Most of you may be knowing that the moon does not possess an atmosphere because the thermal velocity acquired by gas molecules on the moon when heated by the solar radiations is significant compared to the escape velocity on the moon’s surface (2.4 km/s). The escape velocity on the earth’s surface is 11.2 km/s which is much greater than the velocity acquired by oxygen and nitrogen gas molecules on getting heated by solar radiations. (In the case of hydrogen molecules, this is not the case). It is enough that the most probable velocity

[√(2RT/M)] of a gas molecule is in excess of about 20% of the escape velocity, for the molecule to escape to outer space.

Now, consider the following question:

Now, consider the following question:

**The radius of the earth is 6400 km and the acceleration due to gravity on the earth’s surface is 9.8 ms**

(a) 1.59×10

The escape velocity is given by v

^{–2}. The universal gas constant is 8.4 J mol^{–1}K^{–1}. The temperature at which the r.m.s. velocity of oxygen gas molecules becomes equal to the velocity of escape from the surface of the earth is(a) 1.59×10

^{6}K (b) 1.59×10^{5}K (c) 1.59×10^{4}K (d) 1.59×10^{3}K (e) 1.59×10^{2}K_{e}= √(2gR

_{E}) where ‘g’ is the acceleration due to gravity and ‘R

_{E}’ is the radius of the earth.

On substituting for ‘g’ and ‘R

_{E}’, the escape velocity, v

_{e}= 11.2×10

^{3}m/s

The molecular velocity (r.m.s.) is given by v = √(3RT/M) where ‘R’ is universal gas constant, ‘T’ is the temperature (in Kelvin) and ‘M’ is the molar mass of the gas (oxygen in the present case).

Therefore, √(3RT/M) = 11.2×10

^{3}. Substituting for R = 8.4 and M = 0.032 kg, the temperature works out to be 1.59×10

^{5}K.

Now, consider the following question which is based on Kepler’s law:

**A planet moves around the sun. When it is farthest away from the sun at distance r**

(a) r

_{1}, its speed is v_{1}. When it is closest to the sun at distance r_{2}its speed will be(a) r

_{1}v_{1}/r_{2}(b) (r_{1}/r_{2})^{2}v1 (c) √(r_{1}/r_{2}) ×v_{1}(d) r_{2}v_{1}/r_{1}(e) √(r_{2}/r_{1})× v_{1}According to Kepler’s law, the line joining the planet to the sun sweeps out equal areas in equal intervals of time. If we consider a very small time interval δt, the areas swept when the planet is at apogee (farthest away) and at perigee (closest to the sun) will be triangles whose areas are directly proportional to v

_{1}r

_{1}and v

_{2}r

_{2}respectively. [The bases of the triangular areas swept in the time δt are v

_{1}δt and v

_{2}δt and the altitudes are r

_{1}and r

_{2}respectively].

Therefore, from Kepler’s law, r

_{1}v

_{1}= r

_{2}v

_{2}so that v

_{2}= r

_{1}v

_{1 }/r

_{2}

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