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Gravitational force is always attractive, conservative and independent of medium and acts on line foining the centres of the bodies.

**Newton's law of gravitation:**

Force acting between two bodies directly proportional to product of the masses and inversly proportional to square of distance between them.

Given by,

$F=$$r_{2}Gm_{1}m_{2} $ where G= Universal gas constant=$6.67∗10_{-11}Nm_{2}kg_{-2}$

**Gravity** is the phenomenon of attraction of surrounding bodies to centre of particular body and is a vector quantity It holds the atmosphere around earth.

Given by:

$F=mg$

$F=r_{2}GMm $ so, $g=r_{2}GM $

so, $g=34 πGρr$

**Variation of acceleration due to gravity**

1.**with height**

$g_{′}=(R+hR )_{2}∗g$

= $(1−R2h )∗g$ for h< < R

2.**With depth**

$g_{′}=R_{3}GM ∗(R−x)$$=g(1−Rx )$

At center of earth g'=0; x=R

3.**With rotation of earth**

$g_{eff}=g−Rw_{2}cos_{2}θ$

At equator, $θ=0°$ ; $g_{eff}=g−Rw_{2}$

At poles, $θ=90°$; $g_{eff}=g$ There is no effect of rotation at poles

Radius of earth= 6400 Km

**Gravitational intensity**: It is the force experienced by unit point mass placed at that point of consideration with gravitational field $V=mw =−RGM $. It is scalar quantity.

**Gravitational Potential (V): **It is work done to take a unit mass from infinity to a certain point of consideration within gravitational field.field $V=mw =−RGM $. It is scalar quantity.

**Gravitational potential energy: **It is work done to take a body of mass m from infinity to a point of consideration within gravitational field.

$V=−rGmM $

Gravitational potential energy on height h from the surface of earth

$V=−R+hGMm $

Its maximum value is 0 to infinity.

Change in gravitational PE when a body is raised from surface to height h is

$ΔU=−RGMm −(R+h−GMm )$

=$mgh(R+hR )$

Gravitational potential and intensity are related as

$E=−rdV $

$V=∫−E.dr$

**Gravitational field and potential**

1.For point mass at distance r,

$E=r_{2}GM $ and $V=−rGM $

2.For hollow sphere of mass M having radius R

i. Outside point (r>R): $E=r_{2}GM $ and $V=−rGM $

ii. On the surface (r=R): $E=R_{2}GM $ and $V=−RGM $

iii. Inside point (r$E=0$

3.For solid sphere of mass M and radius R

i. Outside point (r>R): $E=r_{2}GM $ and $V=−rGM $

ii. On the surface (r=R): $E=R_{2}GM $ and $V=−RGM $

iii. Inside point (r$E=R_{3}GM ∗r$

4.For ring of radius R, mass M on its axis

$E=(R_{2}+r_{2})_{23}GM ∗r$ $V=(R_{2}+r_{2})_{21}−GM $

#If two point masses $m_{1}$and $m_{2}$are kept at distance r then field intensity becomes zero at a distance

$x_{2}Gm_{1} =(r−x)_{2}Gm_{2} $ $→$ $x=m_{1} +m_{2} m_{1} .r $

**Motion of satellites**

$r_{2}GMm =rmv_{2} $ $→$ $v=(rGM )_{21}$

Time period

$T=v2πr =GM 2π ∗r_{23}$$∝$ $r_{23}$

For nearer satellite, r=R so, $v=(RGR_{2} )_{21}=gR $

and $T=2π(gR )_{21}$= 84.6 min

Geostationary satellite appears stationary with respect to earth and launched on equateral plane (time period t = 24 hours) and height= 36,000 km

**Energy of satellite**

P.E. = $r−GMm $

K.E.=$21 mv_{2}=2rGMm $= binding energy

T.E. = P.E. + K.E.= $2r−GMm $= (PE=2TE=-2KE)

If a body is projected from earth then the maximum height attained by it is

$TE_{initial}=TE_{final}$

or, $21 mv_{2}+(R+GMm )=0+(R+h−GMm )$

or, $2v_{2} =(R+h)hgR $

so, $h=(v2gR )−1R =(vv )_{2}−1R $

For escape velocity, h$→$$∞$ $v_{e}=2gR $ called escape velocity = 11.2 km/s for each

$v_{escape}=2 ∗v_{orbital}$

So, on increasing the orbital velocity of satellite by 41.4% the satellite escapes ( or KE by 100%)