Derive a formula for the escape speed of a hydrogen molecule from the Earth’s at

Question

Derive a formula for the escape speed of a hydrogen molecule from the Earth’s atmosphere

hint#1:
the kinetic energy of the molecule is mv^2/2
potential energy of the molecule in a gravitational field is U=GmM/r where G is Newton’s gravitational constant, m is the mass of the molecule, M is the mass of the Earth and r is the distance between the molecule and the centre of the Earth.
hint#2. The thermal speed of a hydrogen molecule is 3kT/2, where k is Boltzmann’s constant and T is the temperature in deg. Kelvin.).

Compute the escape speed ve at the top of the atmosphere, z 110 km; at what temperature is the thermal speed of hydrogen molecules comparable with ve? Compare the result with the escape speed from the surface of the Moon. The Earth has mass ME = 5.98 x 10^24 kg and radius RE = 6370 km, and the Moon has mass MM = 7.35 x 10^22 kg and radius RM = 1740 km. Boltzmann’s constant is 1.38 x 10-23 J. K-1.

Explanation / Answer

Escape velocity of a body on the earth = 11.2 Km/sec

we know that

PV = nRT

or (1/3)*Mass*V2rms = RT (n=1 for 1 mole)

or 2/3*(11.2)2 = RT

or 83.63/R = T

T = 904.8K

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.