Variation of Acceleration Due to Gravity

Inertial and Gravitational Mass

Mass of a body is the quantity of matter contained in it. The effect of a force applied on a body depends on its mass. It has appeared in two entirely different contexts: Newton’s second law of motion and Newton’s law of gravitation. So meaning of mass is clear if we define mass as (i) inertial mass and (ii) gravitational mass.

Inertial Mass

The inertial mass of a body is the measure of the ability of the body to oppose the change in the velocity i.e. acceleration produced by an external force. It is a measure of the body’s resistance to being accelerated.

As we know,

$$F=ma \: \text {and is}\: a= 1 \text {then}\: F = m $$

Gravitational Mass

Gravitational mass is defined by the Newton’s law of gravitation. The gravitational pull of the earth of mass M and radius R on a body of mass m is given as

\begin{align*} F &= \frac {GMm}{R^2} \\ \text {or,} m &= \frac {F\times R^2}{GM} \\ \end{align*}

The mass m of the body, in this case, is the gravitational mass of the body. This relation shows that the gravitational force increases if the mass of the object increases i.e. the measure of gravitational force on the body is called gravitational mass.

Difference between Inertial Mass and Gravitational Mass

Weight

The weight of an object is defined as the gravitational force with which the object is attracted by the earth towards its centre. If m be the mass of the object and g be the acceleration due to gravity, then the weight of the object at a certain place is given by

$$ W = mg $$

It is a vector quantity being a force.

Difference between Mass and Weight

Variation of Acceleration Due to Gravity

Acceleration due to gravity varies from place to place. These are discussed below.

1. Shape of the earth

   The earth is not perfectly spherical. It is flattered at the pole and bulged at the equator. So the equatorial radius R_o is more than the polar radius R_p.
   \begin{align*} \text {As} \: g &= \frac {GM}{R^2} \\ \text {So,} \: g &\propto \frac {1}{R^2} \\ \text {As} \: R_e >R_p \\ \therefore g_p >g_e \\ \end{align*}
   So, the acceleration due to gravity is maximum at the pole and minimum at the equator.
   
   

2. Due to height

   Let an object of mass m is at point P at the height h from the earth’s surface as shown in the figure. Let M be mass of the earth and R its radius. Acceleration due to the gravity, g at the earth’s surface is
   

   \begin{align*} g &= \frac {GM}{R^2}\\ \end{align*} The acceleration due to gravity g’ at P is \begin{align*}g’ &= \frac {GM}{(R+h)^2} \dots (iii) \\ \end{align*} Combining equations} (I) {and} (ii)we get\begin{align*}\frac {g’}{g} &= \frac {GM}{(R+h)^2} \times \frac {R^2}{GM} = \frac {R^2}{(R+h)^2} \\ g’ &= g\times \frac {R^2}{(R+h)^2} \dots (iii)\\ \end{align*}
   Since (R+h)>R, we have g’
   \begin{align*} g’ &= \frac {g}{\left (\frac {R+h}{R} \right )} = g \left ( 1+ \frac hR \right )^{-1} \\\end{align*}

   Expanding ( 1+ \(\frac {h}{R })^{-1} \) by binomial theorem and neglecting the higher power of h/R which has small value, we get \begin{align*} \\ g’ &= g\left ( 1-\frac {2h}{r} \right ) \dots (iv) \\ \end{align*}

   
   

3. Variation of g with depth

   Let us consider the earth to be a perfect sphere of radius R of uniform density ρ.
   \begin{align*} \text {Mass of the earth M} &= \frac 43 \pi R^2 \rho . \\ \text {Acceleration due to gravity on earth’s surface} \\ g &= \frac {GM}{R^2} = \frac {G\frac 43 \pi R^3 \rho }{R^3} \\ \text {or,} \: g &= \frac 43 G\pi R\rho \dots (v) \\ \end{align*}
   If we go to a point at a depth of x from the earth’s surface, as shown in the figure, then acceleration due to gravity at point P is due to the sphere of radius (R-x).
   \begin{align*} \text {Mass of this sphere of radius} (R-x) is M’ = \frac 43 \pi (R-x)^3\rho \\ \text {and acceleration due to gravity at P is } \\ g’ &= \frac {GM’}{R^2} = \frac {G\frac 43 \pi (R-x)^3 \rho }{(R-x)^3} \\ \text {or,} \: g’ &= G\frac 43 \pi (R-x) \rho \dots (vi) \\\text {Dividing equations} (I) \text {by} (ii), \text {we get} \\ \frac {g’}{g} &= \frac { G\frac 43 \pi (R-x) \rho}{ G\frac 43 \pi R\times \rho } = \frac {R-x}{R} = \left ( 1- \frac xR \right ) \\ \therefore g’ &= g\left ( 1- \frac xR \right ) \dots (vii) \\ \text {As} \: \left ( 1- \frac xR \right ) \\ \end{align*}

4. Variation of g due to rotation of the earth or effect of latitude.

   

   Let us consider an object of mass m at a point P on the earth surface at latitude of Ф. When the earth is not rotating its weight is , W= mg acts along the radius of the earth towards the centre.

   When the earth rotates with angular velocity () on its axis, the object at P will also rotate about the centre C of the radius of the earth. So, the object will experience the centrifugal force, and the object is under the action of two forces centrifugal force and its weight.

   The resultant of two forces is shown by PB in the figure which is apparent weight of mg of the object. By using parallelogram law of vector addition,
   \begin{align*} PB^2 &= PO^2 + PA^2 + 2PO \times PA \cos (180 - \phi ) \\ \text {or,} \: m^2g’^{2} &= (mg)^2 + (mR\omega ^2 \cos \phi )^2 + 2\times mg \times mR\omega ^2 \cos \phi \times (-\cos \phi ) \\ \text {or,} \: g’ &= g \left [ 1 + \frac {R^2 \omega ^4 \cos ^2 \phi }{g^2} -\frac {2R\omega ^2 \cos ^2 \phi }{g} \right ]^{1/2}\\ \text {Since} \frac {R\omega ^2}{g} \\ \end{align*} is a small quantity, the terms containing the factor\begin{align*} \frac {R^2\omega ^4}{g^2} \text {can be neglected.} \\ \text {Hence,} \: g’ &= g\left [ 1 -\frac {2R\omega ^2 \cos ^2 \phi }{g} \right ]^{1/2}\\ \end{align*}

   Expanding the above equation by using binomial theorem\begin{align*} g’ &= g \left [ 1 -\frac 12 \frac {2R\omega ^2 \cos ^2 \phi }{g} + \dots \right ]\\ \end{align*} Higher powers of \begin{align*}\: \frac {2R\omega ^2 \cos ^2 \phi }{g}\\ \end{align*} can be neglected and\begin{align*} \\ g’ &= g \left [ 1 -\frac {R\omega ^2 \cos ^2 \phi }{g} \right ]\\ \text {or,} \: g’ &= g- R\omega ^2 \cos ^2 \phi \\ \end{align*} This equation shows that due to the rotation of the earth, the acceleration due to gravity decreases. It also shows that the acceleration due to gravity increases as the latitude increases \begin{align*} {At the equator,} \phi = 0, so \cos 0 = 1 \\ \therefore g_e &= g – R\omega ^2 \\ \text {At the poles,} \phi = 90^o , \text {and} \cos 90^o = 0 \\ \therefore g_p &= g \\ \end{align*} There is no effect on the acceleration due to gravity at the poles due to the rotation of the earth