A simple flying maneuver

Suppose that the original lift force was $F_0$ , and that the new lift force (perpendicular to the plane) is $F_l$ . Also, let the angle above horizontal that the wings are at be $\theta$ . Note that the new vertical lift force is $F_l\cos \theta$ . Because the new and original vertical lift force must be equal, we have

$F_0=F_l\cos \theta$ .

Let the new velocity be $v_f=v+\Delta v$ . Because the lift force is proportional to the square of the velocity, we have

$\frac{F_0}{720^2}={F_l}{v_f^2}$

$\frac{F_l\cos \theta}{720^2}={F_l}{v_f^2}$

$v_f^2 \times \cos \theta=720^2$ .

Also, since the plane if flying a banked curve, we have the formula:

$\tan \theta = \frac{v^2}{rg} = \frac{v^2}{8g}$

Unfortunately, we must convert $9.8 \text{m/sec}^2$ into kilometers per hour. There are $1000$ meters in a kilometer and $(3600)^2$ square seconds in a square hour, so we multiply by $\frac{3600^2}{1000}$ and get

$g=127008$ kilometers per square hour.

Therefore,

$\theta = \tan ^{-1}(\frac{v^2}{8(127008)})$

Plugging this into the earlier equation gives us

$v_f^2 \times \cos \tan ^{-1}(\frac{v^2}{8(127008)})=720^2$ .

Solving this equation for $v_f$ gives us $v_f=776.37$ . Therefore,

$\Delta v = v_f-v=\boxed{56.4}$

The relevant forces acting on the aircraft are the lift $F_{L}$ and the weight $F_{g}=m g$ . There are two more forces acting on the plane: thrust and drag. However, they need not be taken into account as they cancel out. When the plane is flying horizontally at $v_{0}=720 km/h$ we have that $F_{g}=mg=F_{L}=k v_{0}^{2}\rightarrow k=\frac{mg}{v_{0}^{2}} .$ When the plane banks, the airspeed must be increased in order to keep the same vertical component of the lift force. From a free body diagram we see that $F_{L}^{2}= F_{g}^{2}+ F_{net}^{2}$ Here $F_{L}= k v^{2}$ while $F_{net}$ is the net force acting on the plane. Clearly, the acceleration of the plane during its turn is centripetal. From Newton's second Law we have that $F_{net}=\frac{ m v^{2}}{R}.$ Combining the above equations we find the velocity of the plane during the turn. $v= \frac{ v_{0}}{(1-\frac{v_{0}^{4}}{g^{2}R^{2}})^{1/4}}=776.4 \textrm{ km/h} \rightarrow \Delta{v}=56.4 \textrm{ km/h} .$