Eighteen years

About eighteen years ago the following problem was posed at the end of a Physics lesson regarding the gravitational field:

“Assuming that the Earth is perfectly spherical and homogeneous (i.e. constant density), prove that if a tunnel was dug following one of the Earth diameters and passing through its centre, a small ball dropped at the surface would describe a harmonic motion”.

Naturally, the Earth is neither a perfect sphere nor its density is constant. However, if this was the case, the acceleration exerted on the small ball by the gravitational force would reduce proportionally to its location, in such a way that it would become 0 at the centre of the Earth. Naming r the direction of the motion and R, the radius of the Earth, it is satisfied:

g(r) = - Cr(t)                                                                                                      
(1),

with the constant C being:

C=4/3*Pi* G*density of Earth                                                                             
(2),

and G, the Universal constant of gravitation, 6.67·10^-11 Nm^-2kg^-2.

Equation (1) -a direct finding from the Gauss’ law, the law of flux, as we learned it in the high school- seems to me now a sufficient proof, because a feature of all harmonic motions is to have its acceleration proportional to the negative of its position vector. Nevertheless, it can be shown that the form of the r(t) is a cosine type of function, which is a harmonic one. An introduction to ordinary differential equations (ODE) is needed to solve the problem, but it is rather straight-forward, only a quick, warming-up exercise –despite the dimensions of the tunnel needed. Realization is a matter of concepts and first principles, and the fact that it took me so many years to understand –even though I did try no more times than fingers I have in my right hand- might give idea of how feeble the fundamentals are normally laid in the soil of the student’s brain.

The starting point is the momentum equation applied to the descending ball. Gravity is the only force. Using Equation (1):

m dv(t)/dt = m g(r)                                                                                             
(3),

where v(t) is the velocity and m, the mass. The velocity and the position are connected through the derivative, and Equation (3) can be arranged to:

d2r(t)/dt2 + C r(t) = 0                                                                                        
(4)

Equation (4) is a linear ODE with constant coefficients and solutions are of the form (heuristic approach):

r(t) of the type exp (at)                                                                                                    
(5),

where a is a parameter. Using Equation (5) in (4), the characteristic equation of the ODE is found:

a^2 +C = 0                                                                                                        
(6)

The solution of (6) is:

a = +/- i*sqrt(C)                                                                                                 
(7)

The function of r(t) is, therefore:

r(t) = A exp (i*t*sqrt(C)) + B exp (-i*t*sqrt(C))                                                   
(8),

where A and B are constants that can be calculated using two initial conditions: r(t) = R and v(t) = 0 at t = 0, respectively. The velocity v(t) is calculated by taking the derivative of Equation (8). It is found, then:

A = B = R/2                                                                                                       
(9)

And:

r(t) = R/2 * (exp (i*t*sqrt(C)) + exp (-i*t*sqrt(C))                                            
(10)

The complex exponentials are:

exp (i*t*sqrt(C) = cos (t*sqrt(C)) + i sin (t*sqrt(C))                                           
(11a)

exp (-i*t*sqrt(C) = cos (t*sqrt(C)) - i sin (t*sqrt(C))                                          
(11b)

Substituting Equations (11a and b) in (10), the imaginary parts are cancelled, and the final solution describing the location of the small ball at any time t is:

r(t) = R cos (t*sqrt(C))                                                                                      
(12)

Equation (12) describes a harmonic motion.

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