# Gravitation

NEET PHYSICS GRAVITATION CONCEPTS

It is the phenomenon of mutual attraction between any two bodies in the universe

The body in the universe attracts other bodies towards itself with a force known as Gravitational Force

Kepler Law

1) First law of orbits

States that all particles move in elliptical orbits with the sun at one of the foci of the ellipse

The closest point near the sun is called perihelion(or)perigee and the farthest distance from the sun is called aphelion(or)apogee

Mathematical Equation is SP + $S^{’}$P is constant

2) Second law of areas

States that the line joining the planet sweeps out equal areas in equal interval of time and areal velocity is constant

Areal velocity is $\frac{dA}{dt}$ = $\frac{1}{2}$ rv = $\frac{ L}{2m}$ = constant

3) Third law of period

States that square of the time period(T) of revolution of a planet around the sun is proportional to the cube of the semi major axis(a) of its elliptical orbit

Mathematical Equation is $T^{2}$ ∝ $a^{3}$

Newton law of Gravitation

States that every particle in the universe attracts every other particle with a force whose magnitude is directly proportional to the product of their masses and inversely proportional to the square of distances between their centers

Mathematical Equation is $\frac{Gm_1m_2}{r^{2}}$

Magnitude is 6.67 X $10^{-11}$

Points Regards about force of gravitation

1) It is universal because it acts for any two bodies anywhere in the space

2) It is always attraction in nature and independent of medium and presence of other bodies

3) It is the central force and acts along the line joining the centers of two bodies

4) It is always conservative in nature as it doesn’t depend upon the path

5) Force of attraction due to the hollow spherical shell of uniform density on a point situated inside it is zero

6) The force between two bodies is gravitational force and force between the earth and other body is called gravity force

Vector form of Gravitation:

F = $\frac{Gm_1m_2}{\frac{r_1-r_2}{|r_1-r_2|^3}}$ and force due to the two particles is equal in magnitude but opposite in direction

Principle of superposition of gravitational forces

It states that the resultant force F due to number of point masses is equal to vector sum of forces exerted by the individual masses on the given particle

$F_{net}$ = $F_{12} + F_{13} + F_{14} + ………………….. + F_{1n}$

The direction of force is equal to direction of distances between the particles

Gravity

It is the force by which earth attracts the body towards its Centre

Acceleration due to gravity (g) of earth

It is the force acting on a body having unit mass placed on or near the surface of the earth or defined as the force exerted by earth on the freely falling body is called acceleration due to gravity

Relation b/w acceleration due to gravity(g) and gravitational constant is g = $\frac{GM}{R^2}$ and by substituting M as 6 x $10^{24}$ and R as 6400 km value of g is 9.8 $\frac{m}{s^2}$

The value of g is independent on mass shape size etc. of the body and depends upon the mass and radius of planet

The value of g on moon is $\frac{1}{6}$ is value of g on earth

1) Due to height on the surface of earth

$g^{‘}$ = $\frac{GM}{R^2(1+\frac{h}{R})^2}$ = g ($\frac{1-2h}{R}$)

Decrease in the value of g is $\frac{2gh}{R}$

Fractional decrease in the value of g is $\frac{2h}{R}$

Percentage decrease in the value of g is $\frac{2h}{R}$100

2) Due to depth below the surface of earth

g’ = $\frac{GM}{R^3 (R-d)}$ = g (1-$\frac{d}{R}$)

Decrease in the value of g is $\frac{g}{R}$d

Fractional decrease in the value of g is $\frac{d}{R}$

Percentage decrease in the value of g is $\frac{d}{R }$x100

3) Due to the shape of the earth

At Poles $g_p$ = $\frac{GM}{R_p^2}$

At Equator $g_{eq}$ = $\frac{GM}{R_{eq}^2}$

As $R_p$ < $R_{eq}$ then $g_p$ > $g_{eq}$

Variation in g due to axial rotation of earth

g= g− $ω^2$R$cos^2$ φ

This gives the acceleration due to gravity due to rotation of earth.

At equator; φ = 90

g=g

At equator; φ = 0

g=g− $ω^2$R

At a latitudinal position ω on earth, where w is rotational velocity of the earth, the variation on gravitational acceleration follow above relationship.

Gravitational Field

It is defined as the space around a body in which a body attracts any other body experience a force of attraction

It is denoted by E or I

Formula is $\frac{F}{M}$

It is a vector and always directed towards the Centre of gravity of body whose gravitational field intensity is to be considered

Gravitational Field Intensity of a Point Mass

$E_a$ = $\frac{-GMm}{r^2}$

Gravitational Field Intensity due to Ring

E(r) = $\frac{GMr}{(R^2 + r^2)^{3/2}}$

E(r) = $\frac{2GM}{3(\sqrt3)R^2}$

Gravitational Field due to Uniform Spherical Shell

Positionof point ‘P’

Inside the spherical shell (r < R) E = 0

On the surface of the spherical shell (r R) E = $\frac{GM}{R^2}$

Outside the spherical shell ( r > R) E = $\frac{GM}{r^2}$

Gravitational Field due to Uniform Solid Sphere

The position of point ‘P’

Inside the uniform solid sphere (r < R) E = $\frac{GMr}{R^3}$

On the surface of the uniform solid sphere (r= R) E = $\frac{GM}{R^2}$

Outside the uniform solid sphere (r > R) E = $\frac{GM}{r^2}$

Gravitational Potential

It is the work done in bringing the unit mass from reference(infinity) to a given point in the gravitational field

It is denoted by V

Formula = $\frac{W}{m}$

It is a scalar quantity

Gravitational Potential of a Point Mass

V = $\frac{– GM}{r}$

Gravitational Potential of a Spherical Shell

Case 1: If point ‘P’ lies Inside the spherical shell (r < R):

As E = 0, V is a constant.

The value of gravitational potential is given by, V =$\frac{-GM}{R}$

Case 2: If point ‘P’ lies on the surface of the spherical shell (r = R):

On the surface of the earth, E = $\frac{-GM}{R^2}$

Using the relation V=−∫E.dr we get,

Gravitational Potential V =$\frac{-GM}{R}$

Case 3: If point ‘P’ lies outside the spherical shell (r > R):

Outside the spherical shell, E = $\frac{-GM}{r^2}$

Using the relation V=−∫E.dr we get,

V = $\frac{-GM}{r}$

Gravitational Potential of a Uniform Solid Sphere

,Case 1: If point ‘P’ lies Inside the uniform solid sphere (r < R):

Inside the uniform solid sphere, E = $\frac{-GMr}{R^3}$

Using the relation V=−∫E.dr

The value of gravitational potential is given by,

V = -GM [$\frac{3R^2 – 2r^2}{R^3}$]

Case 2: If point ‘P’ lies On the surface of the uniform solid sphere ( r = R ):

On the surface of a uniform solid sphere, E = V = $\frac{-GM}{R^2}$

. Using the relation V=−∫E.dr we get,

V = $\frac{-GM}{R}$

Case 3: If point ‘P’ lies Outside the uniform solid sphere ( r > R):

V = $\frac{-GM}{r}$

Case 4: Gravitational potential at the center of the solid sphere is given by V = -$\frac{3}{2} ×\frac{GM}{R}$.

Gravitational Potential Energy

When a body of mass (m) is moved from infinity to a point inside the gravitational influence of a source mass (M) 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

Under the action of gravitational force, the work done is independent of the path taken for a change in position so the force is a conservative force. Besides, all such forces have some potential in them.

The gravitational influence on a body at infinity is zero, therefore, potential energy is zero, which is called a reference point.

Gravitational Potential Energy Formula

The equation for gravitational potential energy is:

GPE = mgh

Gravitational Potential Energy Equation

Since the work done is stored as its potential energy U, therefore gravitational potential energy at a point which is at a distance ‘r’ from the source mass is given by;

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

If a test mass moves from a point inside the gravitational field to the other point inside the same gravitational field of source mass, then the change in potential energy of the test mass is given by;

ΔU = GMm ($\frac{1}{r_i}$ – $\frac{1}{r_f}$)

If $r_i$ > $r_f$ then ΔU is negative.

Expression for Gravitational Potential Energy at Height (h) – Derive ΔU = mgh

If a body is taken from the surface of the earth to a point at a height ‘h’ above the surface of the earth, then $r_i$ = R and $r_f$ = R + h then,

ΔU = GMm [$\frac{1}{R}$ – $\frac{1}{R+h}$]

ΔU = $\frac{GMmh}{R(R + h)}$

When, h ≪ R, then, R + h = R and g = $\frac{GM}{R^2}$

On substituting this in the above equation we get,

Gravitational Potential Energy ΔU = mgh

Note:

The weight of a body at the Centre of the earth is zero due to the fact that the value of g at the Centre of the earth is zero.

At a point in the gravitational field where the gravitational potential energy is zero, the gravitational field is zero.

Escape Velocity

This minimum amount of velocity for which the particle escapes the gravitational sphere of influence of a planet is known as escape velocity $v_e$.

When escape velocity is provided to a body it goes to infinity theoretically.

$V_e$ = $\sqrt{2GR}$

From the above formula, it is clear that escape velocity does not depend upon the m

If the source mass is earth then the escape velocity has a value of 11.2 $\frac{km}{s}$.

If v = $v_e$ the body escapes the planets gravitational sphere of influence,

if 0 ≤ V ≤ $V_e$ the body either falls back onto the earth or continues to orbit around the planet within the sphere of influence of the planet.

Orbital Velocity

The velocity with which the test mass orbits around a source mass is known as orbital velocity $V_o$

Vo = $\sqrt{gR}$

The above formula suggests that the orbital velocity is independent of the mass

Relationship Between Escape And Orbital Velocity

The relationship between escape velocity and orbital velocity can be mathematically represented as:

$V_e$ = $\sqrt{2}V_o$

It shows that escape velocity is 2$\sqrt{2}$times greater than orbital velocity.

Cases

• When the velocities are the same, the object will be in constant orbit and the same elevation.

• If escape velocity is less than orbital then the orbit will diminish which will result in the object crashing.

• If it is more then the object will be free in the orbit and will likely float into space.

Time Period Of A Satellite

The time taken by the satellite to complete one revolution around the earth is known as the time period of a satellite.

The time period of a satellite is the ratio of total distance travelled by the satellite around the earth to the orbital velocity.

T= $\frac{2πr}{v_o}$ and we get T= $\frac{2πr}{\sqrt\frac{GM}{r}}$ and through we get the value of T as

$T^2$ = 4π$\sqrt\frac{r^3}{GM}$

The constant of proportionality in the above equation depends only on the source mass but not on the mass.

Kinetic Energy Of A Satellite

Kinetic energy is the energy possessed by the body in motion (whether translational or rotational or a combination of both) in the case of the satellite orbiting around the earth.

Formula K= $\frac{GMm}{2r}$

Kinetic energy can never be negative for any force.

Potential Energy Of A Satellite

Potential energy is the energy possessed by the body in a particular position. Potential energy changes when the position of the body changes.

The potential energy possessed by the satellite at a distance ‘r’ from the centre of the earth is given by;

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

Total Energy Of A Satellite

The total energy of the satellite is the sum of all energies possessed by the satellite in the orbit around the earth.

As only the mechanical motion of the satellite is considered, it has only kinetic and potential energies.

Total energy of the satellite = kinetic energy of the satellite + potential energy of the satellite

E = – $\frac{GMm}{2r}$

From the above equation,

Total energy of the satellite = – (kinetic energy of the satellite)

Total energy of the satellite = – (kinetic energy of the satellite)

Total energy of the satellite = $\frac{potential energy of the satellite}{2}$

Binding Energy Of A Satellite

Binding energy is the energy required to be given to the satellite to escape the gravitational pull of the earth. It is numerically equal to the negative of total energy possessed by the satellite.

Binding energy of a satellite = $\frac{GMm}{r}$

The binding energy of a system of two bodies is the amount of minimum energy required to separate them by infinite distance or it is simply the amount of energy required to make the potential energy of the system equal to zero, which is numerically equal to the kinetic energy of the body under motion.

Angular Momentum Of a satellite

When a satellite of mass ‘m’ is orbiting with an angular speed on the orbital path of radius ‘r’, its angular momentum is given by

L= m$r^2$ω

ω = $\frac{T}{2π}$ substituting this in the above equation we get

L = $\frac{mr^2T}{2π}$

Using T = $\frac{2π}{\sqrt{GM}}$$r^{3/2}$ we get,

L = m{$\sqrt{GMr}$}

From the above equation, it is clear that the angular momentum of a satellite depends on both the mass of a satellite and mass of earth.

It also depends upon the radius of the orbit of the satellite.