Question

An aircraft flying a heading a 60° at air speed 200 ms-1 moves relative to the ground due east (909) at 225 ms-1, If the plane is flying at constant pressure, what is its rate of change in altitude assuming a steady pressure field and f 104 81? 3.

Answer 1

Given : v_air = 200 m/s , v_ground = 225 m/s , f = 10^-4 s^-1

Solution:

Visual of given situation :

Using cosine rules to find geostrophic wind speed :

v_g² = v_air² + v_ground² -2(v_air)(v_ground)cos(30)

= (200)²+(225)²-2(200)(225)cos(30)

= 12682.71

v_g = $\sqrt$12682.71 = 112.6 m/s

Geopotential height gradient can be determined as :

$v_g = \frac{g}{f} \frac{\partial z}{\partial n}$

$\frac{\partial z}{\partial n} = \frac{fv_g}{g}$

$= \frac{(10^-4)(112.6)}{9.81}$

= 1.148 × 10^-3 mm^-1

Applying sine rule to get the angle ($\alpha$) :

$\frac{v_{air}}{sin\alpha} = \frac{v_g}{sin(30)}$

$\alpha = sin^{-1}(\frac{200sin(30)}{112.6})$

= 62.6°

The component of geopotential height gradient in the direcrion of

v_ground is given by:

$\frac{\partial z}{\partial n}cos(90�+62.6�)$ = -1.0192 ×10^-3 mm^-1

The rate of change of altitude is given by:

-1.0192×10^-3* v_ground

= -1.0192×10^-3×225

= -0.229 ms^-1

Answer : -0.229 ms^-1