- (24pts) Derive an equation for the escape velocity of a particle of mass m from the surface of a spherical planet having a mass m, and a radius R. The escape velocity is the minimum speed by the particle of mass m to take it from the surface of the planet to “infinity. The derivation of the equation is easily accomplished by equating the kinetic energy of the particle of mass m to the gravitational potential energy of the particle at the surface of the spherical planet. Recall that the
gravitational potential energy Gmm, /R and also recall that the gravitational acceleration constant g is g-Gmy/R
Use the following data to calculate the escape velocity of a particle from the surface of the earth and the surface of the planet mars
Planet
R(meters)
g (ms)
Earth Mars
6.32 x 10
9.81 3.73
3.38 x 10
(ii) Use the kinetic theory to determine at what temperature O, would have an average velocity equal to its escape velocity. Do this calculation for the Earth and Mars
(iv) What fraction of O, molecules at 1.25 x 10 K have a speed greater that the escape velocity. Do this calculation for both Earth and Mars.
The Correct Answer and Explanation is :
Let’s break this problem into four steps:
Step 1: Derivation of the Escape Velocity Formula
The escape velocity is the minimum velocity needed for a mass ( m ) to escape the gravitational pull of a planet with mass ( M ) and radius ( R ).
From the conservation of energy principle:
[
\text{Kinetic Energy} + \text{Potential Energy} = 0
]
At the planet’s surface:
[
\frac{1}{2} m v_{\text{esc}}^2 – \frac{GMm}{R} = 0
]
Solving for ( v_{\text{esc}} ):
[
\frac{1}{2} v_{\text{esc}}^2 = \frac{GM}{R}
]
[
v_{\text{esc}} = \sqrt{\frac{2GM}{R}}
]
Using the relation ( g = \frac{GM}{R^2} ), we substitute ( GM = gR^2 ) into the equation:
[
v_{\text{esc}} = \sqrt{2gR}
]
Step 2: Calculating the Escape Velocity for Earth and Mars
We use the given values:
For Earth:
- ( R = 6.32 \times 10^6 ) m
- ( g = 9.81 ) m/s²
[
v_{\text{esc}} = \sqrt{2 \times 9.81 \times 6.32 \times 10^6}
]
For Mars:
- ( R = 3.38 \times 10^6 ) m
- ( g = 3.73 ) m/s²
[
v_{\text{esc}} = \sqrt{2 \times 3.73 \times 3.38 \times 10^6}
]
Step 3: Temperature at Which O₂ Has an Average Velocity Equal to Escape Velocity
The average speed of a gas molecule is given by the root mean square (RMS) velocity:
[
v_{\text{rms}} = \sqrt{\frac{3kT}{m}}
]
Setting ( v_{\text{rms}} = v_{\text{esc}} ):
[
T = \frac{m v_{\text{esc}}^2}{3k}
]
where:
- ( k = 1.38 \times 10^{-23} ) J/K (Boltzmann constant)
- ( m ) is the mass of an ( O_2 ) molecule (approximately ( 5.32 \times 10^{-26} ) kg)
We’ll compute ( T ) for Earth and Mars.
Step 4: Fraction of O₂ Molecules with Speeds Greater than Escape Velocity
Using the Maxwell-Boltzmann distribution, the fraction of molecules with speeds greater than ( v_{\text{esc}} ) is:
[
f = \text{erf} \left( \frac{v_{\text{esc}}}{v_{\text{rms}}} \right) + \frac{2}{\sqrt{\pi}} \left( \frac{v_{\text{esc}}}{v_{\text{rms}}} \right) e^{-\left( \frac{v_{\text{esc}}^2}{v_{\text{rms}}^2} \right)}
]
We will compute this fraction for both Earth and Mars at ( T = 1.25 \times 10^3 ) K.
Now, let’s compute these values.
I have computed the escape velocities for Earth and Mars, the temperature at which oxygen molecules would have an average speed equal to the escape velocity, and the fraction of O₂ molecules escaping at 1250 K. You can review the results in the table provided. Let me know if you need any further explanations!