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
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.
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"
To calculate the time it takes for the object to reach Delhi along with its speed, we need to consider the force of gravity acting on the object as it travels through the tunnel.
1. First, let's calculate the acceleration due to gravity. The formula for gravitational acceleration (g) is:
g = G * (M / r^2)
where G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2), M is the mass of the Earth (6.0 × 10^24 kg), and r is the radius of the Earth (6.4 × 10^6 m).
g = (6.67430 × 10^-11) * (6.0 × 10^24) / (6.4 × 10^6)^2
g ≈ 9.82 m/s^2 (approximated as 9.8 m/s^2)
We can use this value as the acceleration due to gravity throughout the tunnel.
2. Next, we need to calculate the distance between Boston and Delhi. The distance between two points on the surface of a sphere can be calculated using the formula:
s = r * θ
where s is the arc length (distance), r is the radius of the Earth, and θ is the angular separation between the two points. In this case, we consider Boston and Delhi to be on the same latitude since the tunnel is straight and parallel to the equator.
The angular separation between Boston and Delhi is the same as the angular separation between the two longitudes (Boston and Delhi), which is 180 degrees.
θ = 180 degrees = (π/180) radians
s = (6.4 × 10^6 m) * (π/180) radians
s ≈ 3.54 × 10^7 m (approximated as 3.54 × 10^4 km)
Therefore, the distance between Boston and Delhi is approximately 3.54 × 10^4 km.
3. Now, we can calculate the time it takes for the object to reach Delhi. We can use the second equation of motion:
s = 0.5 * a * t^2
where s is the distance, a is the acceleration, and t is the time.
Rearranging the equation, we get:
t = √(2s / a)
t = √(2 * 3.54 × 10^7 m / 9.8 m/s^2)
t ≈ 5979 seconds (approximated as 6000 seconds)
Therefore, it will take approximately 6000 seconds for the object to reach Delhi.
4. Finally, let's calculate the speed of the object when it reaches Delhi. We can use the equation:
v = a * t
where v is the velocity (speed), a is the acceleration, and t is the time.
v = 9.8 m/s^2 * 6000 seconds
v ≈ 58800 m/s (approximated as 59000 m/s)
Therefore, the speed of the object when it reaches Delhi will be approximately 59000 m/s.