## Newton’s law of Gravitation:

1. Everybody in the universe attracts other bodies towards its centre. The force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of distance between their centers

$F \propto \dfrac{m_1 m_2}{r^2}$

$F=G \dfrac{m_1 m_2}{r^2}$

Where $G=6.67×10^{-11} Nm^2/kg^2$ is called universal gravitational constant whose value remain same at all the places and for all bodies but depends on system of units.

2. $\textbf{Gravitational Force:}$

(1) Always attractive, conservative and central force depending upon the masses of the bodies but is independent of medium and presence of any other bodies nearby.

(2) It acts along the line joining the centers of the bodies.

(3) It produces action and reaction and hence obeys Newton’s third law of motion.

(4) It is effective for large distance (Nuclear force is effective only for some very small distance.

(5) Comparison of strength :

It is weaker than nuclear, electric and magnetic forces but stronger than intermolecular force.

(6) It exists between any pair of bodies throughout the universe.

Consequence of gravitational force being a central force: Angular momentum of any heavenly body is constant as torque becomes zero relative to sun.

3. $\textbf{Weight}$

The weight of a body is the total gravitational force exerted on the body by all other bodies in the universe. When the body is near the surface of the earth, we can neglect all other gravitational forces and consider the weight as just the earth’s gravitational attraction. At the surface of the moon we consider a body’s weight to be the gravitational attraction of the moon, and so on.

It is a vector quantity. It is always directed towards the centre of the planet.

It holds the atmosphere around the earth.

Mathematically,

W = mg

Where m = mass of the body

g = acceleration due to gravity on the surface of that heavenly body

for earth g = 9.8$m/s^2$

4. $\textbf{Acceleration due to gravity (g)}$

It is the acceleration produced by Earth or planets towards centre of itself.

It is a vector quantity.

According to Newton’s law of motion ( 2nd law), force of gravity is F=mg………….(1)

And we have, $F=\dfrac{GMm}{r^2}$

Where m is the mass of the body

So, comparing we get, $g=\dfrac{GM}{r^2}$

If a body is at the surface of earth, then

$g=\dfrac{GM}{R^2}$ Where r=R is the radius of this earth

M is the mass of planet or earth

In terms of density:

Also $g=\dfrac{4}{3} \pi G \rho R$

Where $\rho$ is the density of earth/planet The value of g does not depend upon mass size and shape of the body.

## Variation of g:

1. Equatorial radius of the earth is about 21 km more than the polar radius so, according to $g \propto \dfrac{1}{R^2}$ , value of g at poles is greater by $0.018m/s^2$ $(1.8 cm/s^2)$ than that at the equator.

2. $\textbf{Due to rotational motion of the earth:}$

If the earth rotates about an axis with angular velocity(ω), then effective g on a body at latitude θ is given by:

$g_{eff} =g'-Rω^2 cos^2⁡θ$

• At equator; i.e. $θ=0^{\circ}$

$g_{eff}=g-Rω^2$

$Δg=g-g_{eff} =Rω^2=0.034m/s^2$

• At poles,

$θ=90^{\circ}$

i.e $g_{eff} =g-0=g$ (i.e. No rotational effect at poles)

3. $\textbf{Due to height above the earth’s surface:}$

Acceleration due to gravity at a height from the earth’s surface is given as:

Acceleration due to gravity at a height from earth's surface is given by:

$g' = \dfrac{GM}{(R+h)^2}$

where, h = height of the body above the earth's surface

If $h <<< R$ then, $g' = g (1 - \dfrac{2h}{R})$

4. $\textbf{Due to depth (x) below earth surface:}$

Acceleration due to gravity (g) below depth (x) from earth surface is given by:

$g = \dfrac{GM}{R^3} (R-x) = g (1- \dfrac{x}{R}) = \dfrac{g}{R} (R-x)$

The value of g decreases with increase in depth.

At center of earth, $x=R, g=0$

## Gravitational Field Strength/ Gravitational Intensity (E)

1. Gravitational intensity at a point in a gravitational field is force experienced by a unit point mass when placed at that point.

It shows how much a body is attracted by the force. Gravitational field due to the earth at a place on the surface of the earth is constant for any object be it an ant or an elephant. It depends on the mass of the heavenly body (Earth) but not on the mass of body experiencing the force.

$E=F/m= \frac{GM}{r^2}$

-Its SI unit is $Nkg^{-1}$

-It is a vector quantity.

-Its value is zero at infinity.

## Gravitational Potential Energy:

1. Gravitational potential energy is the energy possessed or acquired by an object due to a change in its position when it is present in a gravitational field. In simple terms, it can be said that gravitational potential energy is an energy that is related to gravitational force or to gravity.

The most common example that can help you understand the concept of gravitational potential energy is if you take two pencils. One is placed at the table and the other is held above the table. Now, we can state that the pencil which is high will have greater gravitational potential energy that the pencil that is at the table.

When a body of mass (m) is moved from infinity to a point inside the gravitational influence of a source mass without accelerating it, the amount of work done in displacing it into the source field is stored in the form of potential energy this is known as gravitational potential energy. It is represented with the symbol (U).

We know that the potential energy of a body at a given position is defined as the energy stored in the body at that position. If the position of the body changes due to the application of external forces the change in potential energy is equal to the amount of work done on the body by the forces. The gravitational force is conservative: The work done by gravitational force does not depend on the path taken from the earth’s surface to a certain height be it curved or straight path. The gravitational force on a body at infinity is zero; therefore, potential energy is zero, which is called a reference point.

$U=-\dfrac{GMm}{r}$

Where r = distance of the body from the centre of the earth

At the surface of the earth:

$U=- \dfrac{GMm}{R}$ ; R = Radius of the earth

The change in gravitational potential energy ΔU when that body is raised through height h from earth surface is given as

$ΔU=U-U_0=mgh(\dfrac{R}{R+h})$

For $h<<

$∴ΔU=mgh$

Note: The famous formula of Potential energy i.e. P.E. = mgh that we learnt during school days is change in P.E. of a body where height changed is very small in comparison to the Radius of the earth.

## Gravitational Potential:

1. It is the work done to take a unit mass from infinity to a certain point of consideration within the gravitational field.
2. $V=- \dfrac{GM}{r}$

It is a scalar quantity.

3. Relations between gravitational force (f), intensity (E), potential (V) and gravitational potential energy (U)

(i) Gravitational force F

$F=-dU/dr=mE= \dfrac{GMm}{r^2} =-m \dfrac{dV}{dr}$

(ii) Gravitational intensity E

$E=-dv/dr=F/m= \dfrac{GM}{r^2} = \dfrac{-1}{m} \dfrac{dU}{dr}$