Whenever an object falls toward the earth an acceleration is involved. This acceleration is due to the earth's gravitational pull and hence is called acceleration due to gravity or acceleration due to the force of the earth.

Acceleration due to gravity , `g = (GM_e)/(R_e^2)`

where, `M_e =` mass of the earth

`R_e =` radius of the earth

The value of the g on the surface of the earth is equal to `9.8 m//s^2`

`text(Variation in the value of g)`

1. The value of g is minimum at equator and maximum at poles (it happens due to shape of the earth).

2. The observed value of g at the latitude `lamda` is
`g' = g - R_eomega^2 cos ^2 lamda`
At equator, `lamda = 0^o, cos lamda = cos 0^o = 1, g' =g_e`
`g_e = g - R_e omega^2`
At pole, `lamda =90^o, cos x = 0, g' = g_p`

3. The value of g at height h above the earth's surface decreases `g' = g/(1 + h/R_e)^2 ~~ g (1 - (2h)/R_e)`

4. The value of g at depth d below the earth's surface decreases, `g' = g( 1- d/R_e)`

5. The value of g varies due to non-uniformity of the earth (density of the earth ).

Gravitational field of a given body is the space around it, within which gravitational effect due to that body may be experienced.

`text(Intensity of Gravitational Field)`
The intensity of gravitational field at a point is equal to the force acting on the unit mass at that point. Intensity gravitational field is given by
`(GM)/r^2`
Its SI unit is N/kg

Gravitational potential at a point in gravitational field is equal to the work done in carrying a unit mass from infinity to that point. Gravitational potential due to mass m at a distance r is `V = - (GM)/r`

► `text(Gravitational Intensity and Potential due to a Solid and Spherical Shell)`
`text(Solid Sphere)`

For these calculations, we will assume the sphere is placed at its own centre.

(i) `E_text(outside) = (-GM)/r^2, V_text(outside) = (-GM)/r`

(ii) `E_text(Surface) = (-GM)/R^2, V_(Surface ) = (-GM)/R`

(iii) `E_text(inside) = (-GM)/R^3 r , V_text(inside) = (-GM)/(2R^3) ( 3 R^2 - r^2)`

`text(Spherical Shell)`

(i) `E_text(outside) = (-GM)/r^2, V_text(outside) = (-GM)/r`

(ii) `E_text(surface) = (-GM)/R^2 , V_text(surface) = (-GM)/R`

(iii) `E_text(inside) = 0 , V_text(inside) = (-GM)/R`

where , r is the distance from the centre and R is the radius.

Gravitational potential energy of a body at a point is equal to work done in assembling the system of masses from the infinity to its present configuration.

The gravitational potential energy of masses M and m at a distance r is given by

`U = -(GM_e m)/r`

Potential energy of a particle of mass m on the earth's surface is

`U = -(GM_e m) /R_e`

The heavenly body which revolves around the sun is called planet, e.g. the earth.

`text(Kepler's laws of Planetary Motion)`
To explain the motion of the planets, Kepler formulated the following three laws

`text(1. Law of Orbits)`
Every planet moves in an elliptical orbit around the sun, with sun situated at one of the foci of the ellipse.

`text(2. Law of Areas)`
The line joining a planet to the sun sweeps equal areas in equal intervals of time, howsoever small these time intervals may be. Mathematically,

`(dA)/(dt) = 1/2 rv = ` constant `= L/(2m)`

Where `L =` angular momentum of planet and
m = mass of planet.

`text(3. Law of Periods)`
The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet. Thus,

`T^2 prop r^3` or `(T_1/T_2)^2 = (r_1/r_2)^3`

A satellite is a body which is revolving continuously in an orbit around a comparatively much larger body.
Satellite are of two types
1. `text(Natural Satellite)`
A satellite created by nature is called a natural satellite. e.g. Moon
2. `text(Artificial Satellite)`
A man made satellite is called an artificial satellite. e.g. SPUTNIK-I

Orbital velocity of a satellite is the velocity required to put the satellite into its orbit around the earth.

Orbital velocity `v_o = sqrt((GM)/(R +h))`

where, M = mass of the earth
R = radius of the earth
h = height of the satellite above the earth surface

and R + h = orbital radius of the satellite

If g is the acceleration due to gravity on the earth's surface, then

`g = (GM)/R^2 ` or `gR^2 = GM`

Hence `v_o = sqrt( (gR^2)/(R + h)) = R sqrt (g/(R + h)`

When the satellite revolves close to the surface of the earth, i.e. h = 0, then the orbital velocity will become

`v_o = sqrt(gR)`

As ` g = 9.5 ` m/s² and `R = 6.4 xx 10^6 m` , so

`v_o = sqrt(9.8 xx 6.4 xx 10^6)`
`= 7.92 xx 10^3` m/s
`= 7.92 ` km/s `approx8 km//s`

It is the time taken by a satellite to complete one revolution around the earth. It is given by
`T = text( circumference of the orbit)/text( orbital velocity ) = (2 pi ( R + h))/v_o`

As we know, orbital velocity , `v_o = sqrt((GM)/(R + h) )`

`:. T = (2 pi ( R + h))/sqrt((GM)/(R + h)) = 2 pi sqrt((R + h)^3/(GM) )`
But `g = (GM)/R^2 ` or `gR^2 = GM`
`:. T = 2 pi sqrt((R +h)^3 /(gR^2) )`..................(i)

When the satellite revolves close to the surface of the earth, i.e. h = 0, then the time period will be
`T = 2 pi sqrt (R^3/(GM)) = 2 pi sqrt (R^3 /(gR^2 )) = 2 pi sqrt(R/g) = sqrt(((3 pi) /(G rho ))`
Putting ` g = 9.8 m//s^2` and ` R = 6.4 xx 10^6 m` , we get
`T = 2 pi sqrt(( 6.4 xx 10^6)/9.8 ) = 50 78 s = 84.6` min

The potential energy of a satellite is due to its position w.r.t earth. It appears because of gravitational pull acting on satellite due to the earth.
Potential energy of a satellite, `U = - (GMm)/r`

The kinetic energy of a satellite is due to its orbital motion.
Kinetic energy of a satellite , `K = (GMm)/(2 r)`

The total energy of a satellite revolving around the earth is equal to the sum of potential energy and kinetic energy of the satellite.
Total energy of a satellite , `E = K + U = (GMm)/(2r) - (GMm)/r`
`:. E = - (GMm)/(2r)`

If the satellite is orbiting close to the earth, then `r ~~ R`.
Now, total energy of satellite,
`E = - (GMm)/(2r)`

The energy required to remove the satellite from its orbit around the earth to infinity is called binding energy of the satellite.
The total energy o f a satellite is `- (GMm)/(2r)`

In order to escape to infinity, it must be supplied an extra energy equal to `+ (GMm)/(2r)`

So that, its total energy E becomes equal to zero. Hence, binding energy of a satellite `= (GMm)/(2r)`

Escape velocity is the minimum velocity with which a body must be projected vertically upwards in order that it may just escape from the gravitational field of the earth.
Escape velocity `v_e = sqrt(2gR)`

The value of escape velocity on the earth's surface is 11.6 km/s

Relation between orbital velocity of a satellite and escape velocity is `v_e = sqrt2 v_o`

Mass of a body is the measure of its inertia, greater the mass of the body greater will be its inertia. While the weight of a body at any place is the product of its mass and the gravitational acceleration at that place.

`w = mg`

Mass of a body is always constant but the weight of the body can slightly vary from place to place on the earth. At the pole, the weight of a body will be maximum, whereas at the equator, it is minimum.

`text(Weight of a Body in Elevator or Lift)`
When a person (or any body) is inside art accelerating lift, then there ts a change in its called its apparent weight followed by some conditions

(i) When lift is ascending with an upward acceleration or descending with a downward deceleration, then the body experiences that its weight is increased.

(ii) When lift is descending with a downward acceleration or ascending with an upward deceleration, then the body experiences that its weight is decreased

(iii) When lift is moving downward with the acceleration same as the gravitational acceleration, then the body experiences weightlessness.

`text(Weight of a Body at the Moon)`

As mass and radius of the moon is less than the earth, so the force of gravity at the moon is also less than the earth. Its value at the moon's surface is `g/6`