Thru-Earth Train

A straight tunnel is dug from New York to San Francisco, a distance of 5000 kilometers measured along the surface. A car rolling on steel rails is released from rest at New York, and rolls through the tunnel to San Francisco (see Figure 1.1).

Figure 1.1

a) Neglecting friction and also the rotation of the Earth, how long does it take to get there? Take the gravitational acceleration g = 980 m/s² and the radius of the Earth R = 6400 km.

b) Suppose there is now friction proportional to the square of the velocity (but still ignoring the rotation of the Earth). What is the equation for the phase space trajectory? Introduce suitable symbols for the constant of proportionality and for the mass of the car, and also draw a sketch.

c) We now consider the effects of rotation. Estimate the magnitude of the centrifugal and Coriolis forces relative to the gravitational force (ignore friction). Take New York and San Francisco to be of equal latitude (approximately 40° North).

SOLUTION

a) The radial force acting on a particle at any given point inside the Earth depends only on the mass of the sphere whose radius is at that point (see

Figure 1.2a

Figure 1.2a):

where m is the mass of the car, so

The accelerating force will be

so we have

On the surface of the Earth

resulting in

which describes oscillatory motion of frequency   Half of the period of this oscillation is the time for the train to get to San Francisco

b) If there is friction proportional to the square of velocity we have an additional term

where α is the proportionality coefficient for the friction force. Using p = mẋ we obtain

Or

(see Figure 1.2b).

Figure 1.2b

c) The acceleration due to the centrifugal force is given by the formula

where is defined in Figure 1.2a, and ω = 2π/T is the angular frequency of the Earth’s rotation. So the maximum centrifugal acceleration |a_c|_max is on the surface of the Earth

For New York θ ≈ 40^o yielding

So the centrifugal force is only about 0.3% of the gravitational force. The acceleration due to the Coriolis force is

From (a):

Where ϕ = L/2R = 2500/6400 ≈ 0.39 rad. So,

and |a_Cor|_max = 2ωΩRsinϕ, where Ω is the frequency of oscillation found in (a):  and so  Hence, the Coriolis force is about 5% of the gravitational force.