Let a material point of mass m is at a distance r from the center of the Earth. If r >R (the Earth's radius), then
F = GmM / r ²
where G - gravitational constant , and M - the mass of the Earth.
If r <R, then the Earth’s action exerted on the point can be divided into two parts: the action of the inner sphere ( radius r) and the effect of the external spherical shell .
The spherical layer does not create the gravitational field .
Therefore, the force is connected only with the mass M* inside the sphere of radius r
M * = ρ4πr ³ / 4
where ρ is the density of the Earth.
The force of gravity is
F = - GmM * / r ² = - G4πρmr / 3 .
Motion in a tunnel is similar to the motion of the body suspended from the spring.
The object has been dropped from point A (no initial velocity ) . The motion of the object in the tunnel can be seen as the motion of the projection point, orbiting the Earth at its surface, such as a satellite in a circular orbit around the earth .
http://clowder.net/hop/railroad/tunnel.gif
Therefore, the oscillation frequency of the object in the tunnel is the angular frequency of rotation of the satellite around the Earth. Since the centripetal acceleration of the satellite is
a = ω ² r
and, on the other hand , a = F / m,
the angular frequency is
ω = sqrt (a / R) = sqrt (F / mR) = sqrt (G4πρmR/3mR) = sqrt (G 4πρ / 3)
The oscillation period is therefore equal to
T = 2π / ω = sqrt (3π/ρG), where
ρ=M/V= 3M/4πR³,
where M=6•10²⁴ kg, R= 6.4•10⁶ m
As a result, the time for Boston-Delhi motion is t =T/2, and,
if the velocity in Boston is zero, the velocity in Delhi is zero too.
Imagine a frictionless perfectly straight tunnel which runs from Boston (USA) to Delhi (India). We release (at zero speed) an object in the tunnel in Boston. Assume that the Earth is a perfect sphere with radius 6.4*10^3 km, with a mass of 6.0*10^24 kg and that the mass density is uniformly distributed throughout the Earth.
Question: How much time will it take the object to reach Delhi and what will be its speed when it gets to Delhi (3% accuracy is required)? You should ignore friction due to air-drag
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nice explanation. It should also be noted that the period and velocity are the same for any two points on earth through the tunnel. This concept is called "gravity train"
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