At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere?

(Given : Mass of oxygen molecule (m) = 2.76 × 10⁻²⁶ kg,
Boltzmann's constant k_B = 1.38 × 10⁻²³ JK⁻¹)

Concept:

Escape velocity is the minimum velocity required by any object to overcome the gravitational pull of the planet and escape gravitational field so that the object never falls back to planet.

Formula to calculate escape velocity is \(V=\sqrt{\frac{GM}{R}}\)

where V is escape velocity,

G is universal gravitational constant which is same in whole universe.

M is mass of the planet

R is radius from center of the planet

Another formula to calculate escape velocity is \(V=\sqrt{2gR} \)

where g is acceleration due to gravity to the planet

For surface of earth, g = 9.8 ms^-2.

For earth, escape velocity is 11,186 ms^-1.

The relation between root mean square velocity and temperature is \(Vrms= \sqrt {\frac{{3{K_B}T}}{{{m_{{}}}}}} \)

where Boltzmann's constant is kB 

Vrms is root mean square velocity of the gas molecule

m is mass of the molecule 

T is temperature

To escape from surface of earth, Root mean square velocity = Escape Velocity

hence, \(V=Vrms= \sqrt {\frac{{3{K_B}T}}{{{m_{{}}}}}} \)

Calculation:

Mass of oxygen molecule (m) = 2.76 × 10−26 kg

Boltzmann's constant kB = 1.38 × 10−23 JK−1

V_escape = 11200 m/s

Say at temperature T it attains V_escape 

So, \(\sqrt {\frac{{3{K_B}T}}{{{m_{{o_2}}}}}} \) = 11200 m/s

On solving,

T = 8.360 × 10⁴ K