Class 11 | Physics | chapter 8 | Gravitation,Formula Sheets

All the formulas in one go

Universal Law of Gravitation

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

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

where

G = 6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}

is the universal gravitational constant.

Newton's Law of Gravitation in vector form :

F_{12} = \frac{G m _{1} m _{2}}{r ^{2}} \overset{r_{12}}{^}

and

F_{21} = \frac{G m _{1} m _{2}}{r ^{2}} \overset{r_{21}}{^}

Now,

\overset{r}{^}_{12} = - \overset{r}{^}_{21}

, Thus

F_{21} = \frac{- G m _{1} m _{2}}{r ^{2}} \overset{r}{^}_{12}

Comparing above, we get

F_{12} = - F_{21}

Gravitational Field Gravitational field across mass

m

due to the mass

M

E = \frac{F}{m} = \frac{G M}{r ^{2}}

Gravitational potential:

V = \frac{- G M}{r}

and

E = \frac{- d V}{d r}

Ring

V = \frac{- G M}{( a ^{2} + r ^{2} ) ^{1 / 2}}

and

E = \frac{- G M}{( a ^{2} + r ^{2} ) ^{3 / 2}} \overset{r}{^}

E = \frac{- G M c o s θ}{x ^{2}}

Gravitational field is maximum at a distance,

r = \pm a / 2

and it is

E = \frac{- G M}{3 3 a ^{2}}

Thin Circular Disc

V = \frac{- 2 G M}{a ^{2}} ((a^{2} + r^{2})^{1 / 2} - r)

and

E = \frac{- 2 G M}{a ^{2}} (1 - \frac{r}{( r ^{2} + a ^{2} ) ^{1 / 2}}) = \frac{- 2 G M}{a ^{2}} (1 - c o s θ)

Non conducting solid sphere

(a) Point P inside the sphere.

r \leq a

, then

V = \frac{- G M}{2 a ^{3}} (3 a^{2} - r^{2})

and

E = \frac{- G M r}{a ^{3}}

, and at the centre

V = \frac{- 3 G M}{2 a}

and

E = 0

(b) Point P outside the sphere.

r \geq a

, then

V = \frac{- G M}{r}

and

E = \frac{- G M}{r ^{2}}

Uniform Thin Spherical Shell / Conducting solid sphere

(a) Point P Inside the shell.

r \leq a

, then

V = \frac{- G M}{a}

and

E = 0

(b) Point P outside shell.

r \geq a

, then

V = \frac{- G M}{r}

and

E = \frac{- G M}{r ^{2}}

VARIATION OF ACCELERATION DUE TO GRAVITY: Effect of Altitude

g_{h} = \frac{G M _{e}}{( R _{e} + h ) ^{2}} = g (1 + \frac{h}{R _{e}})^{- 2} \approx g (1 - \frac{2 h}{R _{e}})

when

h ≪ R

Effect of depth

g_{d} = g (1 - \frac{d}{R _{e}})

Effect of the surface of Earth

The equatorial radius is about 21 km longer than its polar radius. We know,

g = \frac{G M _{e}}{R _{e}^{2}}

. Hence,

g_{p o l e} > g_{e q u a t o r}

Here,

M_{e}

is the mass of earth.

R_{e}

is the radius of earth.

h

is the height above the earth surface.

d

is the depth below the earth surface.

G

is the gravitational constant.

g

is the acceleration due to gravity.

(g = 9.81 m s^{- 2})

Variation of g due to Rotation of Earth

g^{'} = g - R w^{2} c o s^{2} θ

At a latitudinal position

θ

on earth, where

w

is rotational velocity of the earth, the variation on gravitational acceleration follow above relationship.

SATELLITE VELOCITY (OR ORBITAL VELOCITY)

v_{o} = (\frac{G M _{e}}{R _{e} + h})^{1 / 2} = (\frac{g R _{e}^{2}}{R _{e} + h})^{1 / 2}

when

h < < R_{e}

, the

v_{o} = g R_{e}

∴ v_{o} = 9.8 \times 6.4 \times 1 0^{6} = 7.92 \times 1 0^{3} m s^{- 1} = 7.92 k m s^{- 1}

Time period of Satellite

T = \frac{2 π ( R _{e} + h )}{( \frac{g R _{e}^{2}}{( R _{e} + h )} ) ^{1 / 2}} = \frac{2 π}{R _{e}} (\frac{( R _{e} + h ) ^{3}}{g})^{1 / 2}

Energy of a Satellite

U = \frac{- G M _{e} m}{r}

K . E . = \frac{G M _{e} m}{2 r}

; then total energy

T . E = P . E + K . E = \frac{G M _{e} m}{2 r}

Kepler's Laws Law of area : The line joining the sun and a planet sweeps out equal areas in equal intervals of time.

Areal velocity = \frac{area swept}{time} = \frac{\frac{1}{2} r ( r d θ )}{d t} = \frac{1}{2} r^{2} \frac{d θ}{d t} = C o n s t a n t

Hence, Law of periods :

\frac{T ^{2}}{R ^{3}} = C o n s t a n t

Facebook SDK (Plugin)

Class 11 | Physics | chapter 8 | Gravitation,Formula Sheets

Made with Love by

Contact form

Facebook SDK (Plugin)

Class 11 | Physics | chapter 8 | Gravitation,Formula Sheets

You may like these posts