## Thursday, March 22, 2007

### Two Questions Involving Gravitation

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:
The radius of the earth is 6400 km and the acceleration due to gravity on the earth’s surface is 9.8 ms–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) 1.59×106 K (b) 1.59×105 K (c) 1.59×104 K (d) 1.59×103 K (e) 1.59×102 K
The escape velocity is given by ve = √(2gRE) where ‘g’ is the acceleration due to gravity and ‘RE’ is the radius of the earth.
On substituting for ‘g’ and ‘RE’, the escape velocity, ve = 11.2×103 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) = 11.2×103. Substituting for R = 8.4 and M = 0.032 kg, the temperature works out to be 1.59×105 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 r1, its speed is v1. When it is closest to the sun at distance r2 its speed will be
(a) r1v1/r2 (b) (r1/r2)2 v1 (c) √(r1/r2) ×v1 (d) r2v1/r1 (e) √(r2/r1)× v1

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 v1r1 and v2r2 respectively. [The bases of the triangular areas swept in the time δt are v1δt and v2δt and the altitudes are r1 and r2 respectively].
Therefore, from Kepler’s law, r1 v 1 = r2 v2 so that v2 = r1v1 /r2