**CBSE NOTES CLASS 11 PHYSICS CHAPTER 8**

**CGRAVITATION**

**Introduction to Motion of Celestial Bodies**

**Ptolemy** proposed a **geocentric** model of motion of celestial bodies, about 2000 years ago that all celestial objects, stars, the sun and the planets, revolved around the earth.

Similar theories were also proposed by Indian philosophers.

**Aryabhatta (5th century A.D.)** had proposed a ‘**heliocentric**’ model in which the Sun was the center around which the planets revolved

A Polish monk named **Nicolas Copernicus** (1473-1543) proposed a model in which the planets moved in circles around a fixed central sun, however the church discredited his model.

**Tycho Brahe** (1546-1601) from Denmark spent his entire lifetime recording observations of the planets with the naked eye.

His compiled data were analysed later by his assistant **Johannes Kepler** (1571-1640) and he propsed three laws that are known as **Kepler’s laws**.

**KEPLER’S LAWS**

**Kepler’s Law of Orbits:**All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse.**Ellipse**Locus of a point, which moves in such a way that the sum of its distances from two fixed points always remains constant. The two fixed points are called foci (plural for focus).

**Kepler’s Law of areas:**The line that joins any planet to the sun sweeps equal areas in equal intervals of time.

- Planets move slower when they are farther from the sun than when they are nearer.
- The law of areas can be derived from the law of conservation of angular momentum which is valid for any
**central force**.A

**central force**is such that the force on the planet is along the vector joining the sun and the planet.Let the sun be at the origin and let the position and momentum of the planet be denoted by r and p respectively. Then the area swept out by the planet of mass m in time interval Δt is ΔA given by

$$\mathrm{\Delta}\overrightarrow{\mathrm{A}}\mathrm{}=\frac{1}{2}\mathrm{}\left(\overrightarrow{\mathrm{r}}\mathrm{}\times \mathrm{}\overrightarrow{\mathrm{v}}\mathrm{\Delta}\mathrm{t}\right)$$

Since v = p/m

**,**we have$$\frac{\mathrm{\Delta}\overrightarrow{\mathrm{A}}}{\mathrm{\Delta}\mathrm{t}}=\frac{1}{2}\mathrm{}\frac{\overrightarrow{\mathrm{r}}\mathrm{}\times \mathrm{}\overrightarrow{\mathrm{p}}}{\mathrm{m}}=\frac{\overrightarrow{\mathrm{L}}}{2\mathrm{m}}$$

Now since L is constant for a central force,

$$\frac{\mathrm{\Delta}\overrightarrow{\mathrm{A}}}{\mathrm{\Delta}\mathrm{t}}=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$$

**Kepler’s 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.*T*^{2}*∝ a*^{3}

**Gravitation**

Gravitation is the force of attraction between any two objects in the universe.

Italian physicist **Galileo** (1564-1642) recognised the fact that all bodies, irrespective of their masses, are accelerated towards the earth with a constant acceleration.

Galileo carried out experiments with bodies rolling down inclined planes and arrived at a value of the acceleration due to gravity.

**Centripetal Force**

The force directed towards the centre of the circle is called centripetal force.

**Universal Law of Gravitation**

Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The force is along the line joining the centres of two objects.

Mathematically,

$$\mathrm{F}\propto {\mathrm{m}}_{1}\times {\mathrm{m}}_{2}\mathrm{}$$

And F ∝ $\frac{1}{{\mathrm{r}}^{2}}$

Therefore, we can write,

$$\mathrm{F}=\mathrm{}\mathrm{G}\frac{\mathrm{}{\mathrm{m}}_{1}\times {\mathrm{m}}_{2}}{{\mathrm{r}}^{2}}$$

Where *G* is the universal gravitational constant, *m*_{1} and *m*_{2} are the masses of two bodies and r is the distance between them.

The SI unit of G is N m^{2 }kg^{-2}.

The value of G is 6.673 × 10^{-11} N m^{2 }kg^{-2}.

- The law is
**universal**in the sense that it is applicable to all bodies, whether the bodies are big or small, whether they are celestial or terrestrial. - In vector form, force on m
_{1}due to m_{2}$${\overrightarrow{\mathrm{F}}}_{12}=\mathrm{G}\frac{\mathrm{}{\mathrm{m}}_{1}\times {\mathrm{m}}_{2}}{{\mathrm{r}}^{2}}{\widehat{\mathrm{r}}}_{12}$$

$\widehat{\mathrm{r}}}_{12$

*m*_{1}to*m*_{2}

- The gravitational force is attractive, i.e., the force
**F**is along –**r**. - The gravitational force F
_{12 }on the body 1 due to 2 and F_{21}on the body 2 due to 1 are related as F_{12}= – F_{21}. It obeys Newton’s third law. - Gravitational force obeys
**law of****superposition**of vector addition. - The law is applicable to
**point**masses, but can be applied to extended bodies using superposition principle and by assuming the resultant force acting on the centre of mass of the body.

**The gravitational force due to a hollow spherical shell**

- The force of attraction between a
**hollow spherical shell**of uniform density and a**point mass situated outside**is just as if the entire mass of the shell is concentrated at the centre of the shell. - The force of attraction due to a
**hollow spherical shell**of uniform density, on a**point mass situated inside**it is zero.

**Finding the Gravitational Constant**

The value of gravitational constant G was first found by English scientist Henry Cavendish in 1798.

**Apparatus**

The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire. Two large lead spheres are brought close to the small ones on opposite sides as shown. The big spheres attract the nearby small ones by equal and opposite force as shown. There is no net force on the bar but only a torque which is equal to F times the length of the bar, where **F** is the force of attraction between a big sphere and its neighbouring small sphere.

The bar AB rotates a little due to the torque. The angle of rotation can be measured experimentally.

Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque. If **θ** is the angle of twist of the suspended wire, the restoring torque is proportional to θ, equal to ** τθ**, where

**is the restoring couple per unit angle of twist.**

*τ*If d is the separation between the centers of the big and its neighbouring small ball, M and m their masses, the gravitational force between the big sphere and its neighouring small ball is,

$$\mathrm{F}=\mathrm{}\mathrm{G}\frac{\mathrm{}\mathrm{M}\times \mathrm{m}}{{\mathrm{d}}^{2}}$$

And torque

$$\mathrm{G}\frac{\mathrm{}\mathrm{M}\times \mathrm{m}}{{\mathrm{d}}^{2}}\mathrm{}\mathrm{L}=\mathrm{\tau}\mathrm{\theta}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{G}=\frac{\mathrm{\tau}\mathrm{}\mathrm{\theta}{\mathrm{}\mathrm{d}}^{2}}{\mathrm{M}\mathrm{}\mathrm{m}\mathrm{}\mathrm{L}}$$

All variables except G are known, hence, we can calculate the value of G.

Value of G = 6.67×10^{-11} N m^{2}/kg^{2}

**Why do we say that Cavendish weighed the earth?**

Acceleration *g *is readily measurable. R_{E} is a known quantity. The measurement of G by Cavendish’s experiment (or otherwise), enables us to estimate M_{E}. This is the reason why there is a popular statement “Cavendish weighed the earth”.

**Acceleration Due to Gravity of the Earth**

The acceleration experienced by the mass **m**, which is denoted by ‘g’ on the surface of earth is given by

$$\mathrm{g}\mathrm{}=\frac{\mathrm{F}}{\mathrm{m}}=\mathrm{}\mathrm{G}\frac{\mathrm{}{\mathrm{M}}_{\mathrm{E}}}{{\mathrm{R}}_{\mathrm{E}}^{2}}$$

Where $\mathrm{R}}_{\mathrm{E}$ = radius of earth and $\mathrm{M}}_{\mathrm{E}$ = mass of earth.

**Acceleration due to gravity above the surface of earth**

Consider a point mass m at a height h above the surface of the earth. If the radius of the earth is R_{E}, its distance from the centre of the earth is (*R*_{E }*+ h*).

If *F(h***)** denotes the the magnitude of the force on the point mass *m* at a height *h* from earth’s surface, we get,

$$\mathrm{F}\mathrm{\left(\mathrm{h}\right)=\mathrm{}\mathrm{G}\frac{\mathrm{}{\mathrm{M}}_{\mathrm{E}}\times \mathrm{m}}{({{\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}^{2}}\mathrm{}}$$

$$\Rightarrow \mathrm{}\mathrm{}\mathrm{g}\left(\mathrm{h}\right)=\mathrm{G}\frac{\mathrm{}{\mathrm{M}}_{\mathrm{E}}}{({{\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}^{2}}=\mathrm{}\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}{{{\mathrm{R}}_{\mathrm{E}}}^{2}}\frac{1\mathrm{}}{{\left(1+\frac{\mathrm{h}}{{\mathrm{R}}_{\mathrm{E}}}\right)}^{2}}$$

$$\mathrm{g}\left(\mathrm{h}\right)=\mathrm{g}\mathrm{}\frac{1\mathrm{}}{{\left(1+\frac{\mathrm{h}}{{\mathrm{R}}_{\mathrm{E}}}\right)}^{2}}$$

$$\mathrm{g}\left(\mathrm{h}\right)=\mathrm{g}\mathrm{}{\left(1+\frac{\mathrm{h}}{{\mathrm{R}}_{\mathrm{E}}}\right)}^{-2}$$

For** **$\frac{\mathrm{h}}{{\mathrm{R}}_{\mathrm{E}}}$ << 1, using binomial expansion, we have,

$$\mathrm{g}\left(\mathrm{h}\right)=\mathrm{}\mathrm{g}\mathrm{}\left(1-\frac{2\mathrm{h}}{{\mathrm{R}}_{\mathrm{E}}}\right)$$

**Acceleration due to gravity below the surface of earth**

Consider a point mass *m *at a depth *d *below the surface of the earth.

The distance from the center of the earth is (*R*_{E}* – d*).

The earth can be thought of as being composed of a smaller sphere of radius (*R*_{E}* – d*) and a spherical shell of thickness *d*.

The force on *m* due to the outer shell of thickness *d* is zero

For smaller sphere of radius (*R*_{E}* – d*), the point mass is outside it and hence, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre.

If *M*_{S} is the mass of the smaller sphere, since mass of a sphere is proportional to be cube of its radius,

$$\frac{{\mathrm{M}}_{\mathrm{S}}}{{\mathrm{M}}_{\mathrm{E}}}=\frac{{\left({\mathrm{R}}_{\mathrm{E}}\mathrm{}\u2013\mathrm{}\mathrm{d}\right)}^{3}}{{{\mathrm{R}}_{\mathrm{E}}}^{3}}$$

$$\Rightarrow \mathrm{}{\mathrm{M}}_{\mathrm{S}}\mathrm{}=\mathrm{}\frac{{{\mathrm{M}}_{\mathrm{E}}\left({\mathrm{R}}_{\mathrm{E}}\mathrm{}\u2013\mathrm{}\mathrm{d}\right)}^{3}}{{{\mathrm{R}}_{\mathrm{E}}}^{3}}$$

Thus the force on the point mass is

$$\mathrm{F}\left(\mathrm{d}\right)=\mathrm{}\mathrm{G}\frac{\mathrm{}{\mathrm{M}}_{\mathrm{S}}\times \mathrm{m}}{{{(\mathrm{R}}_{\mathrm{e}}-\mathrm{d})}^{2}}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{}\mathrm{g}\left(\mathrm{d}\right)=\mathrm{G}\frac{\mathrm{}{\mathrm{M}}_{\mathrm{S}}}{{{(\mathrm{R}}_{\mathrm{e}}-\mathrm{d})}^{2}}$$

$$\Rightarrow \mathrm{g}\left(\mathrm{d}\right)=\mathrm{G}\frac{{\mathrm{M}}_{\mathrm{E}}\left({\mathrm{R}}_{\mathrm{E}}\mathrm{}\u2013\mathrm{}\mathrm{d}\right)}{{{\mathrm{R}}_{\mathrm{E}}}^{3}}\mathrm{}$$

$$\Rightarrow \mathrm{g}\left(\mathrm{d}\right)=\mathrm{g}\mathrm{}\frac{\left({\mathrm{R}}_{\mathrm{E}}\mathrm{}\u2013\mathrm{}\mathrm{d}\right)\mathrm{}}{{\mathrm{R}}_{\mathrm{E}}}$$

$$\Rightarrow \mathrm{g}\left(\mathrm{d}\right)=\mathrm{g}\left(1-\frac{\mathrm{d}}{{\mathrm{R}}_{\mathrm{e}}}\right)$$

**Gravitational Potential Energy**

If the position of the particle changes on account of forces acting on it, the change in its potential energy is equal to the amount of **work done** on the body by the force.

Forces for which the work done is independent of the path are the **conservative** forces. Gravitational force is **conservative**.

The force of gravity is practically a constant equal to **mg**, directed towards the center of the earth.

If we consider a point at a height *h*_{1} from the surface of the earth and another point vertically above it at a height *h*_{2} from the surface, the work done in lifting the particle of mass m from the first to the second position

*W*_{12}* = Force × displacement = mg (h*_{2}* – h*_{1}*)*

If we associate a potential energy **W(h**) at a point at a height **h*** *above the surface such that

*W(h) = m g h + W*_{o}* (where W*_{o}* = constant) ;*

Then,

*W*_{12}* = W(h*_{2}*) – W(h*_{1}*)*

Setting *h *= 0 in the last equation, we get

*W (h = 0) = W*_{o}

*h* = 0 means points on the surface of the earth. Thus, *W*_{o }is the potential energy on the surface of the earth.

Now let us consider points at arbitrary distance from the surface of the earth. The gravitational force is no longer constant.

Gravitational force outside the earth at distance **r **from the centre of earth is,

$$\mathrm{F}\left(\mathrm{r}\right)=\mathrm{}\mathrm{G}\frac{\mathrm{}{\mathrm{M}}_{\mathrm{E}}\times \mathrm{m}}{{\mathrm{r}}^{2}}$$

where *M*_{E} = mass of earth, *m *= mass of the particle and **r*** *its distance from the center of the earth.

If we now calculate the work done in lifting a particle from *r = r*_{1}* to r = r*_{2}* (r*_{2}* > r*_{1}*) *along a vertical path, we get,

$${\mathrm{W}}_{12}\mathrm{}={\int}_{{\mathrm{r}}_{1}}^{{\mathrm{r}}_{2}}\mathrm{G}\frac{\mathrm{}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{{\mathrm{r}}^{2}}\mathrm{d}\mathrm{r}$$

$$=\mathrm{}\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}{\left[-\frac{\mathrm{}1}{\mathrm{r}}\right]}_{{\mathrm{r}}_{1}}^{{\mathrm{r}}_{2}}$$

$$=\mathrm{}-\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}\left[\frac{\mathrm{}1}{{\mathrm{r}}_{2}}-\frac{\mathrm{}1}{{\mathrm{r}}_{1}}\right]$$

We can thus write, for* r > R*_{E}

$$\mathrm{W}\left(\mathrm{r}\right)=\mathrm{}-\frac{\mathrm{}\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{\mathrm{r}}+\mathrm{}{\mathrm{W}}_{1}$$

Hence, *W*_{12}* = W(r*_{2}*) – W(r*_{1}*)*

Setting *r *= infinity in the last equation, we get

*W (∞) = W*_{1}.

Thus, *W*_{1} is the potential energy at infinity.

Only the difference of potential energy between two points has a definite meaning. Actual value at any point has no meaning.

We set *W*_{1} = 0, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point.

- The gravitational potential energy associated with two particles of masses
*m*_{1}and*m*_{2}separated by distance by a distance*r*is given$$\mathrm{V}\left(\mathrm{r}\right)=\mathrm{}-\frac{\mathrm{}\mathrm{G}\mathrm{}{\mathrm{m}}_{1}{\mathrm{m}}_{2}}{\mathrm{r}}$$

(V → 0 as r →∞)

- An isolated system of particles will have the total potential energy that equals the sum of energies for all possible pairs of its constituent particles (Superposition principle).

**Escape Velocity**

**Escape velocity** is the minimum speed needed for an object to escape from the gravitational influence of a massive body.

Suppose the object reached infinity with speed *V*_{f}. The energy of an object is the sum of potential and kinetic energy. If *W*_{1} denotes that gravitational potential energy of the object at infinity, the total energy of the projectile at infinity is

$$\mathrm{E}\left(\mathrm{\infty}\right)\mathrm{}=\mathrm{}{\mathrm{W}}_{1}\mathrm{}+\mathrm{}\frac{\mathrm{m}{{\mathrm{V}}_{\mathrm{f}}}^{2}}{2}$$

If the object was thrown initially with a speed *V*_{i}** **from a point at a distance (h + R_{E}) from the center of the earth (*R*_{E}* *= radius of the earth), its energy initially was

$$\mathrm{E}\left(\mathrm{h}\mathrm{}+\mathrm{}{\mathrm{R}}_{\mathrm{E}}\right)=\mathrm{}-\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{\mathrm{h}\mathrm{}+\mathrm{}{\mathrm{R}}_{\mathrm{E}}}+{\mathrm{W}}_{1}\mathrm{}+\mathrm{}\frac{\mathrm{m}{{\mathrm{V}}_{\mathrm{f}}}^{2}}{2}$$

By principle of conservation of energy, we have

$${\mathrm{W}}_{1}\mathrm{}+\mathrm{}\frac{\mathrm{m}{{\mathrm{V}}_{\mathrm{f}}}^{2}}{2}=\mathrm{}-\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{\mathrm{h}\mathrm{}+\mathrm{}{\mathrm{R}}_{\mathrm{E}}}+{\mathrm{W}}_{1}\mathrm{}+\mathrm{}\frac{\mathrm{m}{{\mathrm{V}}_{\mathrm{i}}}^{2}}{2}$$

$$\Rightarrow -\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{\mathrm{h}\mathrm{}+\mathrm{}{\mathrm{R}}_{\mathrm{E}}}\mathrm{}+\mathrm{}\frac{\mathrm{m}{{\mathrm{V}}_{\mathrm{i}}}^{2}}{2}=\mathrm{}\mathrm{}\frac{\mathrm{m}{{\mathrm{V}}_{\mathrm{f}}}^{2}}{2}$$

The R.H.S. is a positive ( > 0 ) hence the L.H.S must also be positive ( > 0 ), i.e.

$$-\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{\mathrm{h}\mathrm{}+\mathrm{}{\mathrm{R}}_{\mathrm{E}}}\mathrm{}+\mathrm{}\frac{\mathrm{m}{{\mathrm{V}}_{\mathrm{i}}}^{2}}{2}0\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{}\frac{\mathrm{m}{{\mathrm{V}}_{\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}}}^{2}}{2}\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{\mathrm{h}\mathrm{}+\mathrm{}{\mathrm{R}}_{\mathrm{E}}}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{}{{\mathrm{V}}_{\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}}}^{2}=\frac{2\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}{\mathrm{h}\mathrm{}+\mathrm{}{\mathrm{R}}_{\mathrm{E}}}$$

$$\Rightarrow \mathrm{}\mathrm{}{\mathrm{V}}_{\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}}=\sqrt{\frac{2\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}{\mathrm{h}\mathrm{}+\mathrm{}{\mathrm{R}}_{\mathrm{E}}}}$$

If *h *= 0, that is the object is thrown from the surface of the earth, then

$${\mathrm{V}}_{\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}}=\sqrt{\frac{2\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}{\mathrm{}{\mathrm{R}}_{\mathrm{E}}}}$$

$$=\sqrt{\frac{2\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}{\mathrm{}{\mathrm{R}}_{\mathrm{E}}}\times \frac{{\mathrm{R}}_{\mathrm{E}}}{{\mathrm{R}}_{\mathrm{E}}}}=\sqrt{2\mathrm{g}{\mathrm{R}}_{\mathrm{E}}}\mathit{}$$

Escape velocity from Earth’s surface is about **11.186 km/s**.

**Why is there no air on moon?**

The escape speed for the moon is 2.3 km/s, about five times smaller than that on earth. This is the reason that moon has no atmosphere. Gas molecules if formed on the surface of the moon having velocities larger than this will escape the gravitational pull of the moon.

**Earth’s Satellites**

Earth satellites are objects which revolve around the earth.

Their motion is very similar to the motion of planets around the Sun and hence Kepler’s laws of planetary motion are equally applicable to them.

Their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis.

Let us consider a satellite in a circular orbit of a distance (*R*_{E}* + h*) from the centre of the earth, where *R*_{E} = radius of the earth. If *m* is the mass of the satellite and **V** its speed, the centripetal force required for this orbit is,

$${\mathrm{F}}_{\mathrm{C}}=\mathrm{}\frac{\mathrm{}\mathrm{m}{\mathrm{V}}^{2}}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}$$

This centripetal force is provided by the gravitational force,

$${\mathrm{F}}_{\mathrm{g}}=\mathrm{}\mathrm{G}\frac{\mathrm{}{\mathrm{M}}_{\mathrm{E}}\times \mathrm{m}}{{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}^{2}}$$

Equating the two forces,

$$\frac{\mathrm{}\mathrm{m}{\mathrm{V}}^{2}}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}=\mathrm{}\sqrt{\frac{\mathrm{}\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}}$$

$$\Rightarrow \mathrm{V}=\mathrm{}\sqrt{\frac{\mathrm{}\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}}$$

So, time period of revolution,

$$\mathrm{T}\mathrm{}=\mathrm{}\frac{2\mathrm{\pi}\left({\mathrm{R}}_{\mathrm{E}}+\mathrm{h}\right)}{\mathrm{V}}$$

$$\Rightarrow \mathrm{T}=\mathrm{}\frac{2\mathrm{\pi}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}^{3/2}}{\sqrt{\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}}$$

Squaring both sides,

$${\mathrm{T}\mathrm{}}^{2}=\mathrm{}\mathrm{}\frac{4{\mathrm{\pi}}^{2}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}^{3}}{\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}=\mathrm{k}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}^{3}$$

$$\Rightarrow {\mathrm{T}\mathrm{}}^{2}\mathrm{}{\mathrm{r}}^{3}$$

This is the **Kepler’s law of periods**.

If *h* = 0, that is the satellite is very near the earth, we get,

$${\mathrm{V}}_{\mathrm{o}}=\mathrm{}\sqrt{\frac{\mathrm{}\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}{{\mathrm{R}}_{\mathrm{E}}}}=\sqrt{\mathrm{g}{\mathrm{R}}_{\mathrm{E}}}$$

$${\mathrm{T}}_{\mathrm{o}}\mathrm{}=\mathrm{}2\mathrm{\pi}\mathrm{}\sqrt{\frac{{{\mathrm{R}}_{\mathrm{E}}}^{3}}{\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}}=\mathrm{}2\mathrm{\pi}\mathrm{}\sqrt{\frac{{\mathrm{R}}_{\mathrm{E}}}{\mathrm{g}}}\mathbf{}$$

If we substitute the numerical values *g = 9.8 m s*^{-2} and *R*_{E}* = 6400 km*, we get,

$${\mathrm{T}}_{\mathrm{o}}\mathrm{}=\mathrm{}2\mathrm{\pi}\mathrm{}\sqrt{\frac{6.4\mathrm{}\times {10}^{6}}{9.8}}=5079.62\mathrm{}\mathrm{s}\mathrm{}\mathrm{}85\mathrm{}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}$$

**Energy of an orbiting satellite**

The kinetic energy of the satellite in a circular orbit with speed **v*** *is,

$$\mathrm{K}\mathrm{E}\mathrm{}=\frac{1}{2}\mathrm{m}{\mathrm{v}}^{2}=\frac{1}{2}\mathrm{}\frac{\mathrm{}\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}\mathrm{}\mathrm{}$$

Since gravitational potential energy at infinity is zero, the potential energy at distance (*R*_{E }*+ h*) from the center of the earth is,

$$\mathrm{P}\mathrm{E}=\mathrm{}-\frac{\mathrm{}\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}$$

Total energy therefore is,

$$\mathrm{T}\mathrm{E}\mathrm{}=\mathrm{K}\mathrm{E}+\mathrm{P}\mathrm{E}$$

$$=\frac{1}{2}\mathrm{}\frac{\mathrm{}\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{\left({\mathrm{R}}_{\mathrm{E}}+\mathrm{h}\right)}-\frac{\mathrm{}\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{\left({\mathrm{R}}_{\mathrm{E}}+\mathrm{h}\right)}$$

$$=-\frac{1}{2}\mathrm{}\frac{\mathrm{}\mathrm{G}{\mathrm{M}}_{\mathrm{E}}\mathrm{m}}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}$$

**Q: Why is the total energy of a satellite negative?**

If the total energy is positive or zero, the object escapes to infinity. Satellites are always at finite distance from the earth and hence their energies cannot be positive or zero.

**Geostationary satellites**

Satellites in a circular orbits around the earth in the equatorial plane with *T *= 24 hours are called **geostationery Satellites**. Since the earth rotates with the same period, the satellite would appear fixed from any point on earth.

Now,

$$\mathrm{T}\mathrm{}=\mathrm{}\frac{2\mathrm{\pi}{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}^{3/2}}{\sqrt{\mathrm{G}{\mathrm{M}}_{\mathrm{E}}}}$$

$${\mathrm{R}}_{\mathrm{E}}+\mathrm{h}=\mathrm{}{\left(\frac{{\mathrm{T}}^{2}\mathrm{}\mathrm{G}{\mathrm{}\mathrm{M}}_{\mathrm{E}}}{4{\mathrm{\pi}}^{2}}\mathrm{}\right)}^{1/3}$$

For *T* = 24 hours, *h* works out to be 35800 *km*, which is much larger than *R*_{E}.

- Geostationery satellites are used for communication, TV broadcasting etc.
The INSAT group of satellites is geostationary satellites used for telecommunications in India.

**Polar satellites**

These are low altitude (*h ***≈** 500 to 800 km) satellites going around the poles of the earth in a north-south direction whereas the earth rotates around its axis in an east-west direction.

Since its time period is around 100 minutes it crosses any altitude many times a day.

A camera fixed on it can view only small strips of the earth in one orbit. Adjacent strips are viewed in the next orbit, so that in effect the whole earth can be viewed strip by strip during the entire day.

These satellites can view polar and equatorial regions at close distances with good resolution.

Information gathered from such satellites is useful for remote sensing, meterology as well as for environmental studies of the earth.

**Weightlessness**

Weight of an object is the force with which the earth attracts it.

When bodies are falling with the same acceleration as that the acceleration due to gravity, they are said to be in **free fall **and it will feel** weightless. **This phenomenon is called the phenomenon of **weightlessness**.

**Examples of free fall **

- A spring balance falling freely will read weight of an object as zero.
- In a satellite around the earth, every part and parcel of the satellite has an acceleration towards the center of the earth which is exactly the value of earth’s acceleration due to gravity at that position. Thus in the satellite everything inside it is in a state of free fall. Thus, in a satellite, people inside experience no gravity.
- Gravity for us defines the vertical direction and thus for the satellite there are no horizontal or vertical directions, all directions are the same.