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**Gravitation:-**

**Kepler’s first law (law of elliptical orbit):-**A planet moves round the sun in an elliptical orbit with sun situated at one of its foci.**Kepler’s second law (law of areal velocities):-**A planet moves round the sun in such a way that its areal velocity is constant.**Kepler’s third law (law of time period):**- A planet moves round the sun in such a way that the square of its period is proportional to the cube of semi major axis of its elliptical orbit.

T^{2 }∝ R^{3}

Here R is the radius of orbit.

T^{2}= (4π^{2}/GM)R^{3}**Newton’s law of gravitation:-**

Every particle of matter in this universe attracts every other particle with a forcer which varies directly as the product of masses of two particles and inversely as the square of the distance between them.

F= GMm/r^{2}

Here, G is universal gravitational constant. G = 6.67 ´10^{ -11}Nm^{2}/ kg^{2}**Dimensional formula of G:**G = Fr^{2}/Mm =[MLT^{-2}][L^{2}]/[M^{2}] = [M^{-1}L^{3}T^{-2}]**Acceleration due to gravity (g):-**g = GM/R^{2}**Variation of g with altitude:-**g' = g(1- 2h/R), if h<<R. Here R is the radius of earth and h is the height of the body above the surface of earth.**Variation of g with depth:-**g' = g(1- d/R). Here g' be the value of acceleration due to gravity at the depth d.**Variation with latitude:-****At poles:-**θ = 90°, g' = g**At equator:**- θ = 0°, g' = g (1-ω^{2}R/g)

Here ω is the angular velocity.**As g**= GM_{e}/R_{e}^{2}, therefore g_{pole }> g_{equator}**Gravitational Mass:**- m = FR^{2}/GM**Gravitational field intensity:-**

E = F/m

= GM/r^{2}**Weight:**- W= mg**Gravitational intensity on the surface of earth (E**_{s}):-

E_{s}= 4/3 (πRρG)

Here R is the radius of earth, ρ is the density of earth and G is the gravitational constant.**Gravitational potential energy (U):**- U = -GMm/r**(a) Two particles:**U = -Gm_{1}m_{2}/r**(b) hree particles:**U = -Gm_{1}m_{2}/r_{12}– Gm_{1}m_{3}/r_{13}– Gm_{2}m_{3}/r_{23}**Gravitational potential (V):**- V(r) = -GM/r

At surface of earth,

V_{s}= -GM/R

Here R is the radius of earth.**Escape velocity (v**_{e}):-

It is defined as the least velocity with which a body must be projected vertically upward in order that it may just escape the gravitational pull of earth.

v_{e}= √2GM/R

or, v_{e}= √2gR = √gD

Here R is the radius of earth and D is the diameter of the earth.- Escape velocity (v
_{e}) in terms of earth’s density:- v_{e}= R√8πGρ/3 **Orbital velocity (v**_{0}):-

v_{0}= √GM/r

If a satellite of mass m revolves in a circular orbit around the earth of radius R and h be the height of the satellite above the surface of the earth, then,

r = R+h

So, v_{0}= √MG/R+h = R√g/R+h

In the case of satellite, orbiting very close to the surface of earth, then orbital velocity will be,

v_{0 }= √gR**Relation between escape velocity v**- v_{e}and orbital velocity v_{0}:_{0}= v_{e}/√2 (if h<<R)**Time period of Satellite:**- Time period of a satellite is the time taken by the satellite to complete one revolution around the earth.

T = 2π√(R+h)^{3}/GM = (2π/R)√(R+h)^{3}/g

If h<<R, T = 2π√R/g**Height of satellite:**- h = [gR2T2/4π2]1/3 – R**Energy of satellite:-****Kinetic energy, K**= ½ mv_{0}^{2}= ½ (GMm/r)**Potential energy, U**= - GMm/r**Total energy, E**= K+U

= ½ (GMm/r) + (- GMm/r)

= -½ (GMm/r)**Gravitational force in terms of potential energy:**- F = – (dU/dR)**Acceleration on moon:-**

g_{m}= GM_{m}/R_{m}^{2}= 1/6 g_{earth }

Here Mm is the mass of moon and Rm is the radius of moon.**Gravitational field:-**

(a) Inside:-

(b) Outside:-**GRAVITATIONAL POTENTIAL & FIELD DUE TO VARIOUS OBJECTS**

**Projectile:-**

**Projectile fired at angle α with the horizontal:**- If a particle having initial speed u is projected at an angle α (angle of projection) with x-axis, then,**Time of Ascent, t**= (u sinα)/g**Total time of Flight, T**= (2u sinα)/g**Horizontal Range, R**= u^{2}sin^{2}α/g**Maximum Height, H**= u^{2}sin^{2}α/2g**Equation of trajectory, y**= xtanα-(gx^{2}/2u^{2}cos^{2}α)**Instantaneous velocity, V**=√(u^{2}+g^{2}t^{2}-2ugt sinα)

and

β = tan-1(usinα-gt/ucosα)**Projectile fired horizontally from a certain height:-****Equation of trajectory:**x^{2}= (2u^{2}/g)y

Time of descent (timer taken by the projectile to come down to the surface of earth), T = √2h/g

Horizontal Range, H = u√2h/g. Here u is the initial velocity of the body in horizontal direction.**Instantaneous velocity:-**

V=√u^{2}+g^{2}t^{2}

If β be the angle which V makes with the horizontal, then,

β = tan^{-1}(-gt/u)**Projectile fired at angle α with the vertical:-****Time of Ascent, t**= (u cosα)/g**Total time of Flight, T**= (2u cosα)/g**Horizontal Range, R**= u^{2}sin^{2}α/g**Maximum Height, H**= u^{2}cos^{2}α/2g**Equation of trajectory, y**= x cotα-(gx^{2}/2u^{2}sin^{2}α)**Instantaneous velocity, V**=√(u^{2}+g^{2}t^{2}-2ugt cosα) and β = tan^{-1}(ucosα-gt/usinα)**Projectile fired from the base of an inclined plane:-**

Horizontal Range, R = 2u^{2}cos(α+β) sinβ/gcos^{2}α

Time of flight, T = 2u sinβ/ gcosα

Here, α+β=θ

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