Earth's Acceleration Due to Gravity

Earth's Acceleration Due to Gravity

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.

Fig. 8.7 The mass m is in a mine located at a depth d below the surface of the Earth of mass M_E and radius R_E. We treat the Earth to be spherically symmetric.

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 \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyizImkaaa!37AB! \le \) 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 Mr is concentrated at the centre.Thus the force on the mass m at P has a magnitude

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2 % da9maalaaabaGaam4raiaad2gadaqadaqaaiaad2eadaWgaaWcbaGa % amOCaaqabaaakiaawIcacaGLPaaaaeaacaWGYbWaaWbaaSqabeaaca % aIYaaaaaaaaaa!3EFD! F = \frac{{Gm\left( {{M_r}} \right)}}{{{r^2}}}\)                                 (8.9)

We assume that the entire earth is of uniform density and hence its mass is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytamaaBa % aaleaacaWGfbaabeaakiabg2da9maalaaabaGaaGinaiabec8aWbqa % aiaaiodaaaGaamOuamaaDaaaleaacaWGfbaabaGaaG4maaaakiabeg % 8aYbaa!406B! {M_E} = \frac{{4\pi }}{3}R_E^3\rho \) where M_E is the mass of the earth R_E is its radius and and  \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdihaaa!37B6! \rho \) is the density. On the other hand the mass of the sphere M_r of radius r is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI0aGaeqiWdahabaGaaG4maaaacqaHbpGCcaWGYbWaaWbaaSqabeaa % caaIZaaaaaaa!3CDF! \frac{{4\pi }}{3}\rho {r^3}\) and  

hence

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGgb % Gaeyypa0Jaam4raiaad2gadaqadaqaamaalaaabaGaaGinaiaadcha % aeaacaaIZaaaaiaadkhaaiaawIcacaGLPaaadaWcaaqaaiaadkhada % ahaaWcbeqaaiaaiodaaaaakeaacaWGYbWaaWbaaSqabeaacaaIYaaa % aaaakiabg2da9iaadEeacaWGTbWaaeWaaeaadaWcaaqaaiaad2eada % WgaaWcbaGaamyraaqabaaakeaacaWGsbWaa0baaSqaaiaadweaaeaa % caaIZaaaaaaaaOGaayjkaiaawMcaamaalaaabaGaamOCamaaCaaale % qabaGaaG4maaaaaOqaaiaadkhadaahaaWcbeqaaiaaikdaaaaaaaGc % baGaeyypa0ZaaSaaaeaacaWGhbGaamyBaiaad2eadaWgaaWcbaGaam % yraaqabaaakeaacaWGsbWaa0baaSqaaiaadweaaeaacaaIZaaaaaaa % kiaadkhacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 % UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 % caaMc8+aaeWaaeaacaaI4aGaaiOlaiaaigdacaaIWaaacaGLOaGaay % zkaaaaaaa!746E! \begin{array}{l} F = Gm\left( {\frac{{4p}}{3}r} \right)\frac{{{r^3}}}{{{r^2}}} = Gm\left( {\frac{{{M_E}}}{{R_E^3}}} \right)\frac{{{r^3}}}{{{r^2}}}\\ = \frac{{Gm{M_E}}}{{R_E^3}}r\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {8.10} \right) \end{array}\)

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)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2 % da9iaadEeadaWcaaqaaiaad2eadaWgaaWcbaGaamyraaqabaGccaWG % TbaabaGaamOuamaaDaaaleaacaWGfbaabaGaaGOmaaaaaaaaaa!3DF1! F = G\frac{{{M_E}m}}{{R_E^2}}\)                         (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

 \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zaiabg2 % da9maalaaabaGaamOraaqaaiaad2gaaaGaeyypa0ZaaSaaaeaacaWG % hbGaamytamaaBaaaleaacaWGfbaabeaaaOqaaiaadkfadaqhaaWcba % Gaamyraaqaaiaaikdaaaaaaaaa!3FF3! g = \frac{F}{m} = \frac{{G{M_E}}}{{R_E^2}}\)                     (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”.

Acceleration Due to Gravity Below and Above the surface of Earth

Acceleration Due to Gravity Below and Above the surface of Earth

Consider a point mass m at a height h above the surface of the earth as shown in Fig. 8.8(a).The radius of the earth is denoted by R_E.Since this point is outside the earth,

Fig. 8.8 (a) g at a height h above the surface of the earth

its distance from the centre of the earth is (R_E+ h ). If F (h) denoted the magnitude of the force on the point mass m , we get from Eq. (8.5) :

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiaacI % cacaWGObGaaiykaiabg2da9maalaaabaGaam4raiaad2eadaWgaaWc % baGaamyraaqabaGccaWGTbaabaWaaeWaaeaacaWGsbWaaSbaaSqaai % aadweaaeqaaOGaey4kaSIaamiAaaGaayjkaiaawMcaamaaCaaaleqa % baGaaGOmaaaaaaaaaa!43C5! F(h) = \frac{{G{M_E}m}}{{{{\left( {{R_E} + h} \right)}^2}}}\)                       (8.13)

The acceleration experienced by the point mass is F (h )/m \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyyIOlaaa!37BF! \equiv \) g(h ) and we get

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiAaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamOramaa % bmaabaGaamiAaaGaayjkaiaawMcaaaqaaiaad2gaaaGaeyypa0ZaaS % aaaeaacaWGhbGaamytamaaBaaaleaacaWGfbaabeaaaOqaamaabmaa % baGaamOuamaaBaaaleaacaWGfbaabeaakiabgUcaRiaadIgaaiaawI % cacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaaaa!486D! g\left( h \right) = \frac{{F\left( h \right)}}{m} = \frac{{G{M_E}}}{{{{\left( {{R_E} + h} \right)}^2}}}\)                               (8.14)

This is clearly less than the value of g on the surface of earth :\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zaiabg2 % da9maalaaabaGaam4raiaad2eadaWgaaWcbaGaamyraaqabaaakeaa % caWGsbWaa0baaSqaaiaadweaaeaacaaIYaaaaaaaaaa!3D20! g = \frac{{G{M_E}}}{{R_E^2}}\). For h<<R_E, we can expand the RHS of Eq. (8.14) :

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiAaaGaayjkaiaawMcaaiabg2da9maalaaabaGaam4raiaa % d2eadaWgaaWcbaGaamyraaqabaaakeaacaWGsbWaa0baaSqaaiaadw % eaaeaacaaIYaaaaOWaaeWaaeaacaaIXaGaey4kaSIaamiAaiaac+ca % caWGsbWaaSbaaSqaaiaadweaaeqaaaGccaGLOaGaayzkaaWaaWbaaS % qabeaacaaIYaaaaaaakiabg2da9iaadEgadaqadaqaaiaaigdacqGH % RaWkcaWGObGaai4laiaadkfadaWgaaWcbaGaamyraaqabaaakiaawI % cacaGLPaaadaahaaWcbeqaaiabgkHiTiaaikdaaaaaaa!5195! g\left( h \right) = \frac{{G{M_E}}}{{R_E^2{{\left( {1 + h/{R_E}} \right)}^2}}} = g{\left( {1 + h/{R_E}} \right)^{ - 2}}\)

For \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGObaabaGaamOuamaaBaaaleaacaWGfbaabeaaaaGccqGH8aapcqGH % 8aapcaaIXaaaaa!3B8D! \frac{h}{{{R_E}}} < < 1\),using binomial expression,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiAaaGaayjkaiaawMcaaiabgwKiajaadEgadaqadaqaaiaa % igdacqGHsisldaWcaaqaaiaaikdacaWGObaabaGaamOuamaaBaaale % aacaWGfbaabeaaaaaakiaawIcacaGLPaaaaaa!4238! g\left( h \right) \cong g\left( {1 - \frac{{2h}}{{{R_E}}}} \right)\)                                 (8.15)

Equation (8.15) thus tells us that for small heights h above the value of g decreases by a factor (1 -2h / R_E ).

Now, consider a point mass m at a depth d below the surface of the earth (Fig. 8.8(b)), so that its distance from the centre of the earth is

(R_Ed ) as shown in the figure. The earth can be thought of as being composed of a smaller sphere of radius (R_E – d ) and a spherical shell of thickness d. The force on m due to the outer shell of thickness d is zero because the result quoted in the previous section. As far as the smaller sphere of radius ( R_E – d ) is concerned, the point mass is outside it and hence according to the result quoted earlier, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre.

If M_s is the mass of the smaller sphere, then,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytamaaBa % aaleaacaWGZbaabeaakiaac+cacaWGnbWaaSbaaSqaaiaadweaaeqa % aOGaeyypa0ZaaeWaaeaacaWGsbWaaSbaaSqaaiaadweaaeqaaOGaey % OeI0IaamizaaGaayjkaiaawMcaamaaCaaaleqabaGaaG4maaaakiaa % c+cacaWGsbWaa0baaSqaaiaadweaaeaacaaIZaaaaaaa!44E9! {M_s}/{M_E} = {\left( {{R_E} - d} \right)^3}/R_E^3\)                       (8.16)

Since mass of a sphere is proportional to be cube of its radius.

Fig. 8.8 ( b) g at a depth d. In this case only the smaller sphere of radius ( R_E – d) contributes to g.

Thus the force on the point mass is 

F(d)=G M_s m / (R_E – d )²            (8.17)

Substituting for M_s from above , we get

F (d) = G M m ( R_E– d ) / R\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaai % aadweaaeaacaaIZaaaaaaa!37AA! _E^3\)       (8.18)

and hence the acceleration due to gravity at a depth d,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamizaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamOramaa % bmaabaGaamizaaGaayjkaiaawMcaaaqaaiaad2gaaaaaaa!3E99! g\left( d \right) = \frac{{F\left( d \right)}}{m}\) is

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGNb % WaaeWaaeaacaWGKbaacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaWG % gbWaaeWaaeaacaWGKbaacaGLOaGaayzkaaaabaGaamyBaaaacqGH9a % qpdaWcaaqaaiaadEeacaWGnbWaaSbaaSqaaiaadweaaeqaaaGcbaGa % amOuamaaDaaaleaacaWGfbaabaGaaGOmaaaaaaGcdaqadaqaaiaadk % fadaWgaaWcbaGaamyraaqabaGccqGHsislcaWGKbaacaGLOaGaayzk % aaaabaGaeyypa0Jaam4zamaalaaabaGaamOuamaaBaaaleaacaWGfb % aabeaakiabgkHiTiaadsgaaeaacaWGsbWaaSbaaSqaaiaadweaaeqa % aaaakiabg2da9iaadEgadaqadaqaaiaaigdacqGHsislcaWGKbGaai % 4laiaadkfadaWgaaWcbaGaamyraaqabaaakiaawIcacaGLPaaacaaM % c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk % W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaabmaabaGaaGioaiaa % c6cacaaIXaGaaGyoaaGaayjkaiaawMcaaaaaaa!744F! \begin{array}{l} g\left( d \right) = \frac{{F\left( d \right)}}{m} = \frac{{G{M_E}}}{{R_E^2}}\left( {{R_E} - d} \right)\\ = g\frac{{{R_E} - d}}{{{R_E}}} = g\left( {1 - d/{R_E}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {8.19} \right) \end{array}\)

Thus, as we go down below earth’s surface, the acceleration due gravity decreases by a factor (1 - d / R_E ). The remarkable thing about acceleration due to earth’s gravity is that it is maximum on its surface decreasing whether you go up or down.