Radar-Guided Missile Intercept Tracking (Aerospace Engineering) Solution

Problem Overview

We calculate the true horizontal ground speed $\frac{dx}{dt}$ of an inbound supersonic aircraft flying at a fixed altitude of $y = 8.0\text{ km}$ tracked by a stationary ground radar system measuring line-of-sight slant range $z = 10.0\text{ km}$ and range closure rate $\frac{dz}{dt} = -0.40\text{ km/s}$ .

Step-by-Step Mathematical Solution

Step 1: Geometric Formulation & Implicit Differentiation

The spatial configuration forms a right triangle where the base is the horizontal ground coordinate $x(t)$ , the vertical height is the altitude $y$ , and the hypotenuse is the slant range $z(t)$ .

The governing formula is the Pythagorean Theorem:

x^2 + y^2 = z^2

(This formula and its time derivative principles are defined in Related Rates.)

Because the aircraft maintains level flight, the altitude is a static constant:

y = 8.0\text{ km} \implies \frac{dy}{dt} = 0

Substituting $y = 8.0$ into the primary relation before differentiation:

x^2 + 8.0^2 = z^2 \implies x^2 + 64.0 = z^2

Differentiating both sides implicitly with respect to time $t$ using the Chain Rule:

2x \frac{dx}{dt} + 0 = 2z \frac{dz}{dt} \implies x \frac{dx}{dt} = z \frac{dz}{dt}

Isolating the horizontal velocity $\frac{dx}{dt}$ :

\frac{dx}{dt} = \frac{z}{x} \frac{dz}{dt}

Step 2: Snapshot Evaluation of Position and Velocity

Find the snapshot ground position $x$ when $z = 10.0\text{ km}$ : $x^2 + 64.0 = (10.0)^2 = 100.0 \implies x^2 = 36.0 \implies x = 6.0\text{ km}$
Compute the horizontal ground speed $\frac{dx}{dt}$ : $\frac{dx}{dt} = \frac{10.0}{6.0} (-0.40) = \frac{5}{3}\left(-\frac{2}{5}\right) = -\frac{2}{3}\text{ km/s} \approx -0.667\text{ km/s} \quad (-666.7\text{ m/s})$
Convert to kilometers per hour: $v = 0.6667\text{ km/s} \times 3600\text{ s/hr} = 2,400\text{ km/h}$

Step 3: Compute Flight Mach Number

Given the speed of sound $a = 300\text{ m/s} = 0.300\text{ km/s}$ :

M = \frac{v_{\text{true}}}{a} = \frac{0.6667\text{ km/s}}{0.300\text{ km/s}} = \frac{20}{9} \approx 2.22

The target is flying at Mach 2.22 (supersonic).

Engineering Conclusion

In aerospace radar tracking and air defense systems, physical sensors cannot directly measure horizontal ground coordinates; antenna arrays only measure the two-way radio ping propagation time, which maps directly to slant range $z(t)$ , and the carrier frequency Doppler shift, which measures radial velocity $\frac{dz}{dt}$ .

Because $\frac{dx}{dt} = \frac{z}{x}\frac{dz}{dt} = \frac{1}{\cos\theta}\frac{dz}{dt}$ (where $\theta$ is the radar elevation angle), the raw radial velocity is always strictly less than the aircraft's actual ground velocity. At the snapshot moment ( $z = 10\text{ km}, x = 6\text{ km}$ ), the ratio $\frac{z}{x} = \frac{10}{6} = 1.67$ . The radar measures an approach rate of $400\text{ m/s}$ , but the drone is actually closing ground distance at $667\text{ m/s}$ .

If automated missile defense targeting computers naively used the raw sensor rate $\frac{dz}{dt}$ without this related rates geometric conversion layer, anti-missile interceptors would systematically underestimate the target's forward velocity and fire significantly behind the supersonic drone.