CLASS 11 GRAVITATION
GRAVITATION
✤ Gravitational forces: It is the forces of attraction b/w two bodies in Universe.
F ∝ m1m2
F ∝ 1/r²
F = G (m1m2) / r²
Here G is Gravitational Constant
[ G = 6.67 x 10⁻¹¹ Nm²/ Kg² ]
[ Dimensions: M⁻¹L³T⁻² ]
❈ Importance of Gravitational Forces:-
-> It makes us stable on earth
-> It makes the planet stable.
-> It makes land & Sea Breezes.
✤ Force of GRAVITY: It is the force of attraction of earth towards its center.
F = mg
Sl unit: Newton
g = 9.8 m/s2
❄ Relationship blw 'g' & G
Let mass of earth = M
mass of body = m
Radius = R
:. Gravitational force: G. Mm / R² - ①
Force due to Gravity
f = mg - ②
from ① & ②
mg = GMm/R2
g= GM/R2 Here, M=6x1024 kg
R=6400km
= 6400x103m
:. g = 6.67×10-11x 6x1024/ 64×64×1010
= 6.67 x6x103/64x64
= 6670x3/32X64 = 9975/1024
[ g =9.8m/s2 ]
[ 4H =R ]
❈ GRAVITATIONAL FIELD
It is the surface around a planet in which only body get a force of attraction.
❈ GRAVITATIONAL INTENSITY
It is defined at the gravitational force per unit mass –
:. E = F/m {N/m}
We have f=GMm/r2
E = GM/r2
✤ GRAVITATIONAL POTENTIAL ENERGY
It is defined as the amount of near k denote pull a losely against Gravitational force.
Let d w work is done in dn displacement then
dw = fdx
dw = GMm/x2 dx
0∫wdw = ∞∫r GMm dx/x2
W = GMm ∞∫r x-² dx
= GMm ( -1/x )r∞
= -GMm ( 1/r - -1/∞ )
W = -GMm / R
This work is stored in the form of
PE = M = -GMm / r Joules.
❆ Gravitational Potential: It is defined as the work done per unit mass.
V = W / m J/kg , V = -GM / R
❈ Satellite :-
when a small body rotate (orbit) the bigger body then small body is called satellite of bigger body.
mv2/r = GMm / r2
v²=GM / r
v=√GM/r
Hence, v is called orbital velocity
❆ VARIATION IN g:-
1) upward at 'h' height above the surface:-
Let a body is placed at h height above the surface of the earth at which
acc due to gravity is g'
We know, g'=GM/(R+h)²………....①
On the surface of the earth
g=GM/R²………...②
①/②
g'/g = [GM/(R+h)²] / GM/R² = R2 / (R+h)2
g’=g [R /(R+h)]²
:. g'
It shows that acc. decreases, with increasing height.
Let R>>h
g'/g=R²(R+h)-2
= R²R-2(1+h/R) -2
= (1+h/R) -2
|g'=g(1-h/R)|
❈ Cases: At 'd' depth inside the earth
Let at 'd' depth acceleration to gravity is g'
:. g' = GM / (R-d)2 M = Vol. x density
= Gx 4/3 π (R-d)3 . ρ/ (R-d)2 = 4/3 π (R-d)3 x ρ
g' = Gx 4/3 π(R-d) ρ --(1) ρ = Density.
On the surface of the earth
g = GM / R2
= Gx 4/3 πR3 ρ / R2
g = Gx 4/3 πRρ --(2)
g'/g = R-d / R
g'=g(1-d/R) -> this equation also represents g'
Case : At the centre; d=R
:. g' = g (1-R/R)
g' = 0
* At the centre, weight = 0. It is called weightlessness.
✤ CONTINUATION OF SATELLITE:-
❈ ORBITAL VELOCITY:- It doesn't depend up on the mass of
g = GM/R2
GM = gR2
v = √gR2 / R+h
If R >> h
v = √gr
Note: If acc. Due to Gravity as 'h' height &'d' depth become equal find Relation
Sol: ATQ g(1 -2h/R) = g (1 - d/R)
2h/R = d/R
d=2h
❈ TIME PERIOD OF SATELLITE :- It is the time-taken by the satellite to complete one revolution
NOTE: T2 = 4π2r3/GM
M = 4π2r3 / GT2
↓
Mass of Any Planet.
✤ ENERGY OF SATELLITE: Total energy in satellite is equal to the sum of KE & PE.
Now, KE= 1/2 mv²
= 1/2 m (G.M/r)
KE = 1/2 GMM / R-- ① & PE = -1/2 G Mm / r --- (2)
:. Total energy = GMm/2r – GMm/r
µ = - GMM / 2R Joules
:. Here-ve shows that ley this energy is, energy of attraction
✤ Binding energy: Amount of energy required to pull the satellite from its orbit...
BE = +GMm/2R Joules
✤ Escape Velocity: It is the minimum to velocity by vehicle body is thrown from a planet & it does not return itself on the planet for the earth escape velocity is 11.2km/s.
Proof:
Let a body of mass 'm' is thrown upward with Velocity v.
30 k.E=1/2 mv²-①
Let dw work is done in du displacement.
:. dw = fdx
:. dw = GMM/x2 dx
0∫wdw = R∫GMM/x2 dx
W-0 = (-GMm/x)∞R
w= GMM (1/∞ - 1/R)
w = GMm/R (2)
By theorem of energy principles
W= KE
GMm/R = 1/2mv²
Ve= √2GM/R
we have, g = GM/R²
R²g = GM
Ve = √2Rg
Note :-(i) Escape Velocity doesn't depend upon the direction of projection
V₁ = V₂ = V₃
(ii) Escape Velocity depend upon the mass of Planet
❆ RELATION B/W ESCAPE VELOCITY (Ve) & ORBITAL VELOCITY (Vo) :-
We have ; Ve = √2Rh - (1) & Orbital Velocity
Vo = √GR - (2)
(1)/(2) => Ve/Vo = √2gR/√gR
Ve = √2Vo
✤ TYPE OF SPACE
-> Geostationary Satellites :-
-> These satellites always rotating around the earth from w->e direction with same speed
-> They rotate from w->e.
-> height of those satellites is app. 36,000 km above the earth.
USES :-
-> In Communicating radio, T.V. & telephone signals across the world.
-> Geostationary Satellites act as reflectors of such signals.
-> In forecasting weather.
-> In studying of meteorites & upper regions of atmosphere.
-> Polar Satellites :-
-> these satellites rotate around the pole
USES :-
-> Polar satellites provide weather reports
-> They are used in spying work for military purposes
-> British polar satellite first detected hole in @ ozone layer.
-> They are used to study topography of moon, venus & mars.
✤ KEPLER'S LAW: -
Law of Orbit: It states that every planet revolve around the Sun in elliptical orbit & the sun stay one of its forces.
Law of Area: It states that line joining the planet to the sun sweep equal area in equal in equal time.
By 2nd law
ar (SPQ) = ar (SAB)
Law of time: It states that square of the time period of a planet is always proportional to the cube of distance of the planet to the sun.
[ T² ∝ r³ ]
⇒ (T₁/T₂)² = (r₁/r₂ )³
✤ Short Question Summary:
✚ Value of 'g'on the surface of moon is about 1/6th of its value on the surface earth.
✚ The Two factors are ① the Value of acceleration due to Gravity on the Planet ① Surface temperature of the planet
✚ If the earth stops rotating about its axis other than value of 'g' increases by w²R
✚ Due to small value 'g' the escape velocity on the earth, the air cannot escape from the surface of the earth. Hence the earth has no atmosphere.
✚ The speed of satellite during descent is much larger than that during its orbit. As the air resistance is directly proportional to velocity, so heat
produced during descent is very large & the Satellite burns up.
✚ An astronaut in a spacecraft feel weightlessness this is because the orbiting spacecraft along with the astronaut is in a state of free fall
towards the earth.
✚ The mass of the nearly 10% of mass of earth, then both forces will be equal in magnitude as gravitational force b/w two bodies is a mutual
force.
✚ Earth is continuously pulling the moon towards its center, still it does not fall because Gravitational force of attraction due to earth provides the centripetal force, which keeps the moon's orbit around the earth.
✚ Moreover, this gravitational force which acts 1 to the Velocity of the moon.
✚ When moving a merry go around we fall side this is because our weight appears to decrease when we move closer 2.
✚ Increase when we move up.
✚ If the Gravitational force of the earth suddenly becomes zero, the satellite will stop revolving around the earth & it will move in a direction
tangential to its original orbit with a speed with which it was revolving around the earth.
✚ The radii of two planets R & 2R respectively & their densities P/P/2
Respectively,
Then g= GM/R² = G/R² *4πRSp = 4πGrρ
g∝Rρ
∴g₁ / g2 = Rρ /QR(P/2) = 1:1
* * * * * * * * * * * * * * *
