`star` THE GRAVITATIONAL CONSTANT

`star` ACCELERATION DUE TO GRAVITY OF THE EARTH

`star` ACCELERATION DUE TO GRAVITY OF THE EARTH

`color{green} ✍️` The value of the gravitational constant `color{green}{G}` entering the Universal law of gravitation can be determined experimentally and this was first done by English scientist `color{green} {" Henry Cavendish"} in 1798.

`●` The apparatus used by him is schematically shown in figure.8.6

`●` The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire.

`●` Two large lead spheres are brought close to the small ones but on opposite sides as shown. The big spheres attract the nearby small ones by equal and opposite force as shown.

`●` There is no net force on the bar but only a torque which is clearly equal to `F` times the length of the bar,where `F` is the force of attraction between a big sphere and its neighbouring small sphere.

`●` Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque . If `theta` is the angle of twist of the suspended wire, the restoring torque is proportional to `theta`, equal to `tau theta.` Where `tau` is the restoring couple per unit angle of twist. `tau` can be measured independently e.g. by applying a known torque and measuring the angle of twist.

`●` The gravitational force between the spherical balls is the same as if their masses are concentrated at their centres. Thus if d is the separation between the centres of the big and its neighbouring small ball, M and m their masses, the gravitational force between the big sphere and its neighouring small ball is.

`color{blue}{F=G(Mm)/d^2...................(8.6)}`

`●` If `L` is the length of the bar `AB` , then the torque arising out of `F` is F multiplied by `L`. At equilibrium, this is equal to the restoring torque and hence

`color{blue} {G(Mm)/d^2 L = tau theta....................(8.7)}`

`●` Observation of `theta` thus enables one to calculate `G` from this equation. Since Cavendish’s experiment, the measurement of G has been refined and the currently accepted value is

`●` The apparatus used by him is schematically shown in figure.8.6

`●` The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire.

`●` Two large lead spheres are brought close to the small ones but on opposite sides as shown. The big spheres attract the nearby small ones by equal and opposite force as shown.

`●` There is no net force on the bar but only a torque which is clearly equal to `F` times the length of the bar,where `F` is the force of attraction between a big sphere and its neighbouring small sphere.

`●` Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque . If `theta` is the angle of twist of the suspended wire, the restoring torque is proportional to `theta`, equal to `tau theta.` Where `tau` is the restoring couple per unit angle of twist. `tau` can be measured independently e.g. by applying a known torque and measuring the angle of twist.

`●` The gravitational force between the spherical balls is the same as if their masses are concentrated at their centres. Thus if d is the separation between the centres of the big and its neighbouring small ball, M and m their masses, the gravitational force between the big sphere and its neighouring small ball is.

`color{blue}{F=G(Mm)/d^2...................(8.6)}`

`●` If `L` is the length of the bar `AB` , then the torque arising out of `F` is F multiplied by `L`. At equilibrium, this is equal to the restoring torque and hence

`color{blue} {G(Mm)/d^2 L = tau theta....................(8.7)}`

`●` Observation of `theta` thus enables one to calculate `G` from this equation. Since Cavendish’s experiment, the measurement of G has been refined and the currently accepted value is

`color{blue}{G = 6.67xx10^(-11) N m^2//kg^2. ....................(8.8)}`

`color{green} ✍️` The earth can be imagined to be a sphere made of a large number of concentric spherical shells with the smallest one at the centre and the largest one at its surface.

`●` A point outside the earth is obviously outside all the shells. Thus, all the shells exert a gravitational force at the point outside just as if their masses are concentrated at their common centre according to the result stated in section 8.3.

`●` The total mass of all the shells combined is just the mass of the earth. Hence, at a point outside the earth, the gravitational force is just as if its entire mass of the earth is concentrated at its centre. For a point inside the earth, the situation is different. This is illustrated in Fig. 8.7.

`●` Again consider the earth to be made up of concentric shells as before and a point mass m situated at a distance r from the centre. The point `P` lies outside the sphere of radius `r`.

`●` For the shells of radius greater than `r`, the point `P` lies inside. Hence according to result stated in the last section, they exert no gravitational force on mass `m` kept at `P`.

`●` The shells with radius `le r` make up a sphere of radius r for which the point `P` lies on the surface. This smaller sphere therefore exerts a force on a mass `m` at `P` as if its mass `M_r` is concentrated at the centre. Thus the force on the mass `m` at `P` has a magnitude

`color{blue}{F= (Gm(M_r))/(r^2)...........................(8.9)}`

`●` We assume that the entire earth is of uniform density and hence its mass is

`M_E=(4 pi)/3 = R_E^2 rho`

`=>`where `M_E` is the mass of the earth `R_E` is its radius and `rho` is the density. On the other hand the mass of the sphere `M_r` of radius `r` is `(4 pi)/3 rho r^3` and hence

`F= Gm (4 pi)/3 rho r^3/r^2=Gm (M_E)/(R_E^2) (r^3)/(r^2)`

`color{blue}{F=(Gm M_E)/(R_E^3) r.........................(8.10)}`

`●` If the mass `m` is situated on the surface of earth, then `r = R_E` and the gravitational force on it is, from Eq. (8.10)

`color{blue}{F=G(M_E m)/(R_E^2).........................(8.11)}`

`●` The acceleration experienced by the mass `m`, which is usually denoted by the symbol g is related to `F` by Newton’s `2^(nd)` law by relation `F = mg`. Thus

`color{blue}{g=F/m =(Gm_E)/(R_E^2)...........................(8.12)}`

`●` Acceleration `g` is readily measurable. `R_E` is a known quantity. The measurement of `G` by Cavendish’s experiment (or otherwise), combined with knowledge of `g` and `R_E` enables one to estimate `M_E` from Eq. (8.12). This is the reason why there is a popular statement regarding Cavendish : “Cavendish weighed the earth”.

`●` A point outside the earth is obviously outside all the shells. Thus, all the shells exert a gravitational force at the point outside just as if their masses are concentrated at their common centre according to the result stated in section 8.3.

`●` The total mass of all the shells combined is just the mass of the earth. Hence, at a point outside the earth, the gravitational force is just as if its entire mass of the earth is concentrated at its centre. For a point inside the earth, the situation is different. This is illustrated in Fig. 8.7.

`●` Again consider the earth to be made up of concentric shells as before and a point mass m situated at a distance r from the centre. The point `P` lies outside the sphere of radius `r`.

`●` For the shells of radius greater than `r`, the point `P` lies inside. Hence according to result stated in the last section, they exert no gravitational force on mass `m` kept at `P`.

`●` The shells with radius `le r` make up a sphere of radius r for which the point `P` lies on the surface. This smaller sphere therefore exerts a force on a mass `m` at `P` as if its mass `M_r` is concentrated at the centre. Thus the force on the mass `m` at `P` has a magnitude

`color{blue}{F= (Gm(M_r))/(r^2)...........................(8.9)}`

`●` We assume that the entire earth is of uniform density and hence its mass is

`M_E=(4 pi)/3 = R_E^2 rho`

`=>`where `M_E` is the mass of the earth `R_E` is its radius and `rho` is the density. On the other hand the mass of the sphere `M_r` of radius `r` is `(4 pi)/3 rho r^3` and hence

`F= Gm (4 pi)/3 rho r^3/r^2=Gm (M_E)/(R_E^2) (r^3)/(r^2)`

`color{blue}{F=(Gm M_E)/(R_E^3) r.........................(8.10)}`

`●` If the mass `m` is situated on the surface of earth, then `r = R_E` and the gravitational force on it is, from Eq. (8.10)

`color{blue}{F=G(M_E m)/(R_E^2).........................(8.11)}`

`●` The acceleration experienced by the mass `m`, which is usually denoted by the symbol g is related to `F` by Newton’s `2^(nd)` law by relation `F = mg`. Thus

`color{blue}{g=F/m =(Gm_E)/(R_E^2)...........................(8.12)}`

`●` Acceleration `g` is readily measurable. `R_E` is a known quantity. The measurement of `G` by Cavendish’s experiment (or otherwise), combined with knowledge of `g` and `R_E` enables one to estimate `M_E` from Eq. (8.12). This is the reason why there is a popular statement regarding Cavendish : “Cavendish weighed the earth”.