08. It is then simply a matter of finding the cube root of r^3, which is 20,427.6255 and subtracting the equatorial radius of mars (3396.2 km) to give us our answer of 17,031.42 km.

Whether satellite

Positioning satelliteGeo synchronous satellite

Launching satellite

Revision

End