`color{green} ✍️` Legend has it that observing an apple falling from a tree, Newton was inspired to arrive at an universal law of gravitation that led to an explanation of terrestrial gravitation as well as of Kepler’s laws. Newton’s reasoning was that the moon revolving in an orbit of radius `R_m` was subject to a centripetal acceleration due to earth’s gravity of magnitude
`color{blue}{a_m=(V^2)/(R_m) =(4 pi^2R_m)/(T^2)..........................(8.3)}`
`=>` where `V` is the speed of the moon related to the time period `T` by the relation ` V =2 π R //T`.
`☞` The time period `T` is about 27.3 days and `R_m` was already known then to be about `3.84 xx 10^8 m`. If we substitute these numbers in equation (8.3), we get a value of am much smaller than the value of acceleration due to gravity `g` on the surface of the earth, arising also due to earth’s gravitational attraction.
`☞` This clearly shows that the force due to earth’s gravity decreases with distance. If one assumes that the gravitational force due to the earth decreases in proportion to the inverse square of the distance from the center of the earth, we will have in agreement with a value of `g ; 9.8 m s^(-2)` and the value of am from Eq. (8.3).
`color{brown}{a_m α R_m^(−2) : g α R_E^(-2)}`
and we get `color{blue}{g/(a_m) =(R_m^2)/(R_E^2) ; 3600.......................(8.4)}`
These observations led Newton to propose the following Universal Law of Gravitation : Every body in the 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 them.
`●` The quotation is essentially from Newton’s famous treatise called ‘Mathematical Principles of Natural Philosophy’ (Principia for short). Stated Mathematically, Newton’s gravitation law reads : The force F on a point mass `m_2` due to another point mass `m_1` has the magnitude
`color{blue}{|F|= G(m_1m_2)/(r^2)..........................(8.5)}`
`●` Equation (8.5) can be expressed in vector form as
`F=G (m_1m_2)/r^2 (-hat r)=-G (m_1m_2)/(r^2) hat r`
`color{blue}{ :. F=-G (m_1m_2)/(|r^3|) hat r}`
`=>` where `G` is the universal gravitational constant, `hat r` is the unit vector from `m_1` to `m_2` and `r=r_2-r_1` as shown in Fig. 8.3.
`●` The gravitational force is attractive, i.e., the force `F` is along `– r`. The force on point mass `m_1` due to `m_2` is of course `– F` by Newton’s third law.
`●` Thus, the gravitational force `F_(12)` on the body 1 due to 2 and `F_(21)` on the body 2 due to 1 are related as `F_(12) = – F_(21)`. Before we can apply Eq. (8.5) to objects under consideration, we have to be careful since the law refers to point masses whereas we deal with extended objects which have finite size.
`●` If we have a collection of point masses,the force on any one of them is the vector sum of the gravitational forces exerted by the other point masses as shown in Fig 8.4.
`●` The total force on `m_1` is
`color{blue}{F_1=(Gm_2m_1)/(r_(21^2)) hat r_(21) + (G m_3m_1)/(r_(31)^2) hat r _(31) + (Gm_4m_1)/(r_(41)^2) hat r_(41)}`
`●` For the gravitational force between an extended object (like the earth) and a point mass, Eq. (8.5) is not directly applicable. Each point mass in the extended object will exert a force on the given point mass and these force will not all be in the same direction.
`●` We have to add up these forces vectorially for all the point masses in the extended object to get the total force. This is easily done using calculus. For two special cases, a simple law results when you do that :
`color{green}{(1) " The force of attraction between a hollow spherical shell of uniform density and a "}`
`color{green}{"point mass situated outside is just as if the entire mass of the shell is concentrated at "}`
`color{green}{"the centre of the shell"}`. Gravitational forces caused by the various regions of the shell have components along the line joining the point mass to the centre as well as along a direction prependicular to this line. The components prependicular to this line cancel out when summing over all regions of the shell leaving only a resultant force along the line joining the point to the centre. The magnitude of this force works out to be as stated above.
`color{green}{(2) " The force of attraction due to a hollow spherical shell of uniform density, on a"}`
`color{green}{ "point mass situated inside it is zero"}`. Qualitatively, we can again understand this result. Various regions of the spherical shell attract the point mass inside it in various directions. These forces cancel each other completely.
`color{green} ✍️` Legend has it that observing an apple falling from a tree, Newton was inspired to arrive at an universal law of gravitation that led to an explanation of terrestrial gravitation as well as of Kepler’s laws. Newton’s reasoning was that the moon revolving in an orbit of radius `R_m` was subject to a centripetal acceleration due to earth’s gravity of magnitude
`color{blue}{a_m=(V^2)/(R_m) =(4 pi^2R_m)/(T^2)..........................(8.3)}`
`=>` where `V` is the speed of the moon related to the time period `T` by the relation ` V =2 π R //T`.
`☞` The time period `T` is about 27.3 days and `R_m` was already known then to be about `3.84 xx 10^8 m`. If we substitute these numbers in equation (8.3), we get a value of am much smaller than the value of acceleration due to gravity `g` on the surface of the earth, arising also due to earth’s gravitational attraction.
`☞` This clearly shows that the force due to earth’s gravity decreases with distance. If one assumes that the gravitational force due to the earth decreases in proportion to the inverse square of the distance from the center of the earth, we will have in agreement with a value of `g ; 9.8 m s^(-2)` and the value of am from Eq. (8.3).
`color{brown}{a_m α R_m^(−2) : g α R_E^(-2)}`
and we get `color{blue}{g/(a_m) =(R_m^2)/(R_E^2) ; 3600.......................(8.4)}`
These observations led Newton to propose the following Universal Law of Gravitation : Every body in the 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 them.
`●` The quotation is essentially from Newton’s famous treatise called ‘Mathematical Principles of Natural Philosophy’ (Principia for short). Stated Mathematically, Newton’s gravitation law reads : The force F on a point mass `m_2` due to another point mass `m_1` has the magnitude
`color{blue}{|F|= G(m_1m_2)/(r^2)..........................(8.5)}`
`●` Equation (8.5) can be expressed in vector form as
`F=G (m_1m_2)/r^2 (-hat r)=-G (m_1m_2)/(r^2) hat r`
`color{blue}{ :. F=-G (m_1m_2)/(|r^3|) hat r}`
`=>` where `G` is the universal gravitational constant, `hat r` is the unit vector from `m_1` to `m_2` and `r=r_2-r_1` as shown in Fig. 8.3.
`●` The gravitational force is attractive, i.e., the force `F` is along `– r`. The force on point mass `m_1` due to `m_2` is of course `– F` by Newton’s third law.
`●` Thus, the gravitational force `F_(12)` on the body 1 due to 2 and `F_(21)` on the body 2 due to 1 are related as `F_(12) = – F_(21)`. Before we can apply Eq. (8.5) to objects under consideration, we have to be careful since the law refers to point masses whereas we deal with extended objects which have finite size.
`●` If we have a collection of point masses,the force on any one of them is the vector sum of the gravitational forces exerted by the other point masses as shown in Fig 8.4.
`●` The total force on `m_1` is
`color{blue}{F_1=(Gm_2m_1)/(r_(21^2)) hat r_(21) + (G m_3m_1)/(r_(31)^2) hat r _(31) + (Gm_4m_1)/(r_(41)^2) hat r_(41)}`
`●` For the gravitational force between an extended object (like the earth) and a point mass, Eq. (8.5) is not directly applicable. Each point mass in the extended object will exert a force on the given point mass and these force will not all be in the same direction.
`●` We have to add up these forces vectorially for all the point masses in the extended object to get the total force. This is easily done using calculus. For two special cases, a simple law results when you do that :
`color{green}{(1) " The force of attraction between a hollow spherical shell of uniform density and a "}`
`color{green}{"point mass situated outside is just as if the entire mass of the shell is concentrated at "}`
`color{green}{"the centre of the shell"}`. Gravitational forces caused by the various regions of the shell have components along the line joining the point mass to the centre as well as along a direction prependicular to this line. The components prependicular to this line cancel out when summing over all regions of the shell leaving only a resultant force along the line joining the point to the centre. The magnitude of this force works out to be as stated above.
`color{green}{(2) " The force of attraction due to a hollow spherical shell of uniform density, on a"}`
`color{green}{ "point mass situated inside it is zero"}`. Qualitatively, we can again understand this result. Various regions of the spherical shell attract the point mass inside it in various directions. These forces cancel each other completely.