UNIVERSE LL W/SAPLINGPLUS MULTI SEMESTER
UNIVERSE LL W/SAPLINGPLUS MULTI SEMESTER
11th Edition
ISBN: 9781319278670
Author: Freedman
Publisher: MAC HIGHER
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Chapter 13, Problem 66Q
To determine

Escape speed on Titan and the value for limiting molecular mass of a gas, which can be retained under the effect of gravitational pull of Titan. Assume that the average atmospheric temperature on Titan is 95 K.

Expert Solution & Answer
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Answer to Problem 66Q

Solution:

2.6 km/s, 2.05×1026 kg

Explanation of Solution

Given data:

The average atmospheric temperature is 95 K.

Formula used:

The formula of the escape velocity is as:

vescape=(2GMR)12

Here, vescape is the escape velocity on Titan, M is the mass of Titan, R is the radius of Titan, and G is the universal gravitational constant having value 6.67×1011 Nm2/kg2.

The requirement for the retention of gas is,

vescape>6(3kTm)12

Here, m is the atomic mass of the gas, T is the temperature of the Titan’s surface, μ is the molecular weight of the gas, and k is Boltzmann constant having value 1.38×1023 m2kg/s2K.

Explanation:

The diameter of Titan is 5150 km and the mass of Titan is 1.34×1023 kg.

Since, radius is half of the diameter, the radius of Titan is,

R=D2

Here, D is the diameter of Titan.

Substitute 5150 km for d.

r=5150 km2(1000 m1 km)=2575 km 

Recall the formula of the escape velocity.

vescape=(2GMR)12

Substitute 2575.5 km for R, 1.34×1023 kg for M and 6.67×1011 Nm2/kg2 for G.

vescape=(2(6.67×1011 Nm2/kg2)(1.34×1023 kg)2575 km)12=(1.79×10132575.5 km(1000 m1 km))12=(6966530.771)12=2.6×103 m/s

Further solve as,

vescape=2.6×103 m/s(1 km1000 m)=2.6 km/s

Recall the requirement for the retention of gas.

vescape>6(3kTm)12

Substitute 1.38×1023 m2kg/s2K for k, 95 K for and 2.6×103 m/s for vescape.

2.6×103 m/s>6(3(1.38×1023 m2kg/s2K)(95 K)m)122.6×103>6(3.933×1021m)12

Squaring both sides.

(2.63×103)2>36(3.933×1021m)m>1.416×10196.91×106m>2.05×1026 kg

Conclusion:

Therefore, the gas with molecular mass greater than 2.05×1026 kg is retained.

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