**Gravitation:**

The force of attraction between any two objects in the upriver is known as gravitation.

The gravitational force;

i. is always attractive in nature.

ii. is mutual.

iii. is very weak.

iv. obeys inverse square law.

v. obeys Newton’s third law of motion.

vi. is independent of medium and chemical composition of the objects.

**Newton’s law of Gravitation:**

Newton’s law of Gravitation states that the gravitational force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Let us consider two objects of masses ‘m_{1}‘ and ‘m_{2}‘ separated by a distance ‘r’. According to Newton’s law of Gravitation

F $\propto $ m_{1}m_{2}………. (i)

F $\propto $ $\frac{1}{{{r}^{2}}}$……………… (ii)

Combining (i) and (ii)

F $\propto $ $\frac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$

$\therefore $ F = $\frac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$, where G is proportionality constant called universal Gravitational constant.

The value of G is 6.67×10^{–11} Nm^{2}kg^{–2}

For m_{1} = m_{2} = 1kg and r = 1m, then we have

F = G

Thus, universal gravitational constant (G) can be defined as the gravitational force between two objects of unit mass whose centers are unit distance apart.

**Gravity and acceleration due to Gravity:**

The force by which the earth attracts any object towards its centre is known as gravity.

The acceleration produced on an object due to the effect of gravity is called acceleration due to gravity (g).

Let us consider the earth to be a complete sphere of radius ‘R’ and mass ‘m’. Let an object of mass ‘m’ be at point ‘P’ above the earth surface at height ‘h’.

According to Newton’s law of Gravitation,

F = G $\frac{Mm}{{{(R+h)}^{2}}}$…..(i)

If the object is falling on the earth surface with acceleration (g), then at p, according to Newton’s second law of Motion,

F = mg………… (ii)

Since (i) and (ii) represent the same force, so we can write

mg = $\frac{GMm}{{{(R+h)}^{2}}}$

$\therefore $ g = $\frac{GM}{{{(R+h)}^{2}}}$

If the object is very close to the earth surface (or object is on the earth surface), then ‘h’ can be neglected.

$\therefore $ g = $\frac{GM}{{{R}^{2}}}$, which is the expression for acceleration due to gravity on the earth surface.

**Variation of acceleration due to gravity ‘g’:**

**1.**** ****Variation of acceleration due to gravity ‘g’ on the surface of earth due to the shape of earth: **

The acceleration due to gravity is given by

g = $\frac{GM}{{{R}^{2}}}$

At pole, g_{P} = $\frac{GM}{{{R}_{p}}^{2}}$…………. (i)

At equator, g_{e} = $\frac{GM}{{{R}_{E}}^{2}}$…………. (ii)

Since R_{P }< R_{E},

$\therefore $ g_{P }> g_{e}

$\therefore $The value of ‘g’ the role is more than that at the equator.

**2.**** ****Variation of acceleration due to gravity ‘g’ with height/altitude (h).**

Let us consider the earth to be a complete sphere of radius ‘R’ and mass ‘M’. Let ‘P’ be any point drove the earth surface at a height of ‘h’ where the deceleration due to gravity is ‘g’. The accelerations due to gravity on the earth surface is given by

g = $\frac{GM}{{{R}^{2}}}$……….. (i)

At point P,

Acceleration due to gravity is given by g’ = $\frac{GM}{{{(R+h)}^{2}}}$……….. (ii)

Dividing (ii) by (i), we get

Or, $\frac{g’}{g}$= ${{\left( \frac{R}{R+h} \right)}^{2}}$

Or, g’ = g${{\left( \frac{R}{R+h} \right)}^{2}}$

Or, g’ = g ${{\left( \frac{R+h}{R} \right)}^{-2}}$

Or, g’ = g ${{\left( 1+\frac{h}{R} \right)}^{-2}}$

Using Binomial expansion and neglecting the terms containing high powers of $\frac{h}{R}$, we get

g’ = g $\left( 1-\frac{2h}{R} \right)$[$\because $Binomial expansion is (1+x)^{–n} = 1 – nx+………..]

$\therefore $ g’ = g $\left( 1-\frac{2h}{R} \right)$ , which is the required expression for acceleration due to gravity at height ‘h’.

At surface, h = 0

$\therefore $ g’ = g.

**3.**** ****Variation of acceleration due to gravity ‘g’ with depth ‘d’. **

Let us consider the earth to be a complete sphere of radius ‘R’ and mass ‘M’. The acceleration due to gravity on the

surface of earth is given by g = $\frac{GM}{{{R}^{2}}}$.

Or, g = $\frac{GV\rho }{{{R}^{2}}}$

Or, g = $\frac{G\rho }{{{R}^{2}}}$$\frac{4\pi {{R}^{3}}}{3}$

Or, g = G$\frac{4\pi R\rho }{3}$…….(i)

Let us a consider which is at depth ‘d’ from the surface of earth where the declaration due to gravity is

g’ = $\frac{GM’}{{{(R-d)}^{2}}}$

Or, g’ = $\frac{GV’\rho }{{{(R-d)}^{2}}}$

Or, g’ =$\frac{G\rho }{{{(R-d)}^{2}}}$$\frac{4\pi }{3}{{(R-d)}^{3}}$

Or, g’ = G $\frac{4\pi (R-d)\rho }{3}$

Dividing (ii) by (i), we get

$\frac{g’}{g}$= $\frac{R-d}{R}$

$\therefore $ g’ = g$\left( 1-\frac{d}{R} \right)$, which is the required repression for acceleration due to gravity at depth (d).

At surface, d = 0

$\therefore $ g’ = g

The acceleration due to gravity in maximum at the surface.

At centre of the earth, d = R

$\therefore $ g’ = 0.

The acceleration due to gravity is minimum (zero) at the centre of the earth.

Thus, acceleration due to gravity decreases with increase in depth.

**4.**** ****Variation of acceleration due to gravity ‘g’ with rotation of earth.**

** Or**

**Variation of acceleration due to gravity ‘g’ with latitude angle.**

Let us consider the earth to be a complete sphere of radius ‘R’ and mass ‘M’. Let an object of mass ‘m’ be placed on the surface of earth where the latitude angle is θ. If the earth is not rotating, the weight of the object (mg) acts along PO. When the earth is rotating with an angular velocity ‘$\omega $’. Then, the object at point p also rotates with the same angular velocity ($\omega $) around the centre ‘C’ and radius ‘r’.

The object ‘m’ experiences centrifugal force ‘m$\omega $^{2}r’ due to the rotation of the earth. The two forces ‘mg’ and ‘m$\omega $^{2}r’

are acting on the same object and their resistant can be obtained by parallelogram law of vector addition,

mg’ = $\sqrt{{{(mg)}^{2}}+{{(m{{\omega }^{2}}r)}^{2}}+2mg.m{{\omega }^{2}}r\text{ }cos(180{}^\circ -\theta )}$

Or, mg’ = $\sqrt{{{m}^{2}}{{g}^{2}}+{{m}^{2}}{{\omega }^{4}}{{r}^{2}}-2{{m}^{2}}g{{\omega }^{2}}rcos\theta )}$

Or, mg’= mg$\sqrt{1+\frac{{{\omega }^{4}}{{r}^{2}}}{{{g}^{2}}}-\frac{2{{\omega }^{2}}rcos\theta }{g}}$

Or, mg’ = mg $\sqrt{1-\frac{2{{\omega }^{2}}Rco{{s}^{2}}\theta }{g}+\frac{{{\omega }^{4}}{{r}^{2}}}{{{g}^{2}}}}$[$\because $ r = R cos$\theta $]

Since the value of $\frac{{{\omega }^{4}}{{r}^{2}}}{{{g}^{2}}}$ is very small. So it can be neglected.

g’ = g${{\left( 1-\frac{2{{\omega }^{2}}R{{\cos }^{2}}\theta }{g} \right)}^{\frac{1}{2}}}$$$

Using Binomial expansion and neglecting higher power terms, we get

g’ = g$\left( 1-\frac{1}{2}.\frac{2{{\omega }^{2}}R{{\cos }^{2}}\theta }{g} \right)$

$\therefore $ g’ = g – $\omega $^{2}R cos^{2}$\theta$

This is the required repressions for acceleration due to gravity ‘g’ due to rotation of the earth (or latitude angle).

**Special Cases**

1. If the earth stops rotating, $\omega $ = 0

$\therefore $ g’ = g

Acceleration due to gravity increases

2. At pole, $\theta $ = 90°

$\therefore $ g’ = g

There is no effect of rotation of earth on the pole.

3. At equation, $\theta $ = 0°.

$\therefore $ g’ = g – $\omega $^{2}R

Acceleration due to gravity decreases the most at the equator.

**Gravitational Field**

The space round the earth up to which the gravitational pull can be experienced is called the gravitational field.

**Gravitation Field Intensity (I).**

Gravitational field intensity at a point within the gravitational field is defined as the force experienced an object of

unit mass. It in devoted by ‘I’ and given by

I = $\frac{Gravitational\text{ }force}{mass}$

It’s unit is N/kg.

Let us consider the earth to be a complete sphere of radius ‘R’ and mass ‘M’. Let an object of mass ‘m’ be at point P

which is at a height ‘h’ from the earth surface, where the gravitational field intensity is to be determined. The

gravitational force at point P is given by

F = $\frac{GMm}{{{r}^{2}}}$

Where, r = R+h

According to the definition of gravitational field intensity,

I = $\frac{F}{m}$

Or, I = $\frac{GMm}{{{r}^{2}}m}$

$\therefore $ I =$\frac{GM}{{{r}^{2}}}$, which is the required expression for gravitational field intensity.

If the object is close to the earth surface,

Then, i.e. h $\to $ 0.

Then, I = $\frac{GM}{{{R}^{2}}}$

**Gravitational Potential (V or v**_{g}**)**

_{g}

Gravitational potential at a point within a gravitational field can be defined as the amount of work done in bringing a body of unit mass from infinity to that point. It is denoted by ‘V’ or ‘V_{g}‘

Let us consider the earth to be a complete sphere of radius ‘R’ mass ‘M’. Let P be a point which is at a distance of ‘r’

from the centre of earth where the gravitational potential is to be determined. The small amount of work done in

bringing the object from A to B through a small displacement ‘dx’ is given by

dw = F_{A} dx

Or, dw = $\frac{GMm}{{{x}^{2}}}$. dx

For unit mass, m = 1 kg

$\therefore $ dw = $\frac{GM}{{{x}^{2}}}$. dx……… (i)

The total amount of work done in bringing the object (m =1kg) from infinity to point ‘P’ can be obtained by integrating

equation (i) from x = $\infty $ to x = r i.e.

Or, $\int\limits_{\infty }^{r}{dw}$ = $\int\limits_{\infty }^{r}{\frac{GM}{{{x}^{2}}}}$ dx

Or, W = GM $\int\limits_{\infty }^{r}{{}}$ x^{-2}. dx

Or, W = GM$\left[ \frac{{{x}^{-2+1}}}{-1} \right]_{\infty }^{r}$

Or, W = GM$\left[ \frac{{{x}^{-1}}}{-1} \right]_{\infty }^{r}$

Or, W = GM$\left[ \frac{-1}{x} \right]_{\infty }^{r}$

Or, W = GM $\left[ \frac{-1}{r}-\frac{-1}{\infty } \right]$

$\therefore $ W = $\frac{-GM}{r}$……….(ii)

According to the definition of V_{g}, above work done (in equation (ii)). Given the gravitational potential at any point p,

i.e.

V_{g} = W = $\frac{-GM}{r}$.

$\therefore $ V_{g} = $\frac{-GM}{R+h}$, which the required expression

**Gravitational Potential Energy (U)**

Gravitational potential energy at a point within a gravitational field can be defined as the amount of work done in bringing a body of given mass from infinity to that point. It is denoted by ‘U’.

Let us consider the earth to be a perfect sphere of radius ‘R’ and mass ‘M’. Let ‘P’ be a point which is at a distance of

‘r’ distance from the centre of the earth, where the gravitational potential energy is to be determined. The small amount

of work done in bringing the object of given mass from A to B through a small displacement ‘dx’ is given by

dw = F_{A}.dx

Or, dw = $\frac{GMm}{{{x}^{2}}}$dx…………………… (i)

The total amount of work done is bringing the object form infinity to point ‘P’ can be obtained by integrating

equation (i) from x = $\infty $ to x = r i.e.

Or, $\int\limits_{\infty }^{r}{dw}$ = $\int\limits_{\infty }^{r}{\frac{GMm}{{{x}^{2}}}}$ dx

Or, W = GMm $\int\limits_{\infty }^{r}{{}}$ x^{-2}. dx

Or, W = GMm$\left[ \frac{{{x}^{-2+1}}}{-1} \right]_{\infty }^{r}$

Or, W = GMm$\left[ \frac{{{x}^{-1}}}{-1} \right]_{\infty }^{r}$

Or, W = GMm$\left[ \frac{-1}{x} \right]_{\infty }^{r}$

Or, W = GMm $\left[ \frac{-1}{r}-\frac{-1}{\infty } \right]$

$\therefore $ W = $\frac{-GMm}{r}$…….(ii)

According to the definition of gravitational potential energy, above work done, gives the gravitational potential energy

at point P, i.e.

U = W = $\frac{-GMm}{r}$

$\therefore $ U = $\frac{-GMm}{r}$Which is the required expression for gravitational potential energy at a point within a gravitational

field.

**Escape Velocity (V**_{e}**) and its Expression:**

_{e}

The minimum velocity with which an object must be thrown vertically upward from the earth surface so that it just escapes the gravitational field of earth and never returns to the earth surface.

Let us consider the earth to be a complete sphere of radius ‘R’ and mass ‘M’. Let A be any point at a distance of x,

where the gravitational force of any object of mass ‘m’ is

F_{A} = $\frac{GMm}{{{x}^{2}}}$

The small amount of work done is moving the object from A to B through a small displacement ‘dx’ is given by

dw = F_{A}×dx

Or, dw =$\frac{GMm}{{{x}^{2}}}$dx………… (i)

Total amount of work done in throwing the object from earth surface to infinity can be obtained by integrating equation

(i) from x = R to x =$\infty $.

i.e. W = $\int\limits_{R}^{\infty }{dw}$

Or, W =$\int\limits_{R}^{\infty }{\frac{GMm}{{{x}^{2}}}}$dx

Or, W = GMm$\int\limits_{R}^{\infty }{{{x}^{-2}}}$dx

Or, GMm$\left[ \frac{{{x}^{-2+1}}}{-1} \right]_{R}^{\infty }$

Or, W = GMm$\left[ \frac{{{x}^{-1}}}{-1} \right]_{R}^{\infty }$

Or, W = GMm$\left[ \frac{-1}{x} \right]_{R}^{\infty }$

Or, W = GMm$\left[ \frac{-1}{\infty }-\frac{-1}{R} \right]$

Or, W =$\frac{GMm}{R}$……….. (ii)

To achieve the above work done, we have to throw the object with the same K.E.

i.e. K. E. = W

or, $\frac{1}{2}$mV_{e}^{2}= $\frac{GMm}{R}$, where V_{e} is escape velocity.

$\therefore $ V_{e} = $\sqrt{\frac{2GM}{R}}$………….. (iii)

Since, g = $\frac{GM}{{{R}^{2}}}$ ⇒ gR = $\frac{GM}{R}$

$\therefore $ V_{e} = $\sqrt{2gR}$ …………. (iv), Which is required for escape velocity. The escape velocity is independent of the mass of the object thrown.

For earth, g = 9.8m/s^{2}, R = 6.24×10^{6}m.

$\therefore $ V_{e} = $\sqrt{2\times 9.8\times 6.4\times {{10}^{6}}}$

Or, V_{e} = 11200 m/s

$\therefore $ V_{e} = 11.2 km/s, which is the escape velocity of the earth.

Q. Show that the escape velocity is independent of the mass thrown.

**Satellite**

An object which revolves in an orbit around a comparatively heavier object at a certain height is called satellite. There

are two types of satellite natural satellite and artificial satellite.

**Orbital Velocity and its Expression**

The minimum velocity required for an object to revolve into a particular orbit at certain height is called orbital

velocity. It is the velocity with which a satellite revolves in a particular orbit.

Let us consider the earth to be a complete sphere of radius ‘R’ and mass ‘M’. Let a satellite of mass ‘m’ revolving

in an orbit of radius ‘r’ around the earth / planet with velocity ‘v’ as shown in figure.

At any instance, the gravitational force of attraction is given by

F_{g} =$\frac{GMm}{{{r}^{2}}}$

This force provides necessary centripetal force $\frac{m{{V}_{0}}^{2}}{r}$,

i.e. F_{g} = F_{c}

$\frac{GMm}{{{r}^{2}}}=\frac{m{{V}_{0}}^{2}}{r}$…………….. (i)

V_{o}^{2} = $\frac{GM}{r}$

Since, g =$\frac{GM}{{{R}^{2}}}$

Or, GM = gR^{2}

Then eq^{n} (i) becomes

V_{0} = $\sqrt{\frac{g{{R}^{2}}}{r}}$

$\therefore $ V_{0} = R $\sqrt{\frac{g}{R+h}}$ [since r = R+h]

Which is the required expression for orbital velocity.

**Time period of satellite:**

Time period (T) = $\frac{2\pi }{\omega }$

Or T = $\frac{2\pi }{{{v}_{0}}/r}$ [since v_{o} = r$\omega $]

Or, T = $\frac{2\pi r}{R}$$\sqrt{\frac{R+h}{g}}$

Or, T = $\frac{2\pi }{R}$$\sqrt{\frac{{{r}^{2}}(R+h)}{g}}$ [since r = R+h]

$\therefore $ T =$\frac{2\pi }{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$, Which is the required expression for time period.

**Height (h) of satellite:**

We have T = $\frac{2\pi }{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$

Or, T^{2} = $\frac{4{{\pi }^{2}}}{{{R}^{2}}}$.$\frac{{{(R+h)}^{3}}}{g}$

Or, (R+h)^{3} = $\frac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}}$

Or, R+h = ${{\left( \frac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right)}^{1/3}}$

$\therefore $ h =${{\left( \frac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right)}^{1/3}}$–R

This is the required expression for height.

**Geostationary Satellite**

A satellite which seems to be stationary over the earth surface at a particular point is called geostationary satellite. It is due to two equal time periods of revolution of satellite and rotation of earth i.e. 24 hours.

**Total Energy associated with a satellite.**

Let us consider the earth to be a complete sphere of radius ‘R’ and mass ‘M’. Let a satellite of mass ‘m’ revolve in an orbit of radius ‘r’ around the earth with orbital velocity ‘V_{o}‘. The total energy of a satellite is the sum of kinetic energy (due to its motion velocity) and gravitational potential energy (due to height). i.e.

E_{total} = K.E. + P.E. …………. (i)

Kinetic energy (K.E.)

K.E. = $\frac{1}{2}m{{V}_{0}}^{2}$………………. (ii)

Or, K.E. = $\frac{1}{2}$m ${{\left( R\sqrt{\frac{g}{R+h}} \right)}^{2}}$

Or, K.E.= $\frac{1}{2}$m$\left( \frac{g{{R}^{2}}}{R+h} \right)$

Or, K.E. = $\frac{1}{2}$m $\frac{GM}{R+h}$ [Since g = $\frac{GM}{{{r}^{2}}}$]

Or, K.E. = $\frac{GMm}{2(R+h)}$

Or, K.E. = $\frac{GMm}{2r}$……… (iii)

Now, the gravitational potential energy of an object at radius ‘r’ is given as

P.E. =$\frac{-GMm}{r}$………. (iv)

Then equation (i) becomes

E_{total} = $\frac{GMm}{2r}$– $\frac{GMm}{r}$

Or, E_{total} = $\frac{-GMm}{2r}$= $\frac{-GMm}{2(R+h)}$

This is the required expression for total energy associated with a satellite. The negative sign shows that the satellite is bound by the force of attraction of the earth. We have to supply energy from outside to make it free.

Also Read: Circular Motion Notes

KMS CAPTAINTHANKYOU SIR.