Here G is universal gravitational Constant

G = 6.67 x 10-11 Nmt2/kg2

Dimensional formula = [M-1 L3 T-2]

Note: This formula is applicable for point mass only.


(1) It is always only attraction type force never repel.

(2) It is independent of medium between particles

(3) It is mutual force

(4) It is applicable for very far distant objects like interplanetary distance as well as for very short distance like inter-atomic distance

(5) It is conservative force means work done by this force is independent of path.

Example: Two particles of equal masses move in a Circle of radius r under the action of mutual gravitational attraction force. Then the speed of each particle if the mass of each particle is m .




Here gravitational force F is centripetal force provider

       Answer (c) 


Space of influence surrounding a mass m in which its gravitational effects are effective is called gravitational field of given mass.

Intensity of gravitational field due to point mass


(1) Shape of earth is perfect sphere

(2) Center of mass of earth is at its center

(3) Uniform mass distribution




Intensity of earth gravitational field at its surface

Now here Eg = g we will use another term specially for earth is acceleration due to gravity Now At earth’s Surface    Direction of Eg or g is always towards the Center of earth Also weight of any object at the surface of earth W = F = m Eg = mg Force acting towards the center of earth Properties of ‘g’

(1) It is independent of mass of any object

(2) It is not a universal constant its value depends on place, position and planet.

NOTE: For example value of g on the surface of moon is approximately g/6 i.e. 1/6 of the value on the earth

Variation of ‘g’

(1) Due to change of planet

As we know at the surface of earth





(2) Due to shape of earth

The earth is not perfectly spherical in shape but is on oblate sphere. It is bulged at the equator and flattened at the pole.

Equatorial radius is 21 km more than polar radius


g ® Minimum at equator

g ® Maximum at poles   

At the surface

gpoles > gequator



R2 = P2 + Q2 + 2 PQ cos a

(mg)2 = (mg)2 + (mrw2)2 + 2 mg mrw2 cos (p - ϕ)






At the equator f= 0°

At the poles g= g - Rw2

f = 90°, g1 = g – Rw2 cos2 90° g = g

4. At the depth h inside the earth

At the surface of earth


At depth h








At the Center of the earth


 g’ = 0 weight less

5.  At h height above the earth’s surface

At the surface of earth


At the h height above






If  h << R

Binomial expansion

(1+ x)n = 1+ nx

If x <<<<<<1

Field due to a Uniform Circular Ring at some Point on its Axis

Field strength at a point P on the axis of a circular ring of radius R and mass M is given by,


This is directed towards the center of the ring. It is zero at the center of the ring and maximum at   (can be obtained by putting ).  Thus, E-r graph is as shown in the maximum value is




(1) Law of orbit: Each planet moves around the sun in an

elliptical orbit with the sun at one of the foci.

Here e is eccentricity of an ellipse





The line joining the sun and planet sweeps out equal area in equal interval of time.


Here L = angular momentum of planet about the sun


Square of time period µ cube of semi major axis of elliptical orbit.


Final Touch Points

Polar and Geostationary Satellites

1. Satellites in low polar orbit pass over the poles. They orbit between 100 km and 200 km above the Earth’s surface, taking around 90 minutes to make each orbit. The earth spins beneath the satellites as it moves, so the satellite can scan the whole surface of the earth. Low orbit polar satellites have uses such as

  • Monitoring the weather.
  • Observing the earth’s surface.
  • Military uses including spying.

2. Geostationary satellites have a different trajectory to polar satellites. They are in orbit above the equator from west to east. The height of their orbit-36,000 km is just the right distance so that it takes them one day (24 hours) to make each orbit. This means that they stay in a fixed position over the earth’s surface. A single geostationary satellite is on a line sight with about 40 percent of the earth’s surface. Three such satellites, each separated by 120 degrees of longitude, can provide coverage of the entire planet. Geostationary satellites have uses such as:

  • Communications-including satellites phones
  • Global positioning or GPS,

Geostationary satellites always appear in the same position when seen from the ground. This is why satellite television dishes can be bolted into one position and do not need to move.

3. Acceleration due to moon’s gravity on moon’s surface is


while acceleration due to earth’s gravity on moon’s surface is approximately  .  This because distance of moon from the earth’s centre is approximately equal to 60 times the radius of earth and

4. Total energy of a closed system is always negative. For example, energy of planet-sun, satellite-earth or electron-nucleus system is always negative..

5. If the law of force obeys the inverse square law


The same is true for electron-nucleus system because there also, the electrostatic force Fe1r2.

6. Trajectory of a body projected from point A in the direction AB with different initial velocities:

Let a body projected from point A with velocity v in the direction AB. For different values of v the paths are different. Here, are the possible cases.

2. If v is not very large the elliptical orbit will intersect the earth and the body will fall back to earth.






(i) If times period of rotation of earth becomes 84.6 min, effective value of g on equator becomes zero or we feel weightlessness on equator.

(ii) Time period of a satellite close to earth’s surface is 84.6 min.

(iii) Time pendulum of a pendulum of infinite length is 84.6 min.

(iv) If a tunnel is dug along any chord of the earth and a particle is released from the surface of earth along this tunnel, them motion of this particle is simple harmonic and time period of this is also 84.6 min.

Note (a) Points (iii) and (iv) come in the chapter of simple harmonic motion.

(b)  is also the time period of small oscillations of a block inside a smooth spherical bowl of radius R.

But this is not 84.6 min because here R is the radius of bowl not the radius of earth.

This expression can be compared with the time period of a pendulum.

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