`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”.
`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”.