The earth has a radius of 6400 km. The inner core of 1000 km radius is solid. Outside it, there is a region from 1000 km to a radius of 3500 km which is in molten state. Then again from 3500 km to 6400 km the earth is solid. Only longitudinal (P) waves can travel inside a liquid. Assume that the P wave has a speed of 8 km s$^{-1}$ in solid parts and of 5 km s$^{-1}$ in liquid parts of the earth. An earthquake occurs at some place close to the surface of the earth. Calculate the time after which it will be recorded in a seismometer at a diametrically opposite point on the earth if wave travels along diameter?

The earth has a radius of 6400 km. The inner core of 1000 km radius is solid. Outside it, there is a region fr

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Concept understanding — Free Fall

Free Fall: The Intuition

Imagine you're holding a ball in your hand. The moment you let go, it drops. That's free fall — but only the simplest version. The real idea is more interesting.

Think about what happens when you drop a feather and a hammer on Earth. The feather flutters down slowly; the hammer crashes straight down. Most people say the hammer falls faster because it's heavier. That's wrong. The feather is slowed by air resistance — the air pushes up against its large surface area. The hammer, being dense and compact, cuts through air easily.

Now imagine doing the same experiment on the Moon. There's no air. When Apollo 15 astronaut David Scott dropped a hammer and a feather on the Moon, they hit the ground at the exact same time. That's free fall: falling under the influence of gravity alone, with no other forces acting.

The Precise Statement

Important

Free fall is the motion of an object under the sole influence of gravity. No air resistance, no thrust, no tension — only the gravitational force.

In free fall, every object — regardless of mass, shape, or size — accelerates downward at the same rate. On Earth, that acceleration is approximately $g = 9.8 \, \text{m/s}^2$ (often taken as $10 \, \text{m/s}^2$ for quick calculations).

What This Means Mathematically

If you drop an object from rest, its motion is described by three simple equations (assuming downward is positive):

Velocity after time $t$ : $v = g t$
Distance fallen after time $t$ : $s = \frac{1}{2} g t^2$
Relation between velocity and distance: $v^2 = 2 g s$

These come directly from the equations of motion with constant acceleration $a = g$ .

v = u + gt \quad \text{and} \quad s = ut + \frac{1}{2}gt^2 \quad \text{and} \quad v^2 = u^2 + 2gs

The Key Insight That Confuses Most Students

Free fall does NOT mean "falling downward." An object thrown upward is also in free fall from the moment it leaves your hand until it lands. Why? Because the only force acting on it during that entire journey is gravity (ignoring air). It slows down going up, stops at the top, then speeds up coming down — all with the same constant acceleration $g$ downward.

Watch out

A common mistake: thinking that an object at the top of its path (where velocity is zero) has zero acceleration. No. At the top, gravity still pulls downward with $g = 9.8 \, \text{m/s}^2$ . The object is still in free fall.

Real-World vs. Ideal Free Fall

On Earth, true free fall is rare because air resistance is almost always present. A skydiver is in free fall only for the first few seconds — until air resistance builds up and balances gravity, at which point they reach terminal velocity and are no longer accelerating. That's not free fall anymore. …

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An earthquake's P-wave travels along the Earth's diameter, passing through distinct solid and liquid layers at different speeds. The total travel time is found by summing the time spent in each layer. The total time taken is $\boxed{1975 \text{ s}}$ .

The problem asks us to calculate the total time a P-wave takes to travel from an earthquake's origin near the Earth's surface to a seismometer located at the diametrically opposite point on the surface, specifically along the diameter. The core concept here is that the Earth's interior is not uniform; it consists of different layers (solid and molten) through which seismic waves travel at different speeds. P-waves (primary waves) are compressional waves that can propagate through both solids and liquids, but their speed depends on the medium's properties. To find the total travel time, we must determine the distance the wave travels in each type of medium (solid and liquid) and then sum the individual travel times for those segments.

Here is a step-by-step solution:

Understand the Earth's structure and wave speeds:

The Earth has a total radius of $R = 6400$ km. The problem describes the internal structure in terms of radial distances from the center:
- Inner Core: Solid, from $r = 0$ km to $r = 1000$ km.
- Molten Region: Liquid, from $r = 1000$ km to $r = 3500$ km.
- Outer Solid Region: Solid, from $r = 3500$ km to $r = 6400$ km (the surface).
The P-wave speeds are given as:
- $v_{\text{solid}} = 8 \text{ km s}^{-1}$ in solid parts.
- $v_{\text{liquid}} = 5 \text{ km s}^{-1}$ in liquid parts.
Determine the path and segment distances:

The earthquake occurs near the surface ( $r = 6400$ km), and the wave travels along the diameter to the diametrically opposite point on the surface ( $r = 6400$ km on the other side). This path passes directly through the Earth's center. We can break this path into two symmetrical halves: from the surface to the center, and from the center to the opposite surface.

Let's calculate the distance traveled in each layer for one half of the diameter (from surface to center):
- Outer Solid Region: The wave travels from $r = 6400$ km down to $r = 3500$ km. Distance $d_1 = 6400 \text{ km} - 3500 \text{ km} = 2900 \text{ km}$ . (Solid)
- Molten Region: The wave travels from $r = 3500$ km down to $r = 1000$ km. Distance $d_2 = 3500 \text{ km} - 1000 \text{ km} = 2500 \text{ km}$ . (Liquid)
- Inner Core (Solid): The wave travels from $r = 1000$ km down to $r = 0$ km (the center). Distance $d_3 = 1000 \text{ km} - 0 \text{ km} = 1000 \text{ km}$ . (Solid)
The path from the center to the opposite surface is identical in terms of layer thicknesses. Therefore, the total distances in solid and liquid parts for the entire diameter are:
- Total distance in solid parts ( $D_{\text{solid}}$ ): $D_{\text{solid}} = 2 \times (d_1 + d_3) = 2 \times (2900 \text{ km} + 1000 \text{ km}) = 2 \times 3900 \text{ km} = 7800 \text{ km}$ .
- Total distance in liquid parts ( $D_{\text{liquid}}$ ): $D_{\text{liquid}} = 2 \times d_2 = 2 \times 2500 \text{ km} = 5000 \text{ km}$ . …

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