Gravitation is an easy chapter of Mechanics in the Physics syllabus of both JEE Main and JEE Advanced. These laws are not new to students as they have read about the story of an apple falling on the head of Sir Issac Newton. The chapter is important, not only because few questions are asked from this chapter, but also because if you understand its concepts then another chapter, named Electrostatics, will be very easy for you as they both have similar concepts.

## Gravitation JEE: Newton’s law of Gravitation

The first part of this chapter is Newton’s law of gravitation. I don’t think I have to explain much about this topic to you. If you know the formulae, you can solve any question related to it. Just be cautious in some cases, like finding gravitational force between two large objects, for example, between planet Earth and the Sun. You have to measure distance from their respective centers as d (distance) will be radius of Earth + radius of Sun + distance between their surfaces. Rest all is easy.

## Gravitation JEE: Gravitational field and intensity

This topic is again simple. You must memorize and derive the formulas of gravitational field of various bodies like uniform sphere, shell etc. The space around a body where the gravitational force exerted by it can be experienced by any other particle is known as the “**gravitational field**” of the body. The strength of this gravitational field is referred to as “**intensity**”, and it varies from point to point. You should note that it is a vector quantity. So while performing addition or subtraction, follow vector rules for the same.

The **gravitational field** at a point P due to a particle of mass m kept at the point O (origin) is given by

= _{g}/m_{o} = Gm/r^{2}

Where = xi + yj + zk represents the position vector of the point P with respect to the source at the origin.

= _{1} + _{2} + _{3} +….+ _{n}

For a continuously distributed mass, the formula changes to =∫d, where d gravitational field intensity due to an elementary mass dm.

Learn about the effect of gravitation field when a body is at some height above the earth surface or below the earth surface. Questions are asked sometimes which use direct formulas from this part.

For example-

If the acceleration due to gravity at the given height is g’, then

=> mg’=GMm/R^{2}(1+h/R)^{2} ….. (1)

=> g’=(GM/R^{2})*(1+h/R)^{-2}

=> g’= g(1-2h/R) {h<<R} ……(2)

Please note that here eq.2 can only be used when h<<R. If h is comparable to R then you have to use eq.1.

In the same way, you can derive the value of **g** for height **h** below the earth’s surface.

## Gravitation JEE: Orbital Velocity

When a satellite revolves in an orbit around a planet, it requires centripetal force to do so. This centripetal force is provided by the gravitational force between the planet and the satellite. Using this concept, you should try to get the formula of orbital velocity by yourself, which is v= (gR)^{1/2}

## Gravitation JEE: Gravitation potential energy

“**Potential energy**” of a system of two masses is defined as the amount of work done in bringing these two masses from infinity to their respective places. Change in gravitational potential energy of a system is defined as the negative of the work done by the gravitational force as the configuration of the system is changed.

Remember and derive the formulas of gravitational potential energy for different bodies. Know about the relation between Gravitational Potential and the Gravitational Field Strength, i.e. E = -dV/dr

Derive the formula of escape velocity by yourself and solve a number of problems for better understanding of various concepts.

Gravitational Potential is a scalar quantity. So, potential at a point due to number of bodies can be found using arithmetic sum of potential by each body at the point.

V = -G(m_{1}/r_{1} + m_{2}/r_{2} + m_{3}/r_{3} +……….)

“**Escape velocity**” on the surface of earth is the minimum velocity given to a body to make it free from the gravitational field, i.e. it can reach an infinite distance from the earth.

Let v_{e} be the escape velocity of the body on the surface of earth.

1/2 mv_{e}^{2} – GMm/R = 0 => v_{e} = √2GM/R.

You should must understand when to use the concept of potential and when to use the gravitational field intensity concepts to solve problems. This will only come when you have a strong grasp on the fundamentals.

## Gravitation JEE: Satellites and Planetary motion

This is the last topic in Gravitation. First come **Kepler’s laws**; you should know about the three laws, especially First and Third law (T^{2} is proportional to R^{3}). Solve questions based on these laws. Then, learn to find out time period and orbital velocity of a satellite revolving at a certain height above the planet.

Know about geostationary satellites and find escape velocities in these cases. Also, try to figure out what will happen when speed of satellite is less than, equal to, or greater than the escape velocity, and the trajectory it follows in those cases. For example, if the speed is greater than the escape velocity then the satellite escapes the gravitational field of Earth along a hyperbolic trajectory.

If v is the velocity given to a satellite and v_{0} represents the velocity of a circular orbit and v_{e} the escape velocity,

i.e. v_{0} = √GM_{e}/(R_{e}+h)

then v_{e} = √2GM_{e}/(R_{e}+h)

## Gravitation JEE: Solved Example

There is a good question which I would like to keep here as it may confuse you in multiple choice questions and is very good for understanding too.

### Question

If we see a satellite from Earth, how long will it take for one revolution?

### Answer

Let us consider a satellite in circular orbit with a time period T_{s}. The Earth also rotates with the time period T_{e} = 24 hrs. If an observer on Earth sees this satellite, the angular velocity of the satellite will be . Hence, the time period of revolution will seem different from T_{s} and will be observed as T_{SE}.

Two cases arise for calculation of T_{SE}:-

- If the satellite and Earth are revolving and rotating respectively in the same direction.

Hence, 2Π/T_{SE}= |2Π/T_{s}– 2Π/T_{e}|

=> T_{SE}= T_{s}T_{e}/|T_{e}– T_{s}| - If the satellite and Earth are revolving and rotating respectively in the opposite direction.

Hence, 2Π/T_{SE}= 2Π/T_{s}+ 2Π/T_{e}

or T_{SE}= T_{s}T_{e}/|T_{s}+ T_{e}|.

This is the summary of the chapter gravitation. All you have to do is to practice questions related to above topics so that you become comfortable with the concepts. Then you will be able to solve its questions in JEE. Best of luck.

but gravitational field is not given in jee mains syllabus released on official site