IIT JEE / CET – Main / Advanced Physics 2013 & 2014 : Physics Free Notes — Gravitation – All the essentials in one page [ PO'PS.Series.3]


Gravity or gravitation is basically a natural phenomenon in which , objects with mass attract each other. Gravitation is one of the four fundamental forces of nature, along with the strong force , electromagnetism and the weak force.   Gravity …..

  1. causes objects with mass to experience weight , and to fall to the surface of the earth / planet.
  2. causes dispersed matter to coalesce, thus accounting for the existence of the earth, planets, suns and most things in the universe.
  3. is responsible for keeping the Earth and the other planets in their orbits around the Sun.
  4. results in the formation of tides.
  5. results in natural fluid convection currents , causing fluid flow to occur under the influence of a density gradient.
  6. heats the interiors of stars & planets to high temperatures ( radioactivity helps in this process ).

Def – . The natural force of attraction exerted by a celestial body, such as Earth, upon objects at or near its surface ( in the gravitational feild ), tending to draw them toward the center of the body. It is the natural force of attraction between any two massive bodies, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Difference between Gravity and Gravitation – Eventhough one may use the two words interchangeably, gravity basically refers to weight and associated facts related to earth , while gravitation ( attraction between masses ) is a universal phenomenon.

Basic calculations in gravity are done using Newton’s laws of universal gravitation. According to Modern Physics ,  gravitation is described using the general theory of relativity, in which gravitation is a consequence of the curvature of space – time which governs the motion of inertial objects.


They are the three laws of planetary motion discovered by Johannes Kepler during the early years of the seventeenth century.The laws opened the way for the development of celestial mechanics, i.e., the application of the laws of physics to the motions of heavenly bodies.

Kepler's laws -Illustration of Kepler's three laws with two planetary orbits. (1) The orbits are ellipses, with focal points ƒ1 and ƒ2 for the first planet and ƒ1 and ƒ3 for the second planet. The Sun is placed in focal point ƒ1. (2) The two shaded sectors A1 and A2 have the same surface area and the time for planet 1 to cover segment A1 is equal to the time to cover segment A2. (3) The total orbit times for planet 1 and planet 2 have a ratio a13/2 : a23/2.

n 1687, Isaac Newton demonstrated that any body, moving in an orbit around another body that attracts it with a force that varies inversely as the square of the distance between them, must move in a conic section. This path will be an ellipse when the velocity is below a certain limit in relation to the attracting force. Thus the first law is a general law that applies to all satellites held in orbit by an inverse-square force.

  1. Law of Orbit : A planet moves in an elliptical orbit around the Sun that is located at one of the two foci of the ellipse. An ellipse is one of the conics curves. ( Note – Perigree is the distance representing the closest/shortest approach of earth/planet with it’s sun. Apogee – is the distance representing the longest approach of earth/planet with it’s sun. )

Heliocentric coordinate system (r, θ) for ellipse. Also shown are: semi-major axis a, semi-minor axis b and semi-latus rectum p; center of ellipse and its two foci marked by large dots. For θ = 0°, r = rmin and for θ = 180°, r = rmax.

r=\frac{p}{1+\varepsilon\, \cos\theta},

where (rθ) are heliocentric polar co-ordinates for the planet, p is the semi-latus rectum, and ε is the eccentricity.

( The latus rectum is the chord parallel to the directrix and passing through the focus (or one one of the two foci , see figure below )

At θ = 0°, perihelion, the distance is minimum

At θ = 90°, the distance is \, p.

At θ = 180°, aphelion, the distance is maximum


The semi – major axis a is the arithmetic mean between rmin and rmax:

\,r_{\rm max}-a=a-r_{\rm min}



The semi – minor axis b is the geometric mean between rmin and rmax:

\frac{r_{\rm max}}b =\frac b{r_{\rm min}}


b=\frac p{\sqrt{1-\varepsilon^2}}.

The semi-latus rectum p is the harmonic mean between rmin and rmax:

\frac{1}{r_{\rm min}}-\frac{1}{p}=\frac{1}{p}-\frac{1}{r_{\rm max}}.

The eccentricity ε is the coefficient of variation between rmin and rmax:


The area of the ellipse is

A=\pi a b\,.

The special case of a circle is ε = 0, resulting in r = p = rmin = rmax = a = b and A = π r2.

2. Law of ( Equal ) AreasThe radius vector of the ellipse (the imaginary line between the planet and the Sun) sweeps out equal areas in equal intervals of time . Consequently , a planet moves more fast when it is closer to the Sun and more slowly when it is farther removed.This law also applies to parabolic & hyperbolic trajectories and is quite similar to the law of conservation of angular momentum. The dynamic nature of this law lies in the fact that , it is not restricted to forces that vary inversely as the square of the distance; rather it is valid for all forces of attraction between the two bodies.

\frac{d}{dt}\left(\frac{1}{2}r^2 \dot\theta\right) = 0,

where \tfrac{1}{2}r^2 \dot\theta is the “areal velocity”.   Hence  1/2 r²ω  is a constant.

Picture 1 of Kepler’s laws

3.Law of Periods : The square of the time for a planet to complete a revolution about the sun is proportional to the cube of semi major axis of the elliptical orbit OR the ratio between the square of a planet’s period (the time required to complete one orbit) to the cube of the mean radius (the average distance from the Sun during one orbit) is a constant.

 {P^2} \propto  {a^3}

where P is the orbital period of planet and a is the semimajor axis of the orbit.

The proportionality constant is the same for any planet around the Sun.

\frac{P_{\rm planet}^2}{a_{\rm planet}^3} = \frac{P_{\rm earth}^2}{a_{\rm earth}^3}.

Derivation :           The centripetal force is balanced by the gravitational force and hence,

mv²/ R = GMm / R²    =>   GM/R = v² .  Now , the planet is moving at a speed of -

v =   orbital circumference / time period = 2πR/T = v.  Therefore,  => GM/R = 4π²R² /T²

or      T²  =  4π²R³ / GM .  For a planet, 4π² /GM is a constant and hence  T² /  R³ = k ( a constant )

Newton’s Law of Gravitation -

Newton’s law of universal gravitation states that every two particles of matter in the universe attract each other with a force that acts in the line joining them, the intensity of which varies as the product of their masses and inversely as the square of the distance between them.Gravitational force F exerted between two particles with masses m1 and m2 separated by a distance d is given by the equation, F=\frac{G m_1 m_2}{d^2} where G is called the universal constant of gravitation.Newton’s law of universal gravitation is effective for very large and very short distances , but not for intermolecular distances.

The Gravitational Constant – ‘ G ‘ -

Eventhough the value of G was determined as early as 1774 , a more accurate one was experimentally validated by Cavendish by his torsion balance in around 1798. The torsion balance , consists of  two small spheres, each of mass m, which are connected by a light rod, suspended in the middle by a thin wire. The deflection caused by bringing the two large spheres each of mass M near the small ones on opposite sides of the rod is measured, and the force is evaluated by observing the period of oscillation of the rod under the influence of the torsion of the wire.

In newtonian gravitation, G [   6.67 × 10−11 SI (mks) units] is an absolute constant, independent of time, place, and the chemical composition of the masses involved.

When m1 = m2 = 1 and r =1 ,  F = G . Hence , Universal Gravitational constant , is equal to the force of attraction acting between two bodies each of unit mass , whose centres are placed unit distance apart. G is a scalar quantity.

Derivation of Newton’s law fromKepler’s law -

Consider a planet of mass m , revolving around it’s sun of mass M , in a nearly circular orbit of radius r with a constant angular velocity of ω and let T be the time period of revolution of the planet around the sun.       ω   = 2π/T       - ( eq.1 )

Considering the centripetal force ( caused by the gravitational attraction between the sun and the planet ) -

F  = mrω²  = mr ( 2π/T )²  =  mr4π² / T² .

Now , according to Kepler’s third law -  T² α r³  or   T²  =  kr³, where k  → constant of proportionality.  Threfore ,

F   =  4π²mr/kr³   =  4π²/K  x  m/r²         - ( eq.2 )

Since 4π²/K is a constant  ,    F  α  m/r²  . Now, according to Newton, the gravitational pull between the sun and the planet is mutual. This means that the force should not only  be directly proportional to the planet’s mass [ F α m ] , but also to that of the sun’s mass , M .

Therefore ,  4π²/K α M   or  4π²/K  = GM  . Substituting this value into (eq.2)   F = GMm/r²

Gravitational Forces    -

  • are independent of the presence or absence of other bodies.
  • are independent of the nature of the intervening medium.
  • are independent of the nature and the size of the bodies,. as long as the mass of the bodies and the distance between their centres doesn’t change.
  • are equal in magnitude and opposite in direction i.e, they obey Newton’s third law of motion and hence form action and reaction pairs.
  • are conservative in nature.
  • are central forces , as they act along the line joining the centres of the two bodies.
  • are always attractive.

Newton’s Law of Gravitation > Vector form -

Consider two particles A and B of masses mi and m2 respectively

Let, Jf2 = displacement vector from A to B, r/j = displacement vector from B to A, F^i = gravitational force exerted on B by A. F^z = gravitational force exerted on A by B. In vector notation, Newton’s law of gravitation is written as follows :

-> mj m2 A

Fi2 = -G——r21 …(1)

where r2X is a unit vector pointing towards A. The negative sign indicates that the direction of F]’2 is opposite to that of r2i. The negative sign also shows that the gravitational force is attractive in nature.

Similarly, F2i = – G —=— r12    , where r12 is a unit vector pointing towards B.

Now, r2l=-rn    . Also, r2j = r12

m! m2 A

F2i = G—— r2i …(2)

Equating (1) and (2), F^ = – F^i    . Therefore, the gravitational forces between two bodies are equal in magnitude and opposite in direction.

Principle of Superposition of Gravitation -

The resultant gravitational force , | F| , acting on a particle due to a number of point masses is equal to the vector sum of the forces exerted by the individual  masses on the given particle .

| F01|   =  Gm0M/ r² x ˆr10 =  Gm0m1/  |r01|²  x ˆr01

|F02|  =  Gm0m2/ |r02|²  x  ˆr02 and   |F0n| = Gm0mn / |r0n|²   x  r0n therefore ,

| F0|  =  Gm0 [  m1ˆro1 / |r01|²   +    m2ˆr02 /   |r02|²    + ........   +   mnˆr0n/  |r0n|²  ]


Indicates an object’s resistance in changing its state of motion when an external force is applied. It is determined by applying a force to an object and measuring the acceleration that results from that force. An object with small inertial mass will accelerate more than an object with large inertial mass when acted upon by the same force. The inertial mass of a body  -

  • is defined by Newton’s second law of motion.
  • is related to it’s own inertia of linear motion.
  • is equal to the magnitude of external force required to produce unit acceleration in the body.
  • is proportional to the quantity of matter it contains.
  • is independent of size, shape and the state ( like temperature ) of the body.
  • is not influenced by the presence or absence of other bodies near it.
  • can be added with other similar bodies by means of algebraic laws.
  • is conserved when it combines with another body in any manner.

Now, by Newton’s second law of motion , F = mia   , so  mi= F/a  where mi refers to inertial mass. Therefore , for unit acceleration mi = F.

Inertial mass of a body changes due to it’s motion. For a body having velocity v , inertial mass is given by -

m  =   m0 /  √  1 – v²/c²

here , m0= rest mass of the body and c = velocity of light in vacuum. Hence , mass changes when the velocity of a body nears that of light.


Gravitational mass of a body refers to it’s interaction with a gravitational feild.  It can be defined as the magnitude of gravitational pull by experienced a body in a graviational feild of unit intensity. For a body of  mass m0,

m0 =   FR² / GM   ( from Newton’s law of gravitation )  . There are two kinds of gravitational mass -

  1. Active graviational mass is a measure of the strength of an object’s gravitational flux (gravitational flux is equal to the surface integral of gravitational field over an enclosing surface). Gravitational field can be measured by allowing a small ‘test object’ to freely fall and measuring its free- fall acceleration. For example, an object in free-fall near the Moon will experience less gravitational field, and hence accelerate slower than the same object would if it were in free-fall near the earth. The gravitational field near the Moon is weaker because the Moon has less active gravitational mass.
  2. Passive gravitational mass is a measure of  the strength of an object’s interaction with a particular gravitational feild. Passive gravitational mass is determined by dividing an object’s weight by its free-fall acceleration. Two objects within the same gravitational field will experience the same acceleration; however, the object with a smaller passive gravitational mass will experience a smaller force (less weight) than the object with a larger passive gravitational mass.

The ( negative ) gravitational mass of a spherical body  -


MASS     IN    PHYSICS -                        [ for reference ]

Properties of mass in Physics.

Dig - How ‘ mass ‘ is defined in Physics .Every object / body having mass  is believed to exhibit all the below stated five properties, However, due to extremely large or extremely small constants, it is generally impossible to verify more than two or three properties for any object.

  1. The Schwarzschild radius (rs) represents the ability of mass to cause curvature in space and time.
  2. The standard gravitational parameter (μ) represents the ability of a massive body to exert Newtonian gravitational forces on other bodies.
  3. Inertial mass (m) represents the Newtonian response of mass to forces.
  4. Rest energy (E0) represents the ability of mass to be converted into other forms of energy.
  5. The Compton wavelength (λ) represents the quantum response of mass to local geometry.


Inertial mass and Gravitational mass -

  1. Both are scalar quantities.
  2. Both are independent of the size and state of a body.
  3. Both are measured in the same units.
  4. Both are equivalent to each other.
  5. Gravitational mass is affected by the presence of other bodies near it , but inertial mass is not.
  6. Inertial mass is measured by the acceleration produced on a body by an applied force ( here gravity has no role). But gravitational mass is a measure of the effect of gravitational pull on a body.
  7. Inertial mass is measured by an inertial balance and gravitational mass is measured by a spring balance.

Equivalence of Gravitational and Inertial Masses ( Galilean equivalence principle or  weak equivalence principle ) -

It refers to the fact that, the gravitational and inertial masses of a body are quite identical.Most important consequences of this equivalence principle applies to freely falling objects ,  in which gravity is the only acting force and all other forces, like friction and air resistance , must be absent or negligible.

Importance -  A more modern version of this principle known as the Einstein equivalence principle or the strong equivalence principle, lies at the heart of the general theorey of relativity  - one of the core and most important theories in Physics . Einstein’s equivalence principle states that within sufficiently small regions of space-time, it is impossible to distinguish between a uniform acceleration and a uniform gravitational field. Thus, the theory postulates that the force acting on a massive object caused by a gravitational field is a result of the object’s tendency to move in a straght line (in other words its inertia) and should therefore be a function of its inertial mass and the strength of the gravitational field.

Let m0 and m’0 be the gravitational masses of two bodies A and B respectively, which are placed at an equal distance of R from another body of gravitational mass C. By Newton’s law of gravitation , the gravitational pull exerted by C on A is -

F = GMm0/R²         and that by C on B is       F’  =  GMm’0/R²    ,  hence -

F/F’          =     m0/m’0 . Now , let m1 and m’1 be the inertial masses of bodies A and B , respectively . Both the bodies are allowed to fall in vaccum from the same height under the influlence of gravity , therefore , F = m1g and  F’  = m’1g

Hence ,   F/F’  =  m1 / m’1 =   m0/m’0 i.e, the gravitational mass of a body is proportional to it’s inertial mass.


It’s basically a model used in physics to explain how gravity exists in the universe . In this model, rather than two particles attracting each other, the particles distort space time by means of their mass, and this distortion is what is perceived and measured as a “ gravitational force” and the area in which this force is experienced is called the “ gravitational feild “ .In such a model one states that matter moves in certain ways in response to the curvature of spacetime. The curious aspect of this model is that , in a sense gravity could be considered a ficitious force .

  1. Gravitational feild in Classical ( Newtonian ) Mechanics : In classical mechanics, this is basically a model , where the field can be determined using Newton’s law of universal gravitation . Consequently, the gravitational field around a single particle is a vector feild consisting at every point of a vector pointing directly towards the particle. The magnitude of the field at every point is calculated on the basis of the force per unit mass on any object at that point in space.
  2. Gravitational feild in Modern ( theory of Relativity ) Physics : In the theory of general relativity , the gravitational field is determined as the solution of Einstein’s feild equations . These equations are dependent on the distribution of matter and energy in a region of space, and unlike Newtonian gravity, it’s not dependent only on the distribution of matter. The fields represent the curvature of space time . General relativity states that being in a region of curved space is equivalent to accelerating up the gradient of the field.

Gravitational field lines - Earth

Gravitational Feild Intensity -

The intensity of gravitational feild of a body at a point in the feild is defined as the force experienced by a body of unit mass placed at that point provided the presence of unit mass does not interfere with the original gravitational feild. It’s a vector quantity and is always directed towards the centre of gravity of the body being considered . It’s denoted by | E |.

Consider a body of mass M with the centre of gravity O . A test mass m0 , placed at a point P in this feild experiences a gravitational force of attraction F . Now , OP = x .  By newton’s law of gravitation ,

F  =  MGm0/ x²        . Intensity of the gravitational feild at point P is -

E  = F / m0 =   [GMm0 / x² ] /m0 =   GM / x²    -    ( eq.1 )

In vector form ,  | E | = – Gm /x²  . ˆx                         -    ( eq . 2)

The negative sign means that the gravitational intensity is an attractive force .  Eq.2 indicates thet gravitational feild intensity is zero at infinite distance from the body . If  the test mass m0 placed at P can move , then it will be accelerated by the force F …….

F  =    m0a         i.e ,   F /m0 =  a    ,  hence   E  =  a        … therefore, the intensity of gravitational feild at a point in the gravitational feild is equal to the acceleration of a test mass placed at thet point .  Let the above situation be that of Earth , so that the point P lies on the earth’s surface – Then   x  =  R ,

E  =  GM / R²   =  g (  where g – acceleration due to gravity on the surface of the earth )

Units of E  =  N /kg   or  m / s²  ( S.I )   dyne / g  or  cm / s²   . Dimensions  :  [  M° L T² ]

The negative energy density of a Gravitational Feild

Gravity and Gravitation -  more details :

Consider the force of attraction between a hollow spherical shell of uniform density and a point mass outside the shell . The force on the point mass will be acting along a line joining it and a region of the hollow shell . This force is , F = GMm / r² , where, r is the distance between the point mass and the centre of the spherical shell. In the above case , one can consider that the entire mass of the spherical shell is concentrated at it’s centre. Now , if this point mass is located inside the spherical shell ( of uniform density ) ,  the gravitational forces acting on the point mass from various small regions of the shell cancel each other . Hence , the net gravitational force on a point mass inside a spherical shell of uniform density would be zero.

Note -  The gravitational force between two bodies will be zero , if the separation between them becomes infinity. Considering the gravity of a body , it will be zero , at the centre of the earth.

Relation between g  and  G :

Consider a body of mass m , placed on the surface of earth , whose mass is M , radius is R and centre is O. Now, by Newton’s law of gravitation,

F  = GMm / R²     From the pull of gravity , F =  mg  . Therefore ,  mg  =  GMm / R²   and hence ,

g  =  GM / R² – we can see that the value of acceleration due to gravity does not depend on the mass , size and shape of the body , but depends on the mass and radius of the earth.

Variations of  g -

variations of g on earth's surface

1 .  Due to altitude  -

Consider the variation of g when a body moves distance upward or downward from the surface of earth.

Let g be the value of acceleration due to gravity at the surface of earth and g’ at a height h above the surface of earth.
If the earth is considered as a sphere of homogeneous composition, then g at any point on the surface of the earth is given by:
Points to note  –
  • At a height equal to the radius of the earth ( when h = R = 6400kms ) , g’ = gR² / (R + R)²   = g / 4 .
  • When using the equation  g’ /g= R² / (R + h)² , note that h must be comparable to the radius of the earth.
  • When using the equation g’ = g ( 1 – 2h/R ) , note that h must be very small when compared to R.

Decrease in the value of g with height h -

g – g’ = 2hg/R  , therefore fractional decrease -  g =  g’ – g/g = 2h/R ,

hence % decrease =  g = [ g - g'/g ] x 100  =  2h/R x 100

. Due to depth  -

g variations due to height & depth

Let earth be a homogenous sphere of radius R and mass M with centre at O and g is the value of acceleration due to gravity at a point A on earth’s surface.

g = GM / R²   . Let ρ be the uniform density of earth’s material , then , M = 4/3 πR²ρ  , therefore ,

g = 4/3 πGRρ       – ( eq.1 )

Now, consider a point B at a depth of d below the earth’s surface where the acceleration due to gravity is g’ .Hence the distance of point B from earth’s centre is ( R – d ). Now , considering earth as a spherical shell of thickness d , the force acting on a mass m at point B is zero ( explained earlier ). Note that B is located outside the surface of the smaller sphere of radius (R – d ). Considering the force acting on m at B , it’s as if the entire mass M’ of the smaller sphere of earth is concentrated at the centre O.

Therefore ,  g’ =   GM’/( R-d)³  and  M’ = 4/3 π(R-d)ρ . Hence g’ = 4/3 πG (R-d) ρ   - (eq.2)

Dividing (eq.2) by (eq.1)  =  g’/g =  R-d/R   =>  g’ = g ( 1 – d/R) ……. hence , the value of g decreases with depth.

As at the centre of the earth , d = R , the value of g’ becomes zero , i.e, acceleration due to gravity is absent at the Earth’s center.This also means that the body will have no weight , eventhough it’s mass doesn’t change.

Note  :   % decrease in the value of g is  g-g’/g x 100  =  d/Rx 100 .

3. Due to Earth’s shape -

Shape of the earth has an effect on g as earth is not a perfect sphere – it’s flattened at the poles and bulged at the equator. Due to this the equatorial radius of earth is around 21 kms greater than the polar radius. Now , g =  GM/R² and hence ,

g α 1/R² . This means that the value of g is greatest at the poles and least at the equator. Hence this equatorial bulge [ along with the centifugal force of earth's rotation conteracting g ] causes sea-level gravitational acceleration to increase from about 9.780 m·s−2 at the equator to about 9.832 m·s−2 at the poles, so an object will weigh about 0.5% more at the poles than at the equator. This translates to a difference of about 1.80 cm s-2. 

Note – the fractional decrease in g for a body , when it is moved from the surface of earth to a height h when  h << R is -

dg /g    =  – 2  ( h/R )


Physics musings / reference

GEOID - For the first time ever , in 2010 , a satellite [ GOCE ] mapped the variations of acceleration due to gravity on earth’s surface to extraordinary detail . The map is known as a geoid – an imaginary global ocean dictated by gravity in the absence of tides and currents. It was first described almost 200 years ago but this is the first time it has been mapped in real-time.

Geoid - The differences in gravitational force are represented using colours - deep blue = -100 metres to deep red = + 100 metres


Gravitational  Potential

It’s the potential associated with gravitational force .  The factors that affect an object’s gravitational potential are its height relative to some reference point, its mass, and the strength of the gravitational field it is in. Defined as the amount of work done in bringing a body of unit mass from infinity to a paticular  point in a gravitational feild without acceleration . It’s a scalar quantity .

V  =   W / m0 , where  W   - the amount of work done  and m0 – mass of the body.

1. Gravitational Potential at a point

Consider earth as a perfect sphere of radius R and mass M whose mass is assumed to be concentrated at it’s centre O . Now , consider any point P , such that OP = r and r > R. Extend OP outwards and take two points on this line such that OA = x and AB = dx.

Now . on an unit mass at point A -    F =  GM/x²   [  as the other is a unit mass ] . The amount of work done in bringing a unit mass without acceleration through a small distance AB = dx is    dW = Fdx  =  GM/x² . dx.

The total work done is found out by integrating the above expression within the limits  x = ∞  to x = r  . This gives W = – GM/r  = Vp where Vp is the gravitational potential at P.   Points to note -

  • If r = ∞ , Vp becomes zero . Therefore , gravitational potential is maximum at infinity.
  • On the surface of the earth , as r = R , Vp becomes  – GM/R .
  • Gravitational potential at a point is always negative.
  • Gravitational potential is a scalar quantity as it has only magnitude and no direction.
  • The unit of gravitational potential is J kg-1 and dimensions are M0L² T-2

2. Gravitational Potential at a point due to Earth

When a body of mass m is brought from infinity to a point at a distance r from the centre of the earth , the work done, W = – GMm/r  [r> R] . We know that gravitational potential V  = W / m  . Therefore ,

V  =     [  - GMm/r ] / m   =   - GM/r    (when  r > R )  =   - GM/R    ( when r = R )

=    - GM[ 3R² - r²/2R³ ]    ( when r < R )      =   3/2 [  - GM/R ]  ( when r = 0 )

Therefore       ,           V centre  =   3/2 V surface .

The potential at point P  ,     Vp  =    - GM [ 3R² - r²/2R² ]

For a Spherical shell of mass M and radius R


For a Spherical shell of mass M and radius R . it’s intensity of gravitational feild  is -

  • when r > R  ,       E     =    – GM/r²
  • when r = R  ,       E     =     – GM/R²
  • when r < R  ,       E     =     0

It’s gravitational potential is   -

  • when r > R ,         V    =    – GM/r
  • when r = R ,         V    =      – GM/R
  • when r < R ,         V    = – GM/R

Gravitational  Feild  and Potential -

The work done by an external agent to move an unit mass from one point to another in the direction of a feild =  – Edr where , E = feild and dr = the infinitesmal distance moved by the unit mass.  Now , dV  =  -Edr  , hence   E  =  – dV / dr .  Therefore , the gravitational feild at any point is equal to the negative gradient of potential at that point .

Gravitational Potential Energy

Trebuchet - an ancient war machine ( shown above a reconstruction made in France ) . It uses the gravitational potential energy of a counterweight to throw heavy projectiles over large distances . [ Of course , you can see similar devices in action in many Hollywood flicks - especially in the battle scenes of Lord of the Rings - LOTR

The working principle of Trebuchet

DefThe Gravitational Potential Energy of a particular body at a point in a gravitational feild of another body is defined as the amount of work done in bringing the given body from infinity to that point without acceleration .

Gravitational potential energy     =  W  +  W0 , where  W = work done in bringing the body to a given point without acceleration and W0 = when the body is having zero potential energy , i.e , when it’s at infinity.

Consider a body of mass m placed at a point P in the gravitational feild of the earth which has a mass M and radius R . The mass of the earth is considered to be concentrated at it’s centre O . Now , note that OP = r such thet r > R  . A line joining OP is drawn and extended , on which two points A and B are noted , such that , OA = x and AB = dx .

Now , at A the body experiences a gravitational force of attraction  =  F  = GMm / x² . The amount of work needed to bring the body through dx without acceleration   =  dW  =  Fdx  = [ GMm/x²] dx .

The total work done in bringing the body from infinity to point P is  W  =  ∫r [ GMm/x²] dx    = – GMm ∫r x² dx  = – GMm [ 1/r - 1/∞ ]

Therefore  ,         W  =  – GMm / r .

Note that this work is stored in the body as it’s gravitational potential energy . Hence ,  U  =  – GMm / r .

Also ,  U = Vp x m  i.e , Gravitational Potential Energy  =  Gravitational potential x mass of the body .

Points to note  [ U = Gravitational potential energy ]     -

  • As the value of r increases , U also increases as it becomes less negative. Hence , at infinity , U becomes zero , ie, it’s maximum value .
  • In bringing a body from infinity to a point , work has to be done by the gravitational feild of the earth . Hence , gravitational force is an attractive force as indicated by the negative sign .
  • U of a point mass m at a distance r such that r > R , from the earth’s centre,   U =  mV  = GMm / r .
  • U of a point mass m relative to infinity at the surface of the earth,  U  =  - Gmm/r .
  • U of a point mass m at the earth’s centre , relative to infinity,  U = mVc  =  –  3/2  GMm/R . Therefore , U at the earth’s centre = 3/2 x U at the earth’s surface.
  • U of an isolated system consisting of two point masses m1 and m2   , U = – G m1m2/ r12 .
  • U of an isolated system consisting of three point masses , m1 ,m2 and m3  = U =  – Gm1m2/r12 + Gm2m3/r23 + Gm1m3/r13.
  • U of an nth point mass in an isolated system consisting of multiple point masses = - [  Gmnm1/r n1 + Gmnm2/r n2 + Gmnm3/r n3 +..... ] .This means that the gravitational potential energy of an isolated system having many point masses is equal to the sum of the gravitational potential energies of all the possible pairs contained in the isolated system. ( The Superposition Principle ).
  • The change in gravitational potential energy when a body of mass m is moved from a point at a distance r1 to another at a distance r2 is given by    2 ΔU = GMm [ 1/r1 - 1/r2] . Note that if r1 > r2, then ΔU becomes negative .

Gravitational Self – Energy -

Def For a body or a system of particles , Gravitational self energy is defined as the work done by an external agent in assembling the body / system of particles from infinitesimal elements / particles that are initially at infinite distances apart .

Gravitational potential energy for n particles at an average distance ‘r’ due to their mutual gravitational attraction is equal to the sum of the potential energy of all pairs of particles .

U  =  – G  ∑all pairs j ≠i mi mj /r0 =    - 1/2 G ∑n i=1 j=1 j ≠1n( m²/r)0 , where n particles , each of mass m are separated from each other by an

average  distance r .    When i = n times and j = ( n-1 ) times , then ,   U  =  – ½ Gn ( n – 1 ) m²/r .

hence ,  Self potential energy ,   U  =  – ½ Gm² / R .


It’s the minimum speed required to propel / project a body from the surface of earth / planet such that it will not return to the earth’s / planet’s surface .  (2) Minimum speed required by a body to escape permanently from the gravitational attraction of the parent object/planet without accelerating further .  Also called  parabolic velocity .

It decreases with altitude and equals the square root of 2 (about 1.414) times the speed needed to maintain a circular orbit at the same altitude. At the surface of Earth, disregarding atmospheric resistance, escape velocity is about 6.96 m/second (11.2 km/second). The atmosphere of Earth is due to the fact that it’s escape velocity is considerably higher than the mean velocity of the gas molecules in its atmosphere . The term velocity in escape velocity is a misnomer as it’s a scalar quantity. At escape velocity the kinetic energy + gravitational potential energy of a body is zero. Escape velocity is the velocity needed by any rocket aiming to bring a satellite beyond earth orbit .

Work done to move the body from the earth’s surface to infinity is  -

W   =  G Mem / Re ,  where  Me and Re are the mass and radius of the earth respectively . Consequently to find the escape velocity -

1/2 mve²  =    G Mem / Re ,  therefore ,  ve =   √ 2 g Re =   11.2 km / s

Points  to note -

  • The actual escape velocity from the earth’s surface will be greater than 11.2 kms/s , if air resistance is also included.
  • When a body is projected from a planet , whose speed is less than the escape velocity , then that body will either move in an orbit around the planet or fall back to the planet .
  • If the body is projected from a planet with a speed greater than the planet’s escape velocity , then that body will escape from the gravitational feild of the planet and move with a velocity v ‘  =  √ v² – ve² , where v’ is the new velocity of the body , v is the initial velocity of the projected body and ve is the escape velocity of the planet .
  • The escape velocity of a body does not depend on the body’s mass or it’s angle of projection .
  • The escape velocity of a body depends on the mass and radius of the planet from which it is projected.


a telecommunication satellite

Orbital   Speed    –

It’s the minimum speed needed to place a satellite in orbit around the earth. Each orbit requires a different minimum speed which is independent of the mass of the satellite .

Considering a satellite in orbit , the centripetal force required to keep the satellite in it’s orbit is provided by the earth’s gravitational pull -

Hence ,  mv²/r   =  GMm/r²      =   v  = √ GM/r ,  now GM = gR² where g – acceleration due to gravity. Therefore ,

v  = √gR²/r        =  R √  ( g / R+h ) . Now , if the satellite is orbiting very close to the earth’s surface such that h << R , then -

vc =  R √  g / R    =    √  g R   =   7.92  km/s .

Points  to  note   -

  • Orbital speed of a satellite depends on the mass and radius of the planet / earth.
  • Orbital speed is directed along the tangent to the orbital path of the satellite at a particular instant .
  • Orbital speed is independent of the mass of the satellite .
  • Orbital speed increases when the orbital radius or the height of the satellite from the earth’s surface decreases.

Suppose vo is the velocity required for a particular orbit , vi is the velocity of projection of a satellite and ve is the escape velocity of earth , then  -

  1. If  vi = ve , the satellite escapes from the earth’s gravitational feild and moves in a parabolic path.
  2. If  vo=  ve , the satellite gets ‘ parked ‘ in the expected orbit and orbits the earth with the earth as the centre of its orbital path.
  3. If  vi <  vo ,the satellite may end up in an elliptical orbit and if the projection is from near the earth’s surface , it will fall back .
  4. If  vi > ve , the satellite escapes from the earth’s gravitational feild and moves in a hyperbolic path.


Defined as the time taken by a satellite to complete one revolution around the earth / planet .

T =  distance covered in one revolution / velocity of the orbit  =  2 π r / v   =  ( 2 π r/R ) ( √ r/g )  =

=   2 π r/R  [ √  ( R + h )²/g ]    -   Eq.1      . Considering earth as a sphere of mean density  ρ ,  -

M = 4/3 π R³ρ   and  g  =    G/ R² ( 4/3 π R³ρ )  . Now , substituting this value in Eq.1 for g  -

T  =  2 π R  ( √ 3 ( R + h )³ / [ 4 π G R ρ ] )  = solving we get =   ( √ 3 π ( R + h )³ / [ G R³ ρ ]  )   – Eq. 2 .

Now , if the satellite is orbiting close to the surface of the earth ,   h + R = R ,

T  =  √ 3  π /G ρ    [ from eq.1 ]   and from eq.2   T =  2 π/ R  / ( √  R³/g )   =   2 π √ R/g .

For earth , this value comes to around  84.6 min .


T  =    2 π r/R  [ √  ( R + h )²/g ]    -   Eq.1  on squaring gives   T²  =  4 π² ( R + h )³/R² g

=>   R + h  =    [ T²  R² g / 4 π² ] 1/3]    =   h  =   [ T²  R² g / 4 π² ]1/3] – R


Consider a satellite of mass m and linear velocity v orbiting  the earth . Then , it’s angular momentum is given by –

L = vr  = mr √ GM/r    = ( squaring and cancelling ) = L  =   √  [ m² / G M r ] .


An orbiting satellite has two forms of energy -

  1. Potential energy due to it’s altitude above the surface of earth and is caused by earth’s gravitational pull on the satellite .
  2. Kinetic energy due to the satellite’s velocity .

Consider a satellite of mass m and velocity v orbiting in an orbital path of radius r , around the earth of mass M and radius R .

Potential energy of satellite ,  U  =  -  G M m /r

Kinetic energy of satellite , K = 1/2  m v²  =  1/2 m [  G M / r ]   . Now , total energy ,  E  = U + K =

=>     E  =  – GMm / 2 r      =  -  G M m / 2 ( R + h ) . Therefore , if the satellite is orbiting near to the earth’s surface , R ≈ r  and hence ,

Total energy , E = – G M m / 2 R

Now , the above equation is for an ideal condition in a circular orbit . If the orbit was elliptical , eventhough the total energy will remain the same and will be negative , both the U and K will vary at different points of the orbit .

Points to note  –

  • The total E of an orbiting satellite will always be negative and never be zero . This is because the satellite ought to be at a finite distance from the earth’s surface. Total energy will be positive / zero only if the orbiting body is at an infinite distance from the earth . But , in such a case the body won’t exist as a satellite and it will escape to infinity .
  • For an orbiting satellite , it’s potential energy is negative and the magnitude of potential energy will be twice the magnitude of it’s positive kinetic energy .


Defined as the energy required to remove a satellite from it’s orbit around the earth to infinity . It is equal to the negative value of the total energy of a satellite in its orbit  -

- E     =   G M m  / 2 r .


Referred to those satellites which revolve around the earth with the same angular speed and direction as is done by the earth around its axis. Conequently such a satellite appears to remain stationary with respect to any point on the surface of the earth .

Naturally , for such a satellite, T = 24 hrs and its speed relative to earth will be zero .

Now  ,   h  =   [ T²  R² g / 4 π² ]1/3] – R     =   36000 km and

orbital speed  =    R √  ( g / R+h )    =   3.1 km / s .

Points   to  note   –

  1. As geostationary satellites remain stationary with respect to points on the surface of the earth , they can be used to transmitt signals of appropriate wavelength between points on the earth’s surface , thereby working as communication devices .
  2. The sense of rotation of the satellite should be the same as that of the earth about its own axis , i.e from west to east .
  3. Geostationary satellites should revolve in an orbit which is coplanar and concentric with the equatorial plane . This is to ensure that the plane of the satellite’s orbit  should be normal to the earths axis of rotation .

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