What is the magnetic field strength is a 45000 kgaircraft is accelerated to 260 kg/hr

Lets calculate the kinetic energy:

ek = (1/2)*45000*(260*1000/3600)^2 = 117.36 x 10^6 J

from the definition of magnetic force we have:
F = Bqv 
where F is the force, B is the magnetic field, q is the charge of the body, v i s velocity of the body.

The aircraft is going through a circular motion with radius equal to the radius of the earth plus the aircraft altitude, lets call all that radious r, therefore the aircraft feels a centripetal force:

F = mv^2/r

hence we can equal the magnetic and centripetal force:

Bqv = mv^2/r

we can solve for the magnetic field:

B = mv/rq

hence the intensity of the strength of the magnetic field is expressed in terms of the amount of charge q contained in the aircraft and the total radius of the circular motion.

B = (45000)(260*1000/3600)/rq

B = (1/rq) * (3.25 x 10^6) Teslas

At maximum height, kinetic energy is converted in potential energy  formula: at max height, ke = ½ * 0.27kg * (16m/s) ² = 34.56 j  so the remaining pe = (41.70 – 34.56)= 7.14 j = 0.27 kg * 9.8m/s² * h  = (9.8/7.14 * 0.27)so max height h = 5.08 or 5 m