For obtaining good score in gravitation you should remember the following:

**(1)**The gravitational attractive force

*F*between two point masses

*m*

_{1}and

*m*

_{2}separated by a distance

*r*is given by

**where**

*F*=*G m*_{1}*m*_{2}/*r*^{2}*G*is the

*gravitational constant*(or, constant of gravitation).

If the masses are

*homogeneous spheres*the above equation still holds; but the distance between the masses is to be taken as the distance between the centres of the spheres.
[The gravitational force on a point mass

*m*due to another point mass_{2}*m*_{1}should be strictly written as**since the force is***F*= –*G m*_{1}*m*_{2}/*r*^{2}*attractive*and hence is directed opposite to the vector distance**r**from*m*_{1}to*m*_{2}].
The gravitational field produced by a point mass

*m*at a point P distant*r*from it is the gravitational force acting on unit mass placed at the point P and is equal to**.***Gm*/*r*^{2}**(2)**Since the weight of a body of mass

*m*is the gravitational force with which the earth of mass

*M*pulls the body, we have

**where**

*mg’*=*GMm*/*r*^{2}*r*is the distance of the body from the

*centre*of the earth and

*g’*is the acceleration due to gravity at the distance

*r*.

Therefore, the acceleration due to gravity (

*g’*) at a distance*r*from the centre of the earth is given by

*g’*=*GM*/*r*^{2}
If the body is on the surface of the earth of radius

*R*, we have the surface value of acceleration due to gravity (*g*) given by

*g*=*GM*/*R*^{2}
Note that the surface value of acceleration due to gravity depends to a small extent on the latitude

*λ*because of the spin motion of the earth and is given by

*g*

_{λ}**=**

*g*–

*ω*

^{2}*R*cos^{2}

*λ*

[

*g**is the surface value of acceleration due to gravity at latitude*_{λ}*λ*,*g*is the gravitational acceleration at the poles where the effect of the spin of the earth is absent,*ω*is the spin angular velocity of the earth and*R*is the radius of the earth].
Since the value of

*λ*is zero at the equator and 90º at the poles, the surface value of acceleration due to gravity is minimum at the equator and maximum at the poles.**(3)**Acceleration due to gravity (

*g’*) at a height ‘

*h*’ is given by

**since**

*g*’ =*GM*/(*R+h*)^{2}*r = R+h*.

If ‘

*h*’ is small compared to the radius ‘*R*’ of the earth,*g’*=*g*(1–2*h*/*R*)**(4)**Acceleration due to gravity (g

*’’*) at a depth ‘

*d*’ is given by

*g’’*=*g*(1–*d*/*R*)**Note that this is true for all values of ‘**

*d*’.

**(5)**

**(**

*Gravitational potential energy**U*) of a mass ‘

*m*’ at a height ‘

*h*’ is given by

*U*= –*GMm*/(*R*+*h*)**This can be written as**

**where ‘**

*U*= –*GMm*/*r**r*’ is the distance from the

*centre*of the earth.

The gravitational potential energy of a mass

*m*_{2}in the gravitational field produced by a mass*m*_{1}is given by*U*= –

*Gm*

_{1}

*m*

_{2}/

*r*where

*r*is the distance between centres of the masses.

You can as well say that the above expression is the gravitational potential energy of a mass

*m*_{1}in the gravitational field produced by a mass*m*_{2}or it is the gravitational potential energy of the pair consisting of the masses*m*_{1}and*m*_{2}.
Three masses can make

*three*independent pairs and there will be*three*terms in the expression for potential energy. Four masses will produce*six*independent*pairs*and hence there will be*six*terms in the expression for potential energy and so on.**(5)**

**(**

*Escape velocity***ve**) from the surface of earth (or any planet or star) is given by

**ve**

**=**

**√(2**

*GM*/*R*) = √(2*gR*)**from a height ‘**

*Escape velocity**h*’ =

**√[2**

*GM*/(*R*+*h*)] = √[2*g’*(*R*+*h*] = √(2*g’r*)

Here

*g’*is the acceleration due to gravity at distance*r*(=*R+h)*from the centre of the earth.**(6)**

**. But K.E. is**

*Kinetic energy (K.E.) and total energy of a satellite are equal in magnitude**positive*where as total energy is

*negative*.

**. (Note that this is true in all cases of central field motion under inverse square law force, as for example, in the case of the electron in the hydrogen atom).**

*The potential energy of a satellite is negative and is equal to twice the total energy*
In the case of a satellite of mass ‘

*m*’ in an orbit of radius ‘*r*’:**Potential energy = –**

*GMm*/*r***Kinetic energy = +**

*GMm*/2*r***Total energy = –**

*GMm*/2*r***(7)**The square of the orbital period of a satellite is directly proportional to the cube of the mean distance from the centre of the earth. (This is in accordance with Kepler’s 3

^{rd}law which states that the square of the orbital period of any planet about the sun is directly proportional to the cube of the mean distance from the centre of the sun).

Therefore,

*T*^{2}α*r*^{3}
In the case of

*circular*orbits*r*is the*radius*of the circle.
In the case of

*elliptical*orbits*r*is the*semi-major axis*of the ellipse.**(8)**Orbital speed ‘v’ of a satellite in an orbit of radius ‘

*r*’ is obtained by equating the centripetal force required for the motion to the gravitational pull. Thus mv

^{2}/

*r*=

*GMm*/

*r*

^{2}

*so that*

**v =**

**√(**where

*GM*/*r*) = √(*g’r*)*g’*is

**the acceleration due to gravity at the orbit and**

*M*is the mass of the earth (or planet).

**(9)**The expression for the orbital period of a satellite is

*T =*2π**√(**

*r*^{3}/*GM*)
[The above expression is obtained from the equation,

*T =*2π*r*/v].**(10)**Since the escape velocity (v

_{e}) and the orbital velocity (v) are given respectively by

**√(2**and

*g’r*)**√(**, we have v

*g’r*)_{e}= (√2) v

Therefore, if the speed of a satellite orbiting the earth is increased to √2 times its normal orbital speed, it will escape into outer space. In other words, if the speed of a satellite is increased by 41.4%, it will escape into outer space.

In the next post we will discuss questions on gravitation. Meanwhile, find some useful multiple choice questions at physicsplus.

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