A Hole Is Drilled Straight Through the Center of the Earth and a Particle Is Dropped Into

A hole is drilled straight through the center of the earth and a particle is dropped into the hole. Neglect rotational effects:

a) Show that the particle's motion, assuming that earth's gravity acts from the exact center of the earth, is simple harmonic motion.

b) Calculate the Period of oscillation.

Answer:

a) To show that the particle's motion, assuming that earth's gravity acts from the exact center of the earth, is simple harmonic motion. R Consider a tunnel drilled due along the diameter of the Earth as shown in figure. O Suppose that at some instant, the particle f y dropped in this tunnel is at point P. The position of the particle as shown in A P B v figure is that position which the particle has reached after crossing the mean position O i.e. the center of the earth. (This is just for convenience and nothing else). The particle is now moving Fig.1: Particle oscillating along the tunnel. downward from point P.

The particle can be assumed to be subjected to the two forces of attraction due to two different parts of the earth, above and below the the line APB. The section of the earth below the curve APB tries to pull the particle downward with force (say) F1 whereas the section above APB tries to pull it upward with force F2. Due to a larger mass on the above side, F2 > F1. As a result, the particle experiences a net/resultant upward force, given by F = F2–F1. With the help of A P B calculus and solid geometry, this resultant or net force can be proved to be equal to the force experienced by the spherical section of the earth of radius y. Fig.2: two sections of the earth, above & The gravitational force F2 is larger than below line APB, outside the spherical section that of F1 only because its mass is greater. Hence it exerts a force of gravitation of larger magnitude. The upper and lower section of the earth about point P outside the spherical section cancel out each other only because the particle is closer to the lower section. Therefore, even though the upper (gray) section is more in mass than the lower section, both the gray regions cancel each other. While doing so, we assume the earth to be a homogeneous one. Hence its density is same throughout so that while oscillating along the tunnel drilled along any diameter of the earth (horizontal, NE-SW etc..) the forces F1 & F2 remain unaltered in their magnitudes. [Direction of the forces will keep on changing with the direction of the tunnel drilled along the diameter of the earth.]

The particle is exactly on the surface of this spherical section of the Earth. [ In short, one can say that the particle experiences an attractive force only due to the spherical section of the earth of radius y. The forces of attraction due to the upper (the region then left between the two surfaces) and lower section cancel out each other, nullify the effect of one another]

Let m – mass of the particle. y – displacement of the particle at some instant t. M – mass of the spherical section of the earth of radius y.  – Density of the spherical section of the earth of radius y.

For earth, density  = 5.51 x 103 kg/m3.

4 The mass of the sphere of radius y is then M  πy3 ρ . ---- (1) 3 The mass of the spherical section of the earth can be assumed to be concentrated at its center. The gravitation force F exerted by this spherical section of mass M is

GMm F   ---- (2) y2 [Negative sign is because the gravitational force is attractive.] As the particle tries to move further, this force F acts on it in the reverse direction, towards the center of the earth. Thus this force tries to restore the particle back to the center of the earth.

Using equation (1) in (2), we get

4 G πy3 ρm F   3 y2  4  i.e. F   πG ρm y  3  4 i.e. F = – ky ---- (3) where k  πG ρm ---- (4) 3 As all the term in the bracket i.e. G, M  and  are constant, k is constant.

 F  –y ---- (5) i.e. The force  – displacement.

The negative sign shows that the force acts in the direction opposite to the direction of displacement of the particle. This is the condition for linear simple harmonic motion.

Thus the particle performs linear simple harmonic motion when dropped in the tunnel drilled along the diameter of the earth.

Calculation of Period of Oscillation of the particle:

From equation (3), F = – ky According to Newton’s second law, force F = ma

 ma = – ky

 k  i.e. acceleration a    y  m  i.e. a = –2y ------(6)

k where ω2  m  4   πG ρm i.e. 3 substituting k from equation (4) ω2    m 4 i.e. ω2  πG ρ 3 4 i.e. ω  πG ρ ------(7) 3

Then Period of oscillation of the particle

2π T  ω 2π T  i.e. 4 πG ρ 3 π T  i.e. 1 G ρ 3 3π i.e. T  ------(8) G ρ

Here, constant of gravitation, G = 6.67 x 10-11 Nm2/kg2 and density of the earth,  = 5.51 x 103 kg/m3.

3 3.14 Then T  6.67 1011  5.51103 9.42  36.7517 108  9.42    4    10  36.7517  i.e. T = 0.506275 x 104 second

i.e. T = 5063 second

i.e. T = 84 min 22.75 second. T = 84.479 min

------Have a nice time.

Mr. Raosaheb Y. Pawar