`●` 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,



`●` 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) :

`color{blue}{F(h)=(GM_E m)/((R_E+h)^2)......................(8.13)}`

`●` The acceleration experienced by the point mass is `F(h)//m equiv g(h)` and we get

`color{blue}{g(h)=(F(h))/m =(GM_E)/((R_E+h)^2)........................(8.14)}`

`●` This is clearly less than the value of `g` on the surface of earth : `g=(GM_E)/(R_E^2)` For , ` h < < R_E` we can
expand the RHS of Eq. (8.14) :

`g(h)=(GM_E)/(R_E^2(1+h//R_E)^2)= g(1+h//R_E)^(-2)`

For `h/(R_E) < < 1` using binomial expression,

`\color{red} ✍️` `color{blue}{g(h) = g( 1-(2h)/(R_E))........................(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_E − d`) as shown in the figure.

`color{red}➢` 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,

`color{blue}{M_s //M_E = ( R_E – d ) 3 // R_E^3 ......................( 8.16)}`

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



`●` Thus the force on the point mass is

`color{blue}{F(d)=G M_s m//(R_E-d)^2.........................(8.17)}`

`●` Substituting for `M_s` from above , we get

`color{blue}{F(d)=G M_E m(R_E-d)//R_E^3.............................(8.18)}`

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

`g(d)=(F(d))/m` is

`g(d)=(F(d))/m =(GM_E)/(R_E^3) (R_E-d)`

`\color{red} ✍️` `color{blue}{g(d)=g(R_E-d)/(R_E)=g(1-d/R_E).......................(8.19)}`

`●` 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.

`color{green} ✍️` If the position of the particle changes on account of forces acting on it, then the change in its potential energy is just the amount of work done on the body by the force.
`●` As we had discussed earlier, forces for which the work done is independent of the path are the conservative forces.

`color{green} ✍️` The force of gravity is a conservative force and we can calculate the potential energy of a body arising out of this force, called the gravitational potential energy.

`●` Consider points close to the surface of earth, at distances from the surface much smaller than the radius of the earth. In such cases, the force of gravity is practically a constant equal to `mg,` directed towards the centre of the earth.
`●` If we consider a point at a height `h_1` from the surface of the earth and another point vertically above it at a height h2 from the surface, the work done in lifting the particle of mass m from the first to the second position is denoted by `W_12`

`W_12= "Force x displacement"`

`color{blue}{=mg(h_2-h_1)....................(8.20)}`

`●` If we associate a potential energy `W(h)` at a point at a height `h` above the surface such that

`color{blue}{W(h)=mgh+W_o...............................(8.21)}`

(where `W_o =` constant) ; then it is clear that

`color{blue}{W_12=W(h_2)-W(h_1).........................(8.22)}`

`●` The work done in moving the particle is just the difference of potential energy between its final and initial positions.
`=>` Observe that the constant `W_o` cancels out in Eq. (8.22). Setting `h = 0` in the last equation, we get `W ( h = 0 ) = W_o` . `h = 0` means points on the surface of the earth.


`\color{green} ✍️` Thus, `W_o` is the potential energy on the surface of the earth. If we consider points at arbitrary distance from the surface of the earth, the result just derived is not valid since the assumption that the gravitational force mg is a constant is no longer valid.
`●` However, from our discussion we know that a point outside the earth, the force of gravitation on a particle directed towards the centre of the earth is

`color{blue}{F=(GM_E m).........................(8.23)}`

`=>` where `M_E =` mass of earth, `m =` mass of the particle and r its distance from the centre of the earth. If we now calculate the work done in lifting a particle from `r = r_1` to `r = r_2 (r_2 > r_1)` along a vertical path, we get instead of Eq. (8.20)

`W_(12) = (r_2)/(r_1) (GMm)/(r^2) dr`

`color{blue}{=-GM_E m 1/(r_2) - 1/(r_1).......................(8.24)}`

`●` In place of Eq. (8.21), we can thus associate a potential energy `W(r)` at a distance `r`, such that

`color{blue}{W(r)=-(G M_E m)/r +W_1,..............................(8.25)}`

`=>` valid for `r > R` , so that once again `W_12 = W(r_2) – W(r_1)`. Setting `r =` infinity in the last equation, we get `W` ( `r =` infinity ) `= W_1` .

`●` Thus, `W_1` is the potential energy at infinity. One should note that only the difference of potential energy between two points has a definite meaning from Eqs. (8.22) and (8.24).
`●` One conventionally sets `W_1` equal to zero, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point.

`color{red}➢` We have calculated the potential energy at a point of a particle due to gravitational forces on it due to the earth and it is proportional to the mass of the particle.
`color{red}➢` The gravitational potential due to the gravitational force of the earth is defined as the potential energy of a particle of unit mass at that point. From the earlier discussion, we learn that the gravitational potential energy associated with two particles of masses `m_1` and `m_2` separated by distance by a distance `r` is given by

`color{blue}{V=-(Gm_1m_2)/r}`

(if we choose `V = 0` as `r ->oo` )

`●` It should be noted that an isolated system of particles will have the total potential energy that equals the sum of energies (given by the above equation) for all possible pairs of its constituent particles. This is an example of the application of the superposition principle.

`color{green} ✍️` If a stone is thrown by hand, we see it falls back to the earth. Of course using machines we can shoot an object with much greater speeds and with greater and greater initial speed, the object scales higher and higher heights.
`●` A natural query that arises in our mind is the following: ‘can we throw an object with such high initial speeds that it does not fall back to the earth?’

`●` The principle of conservation of energy helps us to answer this question. Suppose the object did reach infinity and that its speed there was `V_f`. The energy of an object is the sum of potential and kinetic energy.
`●` As before `W_1` denotes that gravitational potential energy of the object at infinity. The total energy of the projectile at infinity then is

`color{blue}{E(oo)=W_1 +(m V_f^2)/2.......................(8.26)}`

`●` If the object was thrown initially with a speed Vi from a point at a distance `(h+R_E)` from the centre of the earth (`R_E =` radius of the earth), its energy initially was

`color{blue}{E=h+R_E =1/2 mV_t^2-(GmM_E)/((h+R_E))+W_1.......................(8.27)}`

`●` By the principle of energy conservation Eqs. (8.26) and (8.27) must be equal. Hence

`color{blue}{(mV_i^2)/2 -(GmM_E)/((h+R_E))=(mV_f^2)/2...............................(8.28)}`

`●` The R.H.S. is a positive quantity with a minimum value zero hence so must be the L.H.S. Thus, an object can reach infinity as long as `V_t` is such that

`color{blue}{(mV_t^2)/2 - (GmM_E)/((h+R_E)) ge 0.........................(8.29)}`

`●` The minimum value of `V_i` corresponds to the case when the L.H.S. of Eq. (8.29) equals zero.

`●` Thus, the minimum speed required for an object to reach infinity (i.e. escape from the earth) corresponds to

`color{blue}{1/2 m (V_t^2)_(min)=(GmM_E)/(h+R_E)...............................(8.30)}`

`●` If the object is thrown from the surface of the earth, `h = 0`, and we get

`color{blue}{(V_t)_(min) = sqrt((2 GM_E)/(R_E))..................................(8.31)}`

Using the relation `g = GM_E// R_E^2` , we get

`\color{red} ✍️` `color{blue}{(V_t)_(min) = sqrt(2gR_E).....................(8.32)}`

`●` Using the value of `g` and `R_E`, numerically `(V_i)_(min) approx 11.2 km//s`. This is called the escape speed, sometimes loosely called the escape velocity.

`●` Equation (8.32) applies equally well to an object thrown from the surface of the moon with g replaced by the acceleration due to Moon’s gravity on its surface and `r_E` replaced by the radius of the moon.
`●` Both are smaller than their values on earth and the escape speed for the moon turns out to be `2.3 km//s`, about five times smaller. This is the reason that moon has no atmosphere. Gas molecules if formed on the surface of the moon having velocities larger than this will escape the gravitational pull of the moon.