Solutions to Escape Velocity Problems

1)      By the conservation of energy for the orbit:

 

0 =  K + V  = mv² / r +  (-GMm/r) = mv² / r -  GMm / r

 

Then solving for v (the escape velocity);

[2MG/ r]^1/2  =  v

 

Inserting Martian values (for M, r);

 

[2MG/ r]^1/2  =

  [2 (6.4 x 10²³ kg) (6.7 x 10^-11 Nm²/ kg²)/ 3.4 x 10⁶ m ]^1/2

 

=  5.0 x 10³ m/s or 5.0 km/ s

 

2)       The problem basically uses an elementary form of Kepler's  3^rd or harmonic law to obtain Deimos’ mean distance (a) and thereby estimate the value of  the gravitational potential V.  (The mass of Deimos or m = 1.4 x 10¹⁵  kg can be found from a current astronomical data table or via googling)  From the harmonic law we know that the period squared is proportional to the semi-major axis of the orbit cubed:

T² ~ a³

where T is in years and a in astronomical units or AU. (1 AU = 1.496 x 10⁸ km) Note: when we use AU and yrs. we are implicitly comparing the elements of the unknown system (e.g. Deimos  going round Mars) with the known system of the Earth going round the Sun. (For Earth, we know T = 1 year, a = 1 AU.)

    For Deimos  going round Mars, T(D) = 30.3 hours =  1.26 days. Hence, in the specific units we need:  

 

P(D) = 1.26 days/ 365 days/year = 0.0034 yrs.

    The distance or semi-major axis(a1)  for Deimos we call a1, and write:

 

(P1/ 1 yr)²  = k(a1/ 1 AU)³

with k a constant of proportionality that can be set equal to 1. Or, on solving for a1 (and remembering the units used):

a1 =  {(1AU)³  P1]²}^1/3

a1   =  {[0.0034 yrs. ]²}^1/3   

a1   =    3.45 x 10⁶  m

 

The gravitational  Potential V  would be:

V = - GMm/ a1  =

  -  [(6.7 x 10^-11 Nm²/ kg²) (6.4 x 10²³ kg) (1.4 x 10¹⁵  kg) / (3.45 x 10⁶  m) 

 

 =  - 1.8 x 10 ²²  N-m  =  - 1.8 x 10 ²²  J

 

In fact, the value of a1 is a crude approximation to the actual value of Deimos mean distance because of the basic way we computed it from the most elementary form of the 3^rd law. Industrious readers can obtain the actual mean distance via Google and use it to calculate the more accurate value of V!

 

3)      We already have the Martian escape velocity from (1) so just need to compute the mean molecular speed for oxygen. Using information in atomic mass units (from Google) and the mass of 1u of oxygen (1u = 15.994 x 1.67 x 10^-27   kg) we get:

m(O2) =   2u =  5.34 x 10^-26   kg

 

And:    6 [3kT / m]^1/2  = 

 

 6 [3 (1.38 x 10^-23   J/K) (300K)/  5.34 x 10^-26   kg]^1/2

 

Or v(O2) =   3100 m/s =  3.1 km/s 

 

The ratio of the molecular speed to the Martian escape velocity is approx..

 

 (3.1km/s)/ (5.0 km/s) =  0.62

 

From the information provided in the note we read:

“calculations show that if the mean molecular speed is as much as 1/3 the velocity of escape then the planet will lose up to half the gas from its atmosphere (of that composition) in a few weeks”

 

The fraction computed, however, is greater than three fifths – which is nearly 2x the fraction for the half the gas to be lost in a few weeks. We can therefore estimate that at this molecular speed about half the gas will escape within a week to ten days. By the Maxwell –Boltzmann distribution of velocities there will always be some significant fraction of molecules moving faster than the mean speed – hence these will be the ones to escape.