Gravitation

Introduction

Newton at the age of twenty-three is said to have seen an apple falling down from tree in his orchid. This was the year 1665. He started thinking about the role of earth's attraction in the motion of moon and other heavenly bodies.

By comparing the acceleration due to gravity due to earth with the acceleration required to keep the moon in its orbit around the earth, he was able to arrive the Basic Law of Gravitation.          

Newton's law of Gravitation

Newton's law of gravitation states that every body in this universe attracts every other body with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. The direction of the force is along the line joining the particles.

Thus the magnitude of the gravitational force F that two particles of masses \({{m}_{1}}\) and \({{m}_{2}}\) are separated by a distance r exert on each other is given by \(F\propto \frac{{{m}_{1}}\,{{m}_{2}}}{{{r}^{2}}}\)

or  \(F=G\frac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\)

Vector form : According to Newton's law of gravitation

\({{\overset{\to }{\mathop{F}}\,}_{12}}=\frac{-G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}{{\hat{r}}_{21}}\)  \(=\frac{-\,G{{m}_{1}}{{m}_{2}}}{{{r}^{3}}}{{\overset{\to }{\mathop{r}}\,}_{21}}=\frac{-G{{m}_{1}}{{m}_{2}}}{|{{\overset{\to }{\mathop{r}}\,}_{21}}{{|}^{3}}}{{\overset{\to }{\mathop{r}}\,}_{21}}\)

Here negative sign indicates that the direction of \({{\overset{\to }{\mathop{F}}\,}_{12}}\) is opposite to that of \({{\hat{r}}_{21}}\).

Similarly \({{\overset{\to }{\mathop{F}}\,}_{21}}=\frac{-G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}{{\hat{r}}_{12}}\)\(=\frac{-\,G{{m}_{1}}{{m}_{2}}}{{{r}^{3}}}{{\overset{\to }{\mathop{r}}\,}_{12}}=\frac{-\,G{{m}_{1}}{{m}_{2}}}{|{{\overset{\to }{\mathop{r}}\,}_{12}}{{|}^{3}}}{{\overset{\to }{\mathop{r}}\,}_{12}}\)

\(=\frac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}{{\hat{r}}_{21}}\)  [\(\because \,{{\hat{r}}_{12}}=-{{\hat{r}}_{21}}]\)

\(\therefore\) It is clear that \({{\overset{\to }{\mathop{F}}\,}_{12}}\)= – \({{\overset{\to }{\mathop{F}}\,}_{21}}\). Which is Newton's third law of motion.

Here G is constant of proportionality which is called 'Universal gravitational constant'.

If \({{m}_{1}}={{m}_{2}}\,\, and \,\,r=1\,\, then \,\,G=F\)

i.e.  universal gravitational constant is equal to the force of attraction between two bodies each of unit mass whose centres are placed unit distance apart.

(i) The value of G in the laboratory was first determined by Cavendish using the torsional balance.

(ii)The value of G is 6.67\(\times\)10^–11 N–m² kg^–2 in S.I. and 6.67\(\times\)10^–8 dyne- cm²-g^–22 – t–––– in C.G.S. system.

(iii) Dimensional formula \([{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]\).

(iv) The value of G does not depend upon the nature and size of the bodies.

(v) It also does not depend upon the nature of the medium between the two bodies.

(vi) As G is very small, hence gravitational forces are very small, unless one (or both) of the mass is huge.

Properties of Gravitational Force

(1) It is always attractive in nature while electric and magnetic force can be attractive or repulsive.

(2) It is independent of the medium between the particles while electric and magnetic force depend on the nature of the medium between the particles.

(3) It holds good over a wide range of distances. It is found true for interplanetary to inter atomic distances.

(4)It is a central force i.e. acts along the line joining the centres of two interacting bodies.

(5) It is a two-body interaction i.e. gravitational force between two particles is independent of the presence or absence of other particles; so the principle of superposition is valid i.e. force on a particle due to number of particles is the resultant of forces due to individual particles i.e.\(\overset{\to }{\mathop{F}}\,={{\overset{\to }{\mathop{F}}\,}_{1}}+{{\overset{\to }{\mathop{F}}\,}_{2}}+{{\overset{\to }{\mathop{F}}\,}_{3}}+........\)

While nuclear force is many body interaction

(6) It is the weakest force in nature : As F_nuclear > F _{electromagnetic} > F _{gravitational} .

(7) The ratio of gravitational force to electrostatic force between two electrons is of the order of \({{10}^{-43}}\).

(8) It is a conservative force i.e. work done by it is path independent or work done in moving a particle round a closed path under the action of gravitational force is zero.

(9) It is an action reaction pair i.e. the force with which one body (say earth) attracts the second body (say moon) is equal to the force with which moon attracts the earth. This is in accordance with Newton's third law of motion.

Note :The law of gravitation is stated for two point masses, therefore for any two arbitrary finite size bodies, as shown in the figure, It can not be applied as there is not unique value for the separation.

But if the two bodies are uniform spheres then the separation r may be taken as the distance between their centres because a sphere of uniform mass behave as a point mass for any point lying outside it.

Acceleration Due to Gravity

The force of attraction exerted by the earth on a body is called gravitational pull or gravity.

We know that when force acts on a body, it produces acceleration. Therefore, a body under the effect of gravitational pull must accelerate.

The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity, it is denoted by g.

Consider a body of mass m is lying on the surface of earth then gravitational force on the body is given by

\(F=\frac{GMm}{{{R}^{2}}}\)                                            …(i)

Where M = mass of the earth and R = radius of the earth.

If g is the acceleration due to gravity, then the force on the body due to earth is given by

Force = mass \(\times\) acceleration

or       F = mg                                        …(ii)

From (i) and (ii) we have mg \(=\frac{GMm}{{{R}^{2}}}\)

\(\therefore g=\frac{GM}{{{R}^{2}}}\)                                             …(iii)

\(\Rightarrow g=\frac{G}{{{R}^{2}}}\left( \frac{4}{3}\pi {{R}^{3}}\rho  \right)\)

[\(As mass (M) = volume (\frac{4}{3}\pi {{R}^{3}}) \times density (\rho)\)]

\(\therefore g=\frac{4}{3}\pi \rho GR        \)                                           …(iv)

(i)From the expression \(g=\frac{GM}{{{R}^{2}}}=\frac{4}{3}\pi \rho GR\) it is clear that its value depends upon the mass radius and density of planet and it is independent of mass, shape and density of the body placed on the surface of the planet. i.e. a given planet (reference body) produces same acceleration in a light as well as heavy body.

(ii)The greater the value of \((M/{{R}^{2}})\) or \(\rho R,\) greater will be value of g for that planet.

(iii) Acceleration due to gravity is a vector quantity and its direction is always towards the centre of the planet.

(iv) Dimension [g] = [LT^–2]

(v) it’s average value is taken to be 9.8 m/s² or 981 cm/sec² or 32 feet/sec², on the surface of the earth at mean sea level.

(vi) The value of acceleration due to gravity vary due to the following factors :

(a) Shape of the earth,

(b) Height above the earth surface,

(c) Depth below the earth surface and

(d) Axial rotation of the earth.

Variation in g Due to Shape of Earth

Earth is elliptical in shape. It is flattened at the poles and bulged out at the equator. The equatorial radius is about 21 km longer than polar radius, from \(g=\frac{GM}{{{R}^{2}}}\)

At equator  \({{g}_{e}}=\frac{GM}{R_{e}^{2}}\)            …(i)    

At poles  \({{g}_{p}}=\frac{GM}{R_{p}^{2}}\)               …(ii)

From (i) and (ii) \(\frac{{{g}_{e}}}{{{g}_{p}}}=\frac{R_{p}^{2}}{R_{e}^{2}}\)

Since \({{R}_{equator}}>{{R}_{pole}}\)  

\(\therefore \)\({{g}_{pole}}>{{g}_{equator}}\) and \({{g}_{p}}={{g}_{e}}+0.018\,\,m{{s}^{-2}}\)

Therefore the weight of body increases as it is taken from equator to the pole.

Variation in g With Height

Acceleration due to gravity at the surface of the earth

\(g=\frac{GM}{{{R}^{2}}}3\)                                       …(i)

Acceleration due to gravity at height h from the surface of the earth   

\(g'=\frac{GM}{{{(R+h)}^{2}}}\)                                   …(ii)

From (i) and (ii) \(g'=g{{\left( \frac{R}{R+h} \right)}^{2}}\)    …(iii)

=\(g\frac{{{R}^{2}}}{{{r}^{2}}}\)                                            …(iv)                      

[As r = R + h]

(i) As we go above the surface of the earth, the value of g decreases because \({g}'\propto \frac{1}{{{r}^{2}}}\).

(ii) If \(r=\infty\) then \({g}'=0\), i.e., at infinite distance from the earth, the value of g becomes zero.

(iii) If \(h< i.e., height is negligible in comparison to the radius then from equation (iii) we get

 \({g}'=g{{\left( \frac{R}{R+h} \right)}^{2}}=g{{\left( 1+\frac{h}{R} \right)}^{-2}}=g\left[ 1-\frac{2h}{R} \right]\)                                [As \(h<]

(iv)If \(h< then decrease in the value of g with height :

Absolute decrease \(\Delta g=g-{g}'=\frac{2hg}{R}\)

Fractional decrease \(\frac{\Delta g}{g}=\frac{g-{g}'}{g}=\frac{2h}{R}\)

Percentage decrease \(\frac{\Delta g}{g}\times 100\%=\frac{2h}{R}\times 100\%\)

Variation in g With Depth

Acceleration due to gravity at the surface of the earth

\(g=\frac{GM}{{{R}^{2}}}=\frac{4}{3}\pi \rho GR\)              …(i)

Acceleration due to gravity at depth d from the surface of the earth

\({g}'=\frac{4}{3}\pi \rho G(R-d)\)               …(ii)

From (i) and (ii)  \({g}'=g\left[ 1-\frac{d}{R} \right]\)

(i) The value of g decreases on going below the surface of the earth. From equation (ii) we get \({g}'\propto (R-d)\).

So it is clear that if d increase, the value of g decreases.

(ii) At the centre of earth \(\frac{g\,{{R}^{2}}{{T}^{2}}}{4{{\pi }^{2}}}={{\left( R+h \right)}^{3}} \therefore \ \ {g}'=0\), i.e., the acceleration due to gravity at the centre of earth becomes zero.

(iii) Decrease in the value of g with depth

Absolute decrease \(\Delta g=g-{g}'=\frac{dg}{R}\)

Fractional decrease \(\frac{\Delta g}{g}=\frac{g-{g}'}{g}=\frac{d}{R}\)

Percentage decrease\(\frac{\Delta g}{g}\times 100\%=\frac{d}{R}\times 100\%\)

(iv)The rate of decrease of gravity outside the earth (\(\text{if}\,\ h<) is double to that of inside the earth.

Variation in g Due to Rotation of Earth

As the earth rotates, a body placed on its surface moves along the circular path and hence experiences centrifugal force, due to it, the apparent weight of the body decreases.

Since the magnitude of centrifugal force varies with the latitude of the place, therefore the apparent weight of the body varies with latitude due to variation in the magnitude of centrifugal force on the body.

If the body of mass m lying at point P, whose latitude is \(\lambda\), then due to rotation of earth its apparent weight can be given by \(\frac{g\,{{R}^{2}}{{T}^{2}}}{4{{\pi }^{2}}}={{\left( R+h \right)}^{3}}\)

or  \(m{g}'=\sqrt{{{(mg)}^{2}}+{{({{F}_{c}})}^{2}}+2mg\ {{F}_{c}}\ \cos (180-\lambda )}\)

\(\Rightarrow m{g}'=\sqrt{{{(mg)}^{2}}+{{(m{{\omega }^{2}}R\cos \lambda )}^{2}}+2mg\ m{{\omega }^{2}}R\cos \lambda \ (-\cos \lambda )}\)

[As \({{F}_{c}}=m{{\omega }^{2}}r=m{{\omega }^{2}}R\,\cos \lambda \)]

By solving we get \({g}'=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda \)

Note: The latitude at a point on the surface of the earth is defined as the angle, which the line joining that point to the centre of earth makes with equatorial plane. It is denoted by \(\lambda\).

For the poles \(\lambda ={{90}^{o}}\) and for equator \(\lambda ={{0}^{o}}\)

(i) Substituting \(\lambda ={{90}^{o}}\) in the above expression we get \(=2\pi \,\,\sqrt{\frac{{{(R+h)}^{3}}}{g{{R}^{2}}}}\)

\(\therefore{{g}_{pole}}=g\)                                  …(i)

i.e., there is no effect of rotational motion of the earth on the value of \(\) at the poles.

(ii) Substituting \(\lambda ={{0}^{o}}\) in the above expression we get \(T=2\pi \,\sqrt{\frac{{{r}^{3}}}{GM}}\)

\(\therefore{{g}_{equator}}=g-{{\omega }^{2}}R\)                        …(ii)

i.e., the effect of rotation of earth on the value of \(g\) at the equator is maximum.

From equation (i) and (ii)

\({{g}_{pole}}-{{g}_{equator}}=R{{\omega }^{2}}=0.034m/{{s}^{2}}\)

(iii) When a body of mass  is moved from the equator to the poles, its weight increases by an amount

\(m({{g}_{p}}-{{g}_{e}})=m{{\omega }^{2}}R\)

(iv) Weightlessness due to rotation of earth : As we know that apparent weight of the body decreases due to rotation of earth. If \(\omega\) is the angular velocity of rotation of earth for which a body at the equator will become weightless

\({g}'=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda \)

\(\Rightarrow0=g-{{\omega }^{2}}R{{\cos }^{2}}{{0}^{o}}\)     [As \(\lambda ={{0}^{o}}\) for equator]

\(\Rightarrow g-{{\omega }^{2}}R=0\)

\(\therefore T=\,\infty \)

or time period of rotation of earth \(T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{R}{g}}\)

Substituting the value of \(R=6400\times {{10}^{3}}m\)  and  \({{\omega }_{S}}={{\omega }_{E}}\) we get

\(\omega =\frac{1}{800}=1.25\times {{10}^{-3}}\frac{rad}{sec}\) and  \(T=5026.5\ sec=1.40\ hr\)

Note: This time is about \(\left[ \text{As}\,\,\,T=\frac{2\pi }{\omega } \right]\) times the present time period of earth. Therefore if earth starts rotating 17 times faster then all objects on equator will become weightless.

If earth stops rotation about its own axis then at the equator the value of \(g\) increases by \({{\omega }^{2}}R\) and consequently the weight of body lying there increases by \(m{{\omega }^{2}}R\).

After considering the effect of rotation and elliptical shape of the earth, acceleration due to gravity at the poles and equator are related as

\({{g}_{p}}={{g}_{e}}+0.034+0.018m/{{s}^{2}}\) \(\therefore \ {{g}_{p}}={{g}_{e}}+0.052m/{{s}^{2}}\)

Mass and Density of Earth

Newton’s law of gravitation can be used to estimate the mass and density of the earth.

As we know \(g=\frac{GM}{{{R}^{2}}}\), so we have \(M=\frac{g{{R}^{2}}}{G}\)

\(\therefore M=\frac{9.8\times {{(6.4\times {{10}^{6}})}^{2}}}{6.67\times {{10}^{-11}}}=5.98\times {{10}^{24}}kg\approx {{10}^{25}}kg\)

and as we know \(g=\frac{4}{3}\pi \rho GR\), so we have \(\rho =\frac{3g}{4\pi GR}\)

\(\therefore\rho =\frac{3\times 9.8}{4\times 3.14\times 6.67\times {{10}^{-11}}\times 6.4\times {{10}^{6}}}=5478.4\ kg/{{m}^{3}}\)

Inertial and Gravitational Masses

(1) Inertial mass : It is the mass of the material of the body, which measures its inertia.

If an external force F acts on a body of mass m_i, then according to Newton’s second law of motion

\(F={{m}_{i}}a\) or \({{m}_{i}}=\frac{F}{a}\)

Hence inertial mass of a body may be measured as the ratio of the magnitude of the external force applied on it to the magnitude of acceleration produced in its motion.

(i) It is the measure of ability of the body to oppose the production of acceleration in its motion by an external force.

(ii) Gravity has no effect on inertial mass of the body.

(iii) It is proportional to the quantity of matter contained in the body.

(iv) It is independent of size, shape and state of body.

(v) It does not depend on the temperature of body.

(vi) It is conserved when two bodies combine physically or chemically.

(vii) When a body moves with velocity \(v\), its inertial mass is given by

\(m=\frac{{{m}_{0}}}{\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}}\), where m₀ = rest mass of body, c = velocity of light in vacuum,

(2) Gravitational Mass : It is the mass of the material of body, which determines the gravitational pull acting upon it.

If M is the mass of the earth and R is the radius, then gravitational pull on a body of mass \({{m}_{g}}\) is given by

\(F=\frac{GM{{m}_{g}}}{{{R}^{2}}}\) or \({{m}_{g}}=\frac{F}{GM/{{R}^{2}}}=\frac{F}{I}\)

Here \({{m}_{g}}\) is the gravitational mass of the body, if \(I=1\) then \({{m}_{g}}=F\)

Thus the gravitational mass of a body is defined as the gravitational pull experienced by the body in a gravitational field of unit intensity,

(3) Comparison between inertial and gravitational mass

(i) Both are measured in the same units.

(ii) Both are scalar.

(iii) Both do not depend on the shape and state of the body

(iv) Inertial mass is measured by applying Newton’s second law of motion where as gravitational mass is measured by applying Newton’s law of gravitation.

(v) Spring balance measure gravitational mass and inertial balance measure inertial mass.

(4) Comparison between mass and weight of the body

Gravitational Field

The space surrounding a material body in which gravitational force of attraction can be experienced is called its gravitational field.

Gravitational field intensity : The intensity of the gravitational field of a material body at any point in its field is defined as the force experienced by a unit mass (test mass) placed at that point, provided the unit mass (test mass) itself does not produce any change in the field of the body.

So if a test mass \(m\) at a point in a gravitational field experiences a force \(\overrightarrow{F}\) then

\(\overrightarrow{I\,}=\frac{\overrightarrow{F}}{m}\)

 (i) It is a vector quantity and is always directed towards the centre of gravity of body whose gravitational field is considered.

(ii) Units : Newton/kg or m/s²

(iii) Dimension : [M⁰LT^–2]

(iv) If the field is produced by a point mass \(T\propto {{r}^{\frac{n+1}{2}}}\) and the test mass \(m\) is at a distance \(r\) from it then by Newton’s law of gravitation \(F=\frac{GMm}{{{r}^{2}}}\), then intensity of gravitational field

\(I=\frac{F}{m}=\frac{GMm/{{r}^{2}}}{m}\)

\(\therefore      F\propto \frac{1}{{{r}^{n}}}\)

(v) As the distance \((r)\) of test mass from the point mass \((M)\), increases, intensity of gravitational field decreases

\(I=\frac{GM}{{{r}^{2}}}\); 

\(\therefore I\propto \frac{1}{{{r}^{2}}}\)

(vi) Intensity of gravitational field \(=\frac{2\pi {{\left( {{R}^{3}} \right)}^{1/2}}}{{{\left[ G.\frac{4}{3}\pi {{R}^{3}}\rho  \right]}^{1/2}}}=\sqrt{\frac{3\pi }{G\rho }}\), when \(r=\infty \).

(vii) Intensity at a given point (P) due to the combined effect of different point masses can be calculated by vector sum of different intensities

\(\overrightarrow{{{I}_{net}}}=\overrightarrow{{{I}_{1}}}+\overrightarrow{{{I}_{2}}}+\overrightarrow{{{I}_{3}}}+.......\)

(viii) Point of zero intensity : If two bodies A and B of different masses \(T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \sqrt{\frac{{{R}^{3}}}{GM}}\,\, and ,\,{{m}_{2}}\,\, are \,\,T=84.6\,\text{minute}\ \approx 1.4\ hr\) distance apart.

Let \(P\) be the point of zero intensity i.e., the intensity at this point is equal and opposite due to two bodies \(A\) and \(B\) and if any test mass placed at this point it will not experience any force

(ix) Gravitational field line is a line, straight or curved such that a unit mass placed in the field of another mass would always move along this line. Field lines for an isolated mass \(m\) are radially inwards.

(x) As \(R=6400km\) and also \(g=\frac{GM}{{{R}^{2}}}\therefore \ \ \ I=g\)

Thus the intensity of gravitational field at a point in the field is equal to acceleration of test mass placed at that point.

Gravitational Field Intensity for Different Bodies

(1) Intensity due to uniform solid sphere

(2) Intensity due to spherical shell

(3) Intensity due to uniform circular ring

(4) Intensity due to uniform disc