Tunnel through the Earth

What Happens If We Drop a Ball Through a Tunnel in the Earth?

Imagine creating a tunnel straight through the Earth and dropping a ball or any other object into it. How would gravity affect the ball, and would it emerge on the other side? If so, how long would it take?

Tunnel through the Earth: Let’s Go Inside?

This question is often posed in competitive exams and for general curiosity. This article provides a detailed analysis of the behavior of objects in a tunnel through the Earth. We assume no frictional forces between the Earth’s inner surface and the object, meaning there’s no energy loss. We also assume the object is released from rest at one end of the tunnel (i.e., from the Earth’s surface) and that the Earth’s density is uniform (which it isn’t, actually).

Gravity and Acceleration

At the Earth’s surface, the acceleration due to gravity $g$ is about $9.8 \, \text{m/s}^2$, derived from the formula $g = G \dfrac{M}{R^2}$, where $G$ is the universal gravitational constant ($G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 \cdot \text{kg}^{-2}$), $M$ is the Earth’s mass, and $R$ is the Earth’s radius. As the object descends into the Earth, $g$ decreases linearly according to $g = G \dfrac{M}{R^2} \left(1 – \dfrac{r}{R}\right)$, where $r$ is the depth traveled by the object. Thus, at half the Earth’s radius ($r = R/2$) and at the center of the Earth ($r = R$), gravitational acceleration reduces to half and then to zero, respectively.

Simple Harmonic Motion

The force acting on the object is $F = \text{mass} \times \text{acceleration} = m \times (-g’)$. Here, $g’$ is the reduced gravitational acceleration acting on the object. Since $g’ = g'(r)$, the force can be expressed as $F = m \times -g'(r) = -kr$, where $k$ is a constant depending on $m$, $M$, and $R$. This attractive force, directly proportional to the displacement from the reference point, means the object executes Simple Harmonic Motion (SHM).

The time period $T$ for this motion is given by:

$$T = 2 \pi \sqrt{\dfrac{m}{k}} = 2 \pi \sqrt{\dfrac{mR}{mg}}$$
(Since $-mg = -kR$ and $-mg’ = -kr$).

Thus,
$$T = 2 \pi \sqrt{\dfrac{R}{g}} \approx 5068 \, \text{seconds} \approx 84.47 \, \text{minutes}$$

This tells us the object will emerge on the other side after about 42.235 minutes and return after another 42.235 minutes.

Remark

The above analysis assumes a homogeneous Earth, where density is uniform at $5517 \, \text{kg/m}^3$. However, the Earth is not homogeneous. Its density varies from $3350 \, \text{kg/m}^3$ near the surface to $5560 \, \text{kg/m}^3$ at a depth of 2891 km, then jumps to $9900 \, \text{kg/m}^3$ and increases to $13090 \, \text{kg/m}^3$ at the center. Consequently, the value of $g$ stays around $9.8 \, \text{m/s}^2$ up to 2891 km depth, rises to $10.68 \, \text{m/s}^2$ at that depth, and then decreases to zero at the center (according to Preliminary Reference Earth Models). Using a different approach, I found that the time for a roundtrip is about 5060 seconds for a homogeneous Earth and 458 seconds for an inhomogeneous Earth. The value of 5060 seconds is close to the 5068 seconds obtained in the homogeneous model, which can be easily verified.