Mathematics ACCELERATION DUE TO GRAVITY BELOW AND ABOVE THE SURFACE OF EARTH AND GRAVITATIONAL POTENTIAL ENERGY, ESCAPE SPEED

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star ACCELERATION DUE TO GRAVITY BELOW AND ABOVE THE SURFACE OF EARTH
star GRAVITATIONAL POTENTIAL ENERGY, ESCAPE SPEED

### 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,

● 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.

### GRAVITATIONAL POTENTIAL ENERGY

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.
Q 3200001818

Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square.

Solution:

Consider four masses each of mass m at the corners of a square of side l ; See Fig. 8.9.

We have four mass pairs at distance l and two diagonal pairs at distance sqrt 2 l

Hence,

color{orange} {W(r)=-4 (Gm^2)/l =-2 (Gm^2)/(sqrt 2 l)}

=-(2Gm^2)/l (2 + 1/(sqrt 2)) =-5.41 (Gm^2)/l

The gravitational potential at the centre of the square (r = sqrt 2 l // 2) is U(r) =-4 sqrt 2 (Gm)/l

### ESCAPE SPEED

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.

Q 3260101915

Two uniform solid spheres of equal radii R, but mass M  and 4 M have a centre to centre separation 6 R, as shown in Fig. 8.10. The two spheres are held fixed. A projectile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.

Solution:

The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see Fig. 8.10) is
defined as the position where the two forces cancel each other exactly. If ON = r, we have

color{purple} {(GMm)/(r^2) =(4GMm)/((6R-r)^2)}

(6R – r)^2 = 4r^2

6R – r = ±2r

r = 2R or – 6R.

The neutral point r = – 6R does not concern us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a speed
which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface
of M is

color{green} {E_i = 1/2 mv^2 - (GM m)/R - (4GMm)/(5R)}

At the neutral point N, the speed approaches zero. The mechanical energy at N is purely potential.

color{green} {E_N =- (GMm)/(2R) -(4GMm)/(4R)}

From the principle of conservation of mechanical energy

color{orange} {1/2 v^2 -(GM)/R - (4GM)/(5R) =- (GM)/(2R) - (GM)/R}

or v^2 = (2 G M )/R (4/5 -1/2)

v= ((3GM)/(5R))^(1//2)

A point to note is that the speed of the projectile is zero at N, but is nonzero when it strikes the heavier sphere 4 M. The calculation of this speed is left as an exercise to the students.