I would say pressure, but it might depend on what you mean by stability. Pressure will tell you how many particles and how energized those particles are, the more particles the higher the pressure, the harder it is to change or remove the atmosphere due to the larger mass of it. Also consistent pressure measurements can tell you how stable the atmosphere in a particular region is. So if stability means how consistent are the weather conditions, pressure can be a good indicator too, differences and changing pressures can cause weather catastrophes like tornados and hurricanes.
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To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,


Where,
m = Mass of spacecraft
M = Mass of Earth
r = Radius (Orbit)
G = Gravitational Universal Music
v = Velocity
Re-arrange to find the velocity



PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,


From the speed it is possible to use find the formula, so



Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.
PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is



Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.
Answer:
F = 8.6 10⁻¹² N
Explanation:
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Em₀ = U = q ΔV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v²
Em₀ = Emf
e ΔV = ½ m v²
v =√ 2 e ΔV / m
v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)
v = √(1.8075 10¹⁶)
v = 1,344 10⁸ m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Emo = U = q DV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v2
Emo = Emf
.e DV = ½ m v2
.v = RA 2 e DV / m
.v = RA (2 1.6 10-19 51400 / 9.1 10-31)
.v = RA (1.8075 10 16)
.v = 1,344 108 m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10-19 1,344 108 0.4
F = 8.6 10-12 N