in this these twothings describes

2. GROUP MEMBER
HAFSA QAYYUM
19013323-008
HAMZA NASEER
19013323-016
AYAZ AHMAD
19013323-007
HASSAN ILYAS
19013323-030
NAWAZISH
19013323-032

3. NEWTON LAW OF GRAVITATION:
 NEWTON’S LAW OF GRAVITATION STATES THAT EVERY
POINT MASS IN THE UNIVERSE ATTRACTS EVERY OTHER
POINT MASS WITH A FORCE WHICH IS DIRECTLY
PROPORTIONAL TO THE PRODUCT OF THEIR MASSES &
INVERSELY PROPORTIONAL TO SQURE OF DISTANCE
BETWEEN THEM. THE DIRECTION OF FORCE IS ALONG
THE LINE JOINING THE PARTICLES.

4.  NEWTON’S LAW OF GRAVITATION IS THE INVERSE SQUARE LAW IN WHICH
FORCE IS INVERSELY PROPORTIONAL TO THE SQUARE OF THE DISTANCE
BETWEEN THE BODIES.
CONSIDER TWO MASSES m1 and m2 ARE SEPRATED BY DISTANCE r
F ∝ m1m2
F ∝ 1/r2
ON COMBINING BOTH EQUATIONS:
F ∝ m1m2/r2
F = 𝐺m1m2/r2
WHERE “G” IS PROPORTIONALTY CONSTANT. ITS VALUE IN SI UNIT IS
6.673*10-11 Nm2 /KG2 . THE GRAVITATION FORCE OF ATTRACTION BETWEEN
OBJECTS AROUND US IS VERY SMALL AND WE DO NOT FEEL BECAUSE VALUE OF
“G” IS VERY SMALL

5. CAVENDISH EXPERIMENT:
THE ESSENTIAL PART OF CAVENDISH EXPERIMENT USED TO DETERMINE THE
VALUE OF “G” ARE GIVEN BELOW:
 LIGHT RIGID ROD ABOUT 2-FEET LONG.
 TWO SMALL LEAD SPHERES ATTACHED TO THE ENDS OF ROD.
 SUSPENSION THIN WIRE.
 TWO LARGE LEAD SPHERES.
 LAMP SCALE ARRANGMENT.
PROCEDURE:
CONSIDER TWO IDENTICAL SMALL LEAD BALLS OF MASS m ARE CONNECTED
AT THE ENDS OF A ROD HAVING LENGTH “L” AS SHOWN IN FIGURE. THE ROD
IS ATTACHED WITH A FINE TORSION WIRE AT CENTRE AND HUNGS VERTICALLY
DOWNWARD. NOW TWO IDENTICAL BIG LEAD BALLS EACH OF MASS “M” ARE
BROUGHT NEAR BALLS ON OPPOSITE SIDE HAVING DISTANCE r.

6. THE GRAVITATION FORCE ACTING ON 1ST PAIR IS
F = 𝐺Mm/r2
THE GRAVITATION FORCE ACTING ON 2ND PAIR IS
F = 𝐺Mm/r2
THESE TWO FORCES ARE EQUAL IN MAGNITUDE BUT ACTING IN OPPOSITE DIRECTION AND
HAVE CONSTANT DISTANCE L. THESE FORCES FORMS A COUPLE. THIS COUPLE TENDS TO
ROTATE THE ROD IN CLOCKWISE DIRECTION.
THE MAGNITUDE OF THIS TOUQUE ACTING ON SYSTEM IS GIVEN AS:
𝜏 = COUPLE FORCE * SEPRATION BETWEEN FORCES
𝜏 = 𝐺MmL/r2
THIS TORQUE PRODUCES TWIST IN WIRE WHICH IS DIECTLY PROPORTIONAL TO TWISTING
ANGLE. THE VALUE OF ANGLE CAN BE DETERMINED WITH LAMP SCALE ARRANGEMENT.
𝜏 ∝ 𝜃
𝜏=𝜃C
WHERE C IS PROPORTIONALITY CONSTANT CALLED TWISTING CONSTANT. ON COMPARING
eq(1) AND eq(2)
𝐺MmL/r2 = 𝜃C
G=C𝜃r2 /MmL

7. THE VALUE OF PARAMETERS m, M, L, C & r ARE KNOWN. THE VALUE OF G
CAN B CALCULATED BY MEASURING THE TWISTING ANGLE WITH LAMP
SCALE ARRANGMENT THE VALUE OF G COMES OUT 6.673*10-11 Nm2 KG-2
THE NUMERICAL VALUE OF G IS EXTREMELY SMALL. THE FORCE OF
GRAVITATION ATTRACTION IS ONLY FELT FOR OBJECT WITH HEAVY MASSES.
---------------------------

8. VALUE OF g ON EARTH SURFACE
ASSUME THAT EARTH IS A PERFECT SPHERE OF RADIUS Re HAVING CONSTANT
DENSITY 𝜌. NOW PLACE A BODY OF MASS m ON EARTH SURFACE HAVING MASS
Me AS SHOWN IN FIG.
THE GRAVITATION FORCE BETWEEN EARTH AND BODY IS
THE WEIGHT FORCE ACTING ON BODY IS
F = mg
ON COMPARING BOTH EQUATION
mg = G Me m/Re
2
g = G Me /Re
2 ------------------(1)
THIS IS THE VALUE OF g ON EARTH SURFACE. IT IS CLEAR THAT ACCELERTION DUE
TO GRAVITY DOES DEPEND ON MASS m OF THE OBJECT. IT ONLY DEPENDS ON
THE MASS OF THE EARTH & DISTANCE BETWEEN BODY & CENTRE OF THE EARTH.
THE VALUE OF g ON MOON SURFACE IS
gm = GMm /Rm
2 -----------------(2)

9. DIVIDE EQ(1)&EQU(2)
g/gm = (GMe /Re
2 )(Rm
2 /GMm ) = (Me /Mm )*(Rm /Re)2
WE KNOW THAT MASS OF EARTH IS 100 TIMES MORE THAN MASS OF MOON
(Me =100Mm) & ITS RADIUS IS FOUR TIMES THAT OF THE MOON(Re = 4Rm).
g/gm = 100*(1/4) = 6
g = 6 gm
NUMERICAL VALUE OF ACCELERATION DUE TO GRAVITY
THE VALUE OF ACCELERATION DUE TO GRAVITY ON EARTH IS 9.8 m/s2 . IT MEANS
WHEN A BODY FALLS FREELY TOWARDS THE EARTH ITS VELOCITY INCREASE AT THE
RATE OF 9.8 m/s DURING ITS MOTION. SIMILARLY WHEN AN OBJECT IS PROJECTED
VERTICALLY UPWARDS ITS VELOCITY DECREASES AT THE RATE OF 9.8 m/s AND
EVETUALLY VELOCITY BECOMES ZERO.
MAXIMUM VERTICAL HEIGHT
THE HEIGHT AT WHICH THE VELOCITY OF AN OBJECT MOVING AGAINST GRAVITY
BECOMES ZERO IS CALLED MAXIMUM VERTICAL HEIGHT ATTAINED BY THE OBJECT.
WHEN VELOCITY BECOMES ZERO AT MAX. HEIGHT, THE OBJECT STARTS FALLING
DOWNWARD.

10. VARIATION OF ACCELERATION DUE TO GRAVITY
WE KNOW THAT FORCE OF GRAVITY VARIES FROM PLACE TO PLACE ON THE SURFACE
OF EARTH. THERE ARE TWO REASONS BEHIND THIS VARIATION CALLED SHAPE OF
EARTH AND ROTAIO OF EARTH.
THE EARTH IS NOT A PERFECT SPHERE BUT BULGES AT AQUATOR. THE DISTANCE OF A
BODY FROM CENTRE OF THE EARTH WILL CHANGE WHEN IT IS TAKEN FROM POLE TO
EQUATOR. CONCEQUENTLY, GRAVITATION FORCE ALSO VARIES.
DENSITY OF EARTH
THE VOLUME OF SPHERICAL EARTH HAVING Re = 6400 KM IS
V = 4/3 R3
e
V = 4/3 (6400 * 103)3 = 1.09 * 1021 m3
THE DENCITY OF UNIFORM SPHERICAL EARTH HAVING MASS 6 * 1024 Kg IS GIVEN AS
ρ = Me / Ve
ρ = 6 * 1024 / 1.09 * 1021 = 5.5 * 103 Kg/m3
THE DENSITY OF WATER IS 103 Kg/m3 . THE DENSITY OF EARTH IS 5.5 TIMES MORE
THEN THE DENSITY OF WATER.
---------------------------------------------

11. KEPLERS LAW OF AREA FOR PLANETRY MOTION
“A LINE JOINING THE SUN & PLANET SWEEPS OUT EQUAL AREAS IN
EQUAL INTERVAL OF TIME IS CALLED KEPLERS LAW OF AREA”
PROFF:
CONSIDER A PLANET HAVING MASS mP IS REVOLVING AROUND THE SUN
WHICH IS AT FOCUS OF AN ELLIPTICAL ORBIT AS SHOWN IN DIAGRAM. THE LINE
JOINING THE SUN & PLANET HAS LENGTH r & MAKES ANGLE 𝜃 WITH SEMI-MAJOR
AXIS. THE PLANET MOVES NEW POSITION & LINE JOINING THE SUN & PLANET
MAKES ANGLE (𝜃 + ∆𝜃) WITH SEMI MAJOR AXIS.

12. THE AREA SWEPT BY LINE JOINING THE SUN & PLANET IS GIVEN AS:
AREA SWEPT=AREA SPHERICAL SHADED TRIANGLE
∆A=1/2 (BASE)(HEIGHT)
∆A=1/2 r(S) [FOR SMALL ANGLE CHORD ≈ARC]
∆A=1/2 r(r∆𝜃) = 1/2 r2 ∆𝜃
BY DIVIDING ∆𝑡 ON BOTH SIDES
∆A/ ∆𝑡 = ½ r2 ∆𝜃 / ∆𝑡
BY TAKING LIMIT ON BOTH SIDES
lim
∆𝑡→0
∆A/ ∆𝑡 = ½ r2 lim
∆𝑡→0
∆𝜃 / ∆𝑡
dA/dt = ½ r2 ω ---------------(1)
THE ANGULAR MOMENTUM OF THE SYSTEM IS GIVEN AS:
L = r * P = r(mP v)sin𝜃 n
L = mP r V sin 90° n = mP rV
L = mP r (rω) = mp r2 ω
r2 ω=L/mp --------------------------(2)

13. PUTTING EQU (2) IN EQU (1)
dA/dt =L/2mP
THE TOTAL ANGULAR MOMENTUM OF ISOLATED SYSTEM IS CONSTANT ACCORDING
TO LAW OF CONSERVATION OF ANGULAR MOMENTUM. THE MASS OF A PLANET IS
ALSO CONSTANT.
dA/dt = CONSTANT
ITS MEANS AREA dA SWEPT BY A LINE JOINING THE SUN & PLANET IN TIME dt IS
CONSTANT HENCE KEPLERS LAW OF AREA PROVED.
THE KEPLERS LAW OF EQUAL AREAS DESCRIBES THE SPEED AT WHICH ANY GIVEN
PLANET WILL ORDITING THE SUN. THE SPEED AT WHICH ANY PLANET MOVES
THROUGH SPACE IS CONSTANTLY CHANGING. A PLANET MOVES FASTEST WHEN IT
IS CLOSEST TO THE SUN & SLOWEST WHEN IT IS FURTHEST FROM THE SUN.
---------------------------------------

14. KEPLERS LAW OF
PERIODS:THE SQUARE OF TIME PERIOD OF A PLANET IS DIRECTLY PROPORTIONAL TO CUBE
OF THE MEAN DISTANCE BETWEEN SUN AND PLANET IS CALLED KEPLERS LAW OF
PERIODS. THE RATIO T2 /r3 IS CONSTANT FOR ALL PLANETS ORBITING THE SUN.
PROOF:
CONSIDER A PLANET OF MASS mp IS REVOLVING ARROUND SUN HAVING MASS ms
IN ELIPTICAL ORBIT AS SHOWN IN DIAGRAM. THE MEAN DISTANCE BETWEEN
PLANET AND SUN IS r.
THE GRAVITATIONAL FORCE BETWEEN SUN AND PLANET IS
F = G mp ms /r2 ----------(1)
THIS GRAVITATIONAL FORCE PROVIDES CENTRIPETSAL FORCE TO PLANET TO MOVE
IN ORBITS.
F = mp V2 /r = mp (r𝜔)2 /r = mp r𝜔2
F = mp r𝜔2 -------------------(2)

15. ON COMPARING EQU (1) & (2)
mp r𝜔2 = G mp ms /r2
𝜔2 = G ms /r3 -----------------(3)
THE TIME NEEDED BUY THE PLANET TO COMPLETE ONE ROTATIOIN ARROUND SUN
HAVING MEAN DISTANCE r IS CALLED TIME PERIOD.
𝜔 = 2𝜋 / T
𝜔2 = 4 𝜋2 / T2 ------------------(4)
0M COMPARING EQU (3) & (4)
4 𝜋2 / T2 = G ms /r3
T2 = (4𝜋2 /Gms ) r3
T2 ∝ r3
THIS RELATION STATES THAT THE SQURE OF TIME OF A PLANET IS DIRECTLY
PROPORTIONAL TO CUBE OF THE MEAN DISTANCE BETWEEN SUN & PLANET. HENCE
LAW OF PERIODS IS PROVED.