| Mass (m) | Weight (W) |
| It is a quantity of matter contained in a body. | It is the attractive force exerted by earth on any body. |
| Its value does not change with g | more...
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 \[\rho =\frac{3g}{4\pi GR}\]
\[\therefore \] \[\rho =\frac{3\times 9.8}{4\times 3.14\times 6.67\times {{10}^{-11}}\times 6.4\times {{10}^{6}}}=5478.4\ kg/{{m}^{3}}\]
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 :
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 (\[\text{if}\,\ h<<R\]) is double to that of inside the earth.
Acceleration due to gravity at the surface of the earth
\[g=\frac{GM}{{{R}^{2}}}\] ...(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 \[h<<R\] 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 \[h<<R\]]
(iv) If \[h<<R\] 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 \[\frac{\Delta g}{g}\times 100%=\frac{2h}{R}\times 100%\]
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.
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 x 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 \[\left( \frac{4}{3}\pi {{R}^{3}}\times \right)\] 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 \[[g]=[L{{T}^{-2}}]\]
(v) it's average value is taken to be \[9.8\,\,m/{{s}^{2}}\] or \[981\,\,cm/{{\sec }^{2}}\] or \[32\,\,feet/{{s}^{2}},\] 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.
(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 Fnuclear > 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 :
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.
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 \[\text{6}\text{.67}\times \text{1}{{\text{0}}^{-11}}\,N-{{m}^{2}}\,\,k{{g}^{-2}}\] in S.I. and \[\text{6}\text{.67}\times \text{1}{{\text{0}}^{-8}}\,dyne-c{{m}^{2}}-{{g}^{2}}\] 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.
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. Current Affairs CategoriesArchive
Trending Current Affairs
You need to login to perform this action. |
