1) By the

**for the orbit:***conservation of energy*
0 = K
+ V = mv

^{2 }/ r + (-GMm/r) = mv^{2 }/ r - GMm / r
Then solving
for v (the escape velocity);

[2MG/
r]

^{1/2}= v
Inserting
Martian values (for M, r);

[2MG/
r]

[2 (6.4 x 10

^{1/2}=[2 (6.4 x 10

^{23}kg) (6.7 x 10^{-11}Nm^{2}/ kg^{2})/ 3.4 x 10^{6}m ]^{1/2}
= 5.0 x 10

^{3}m/s or 5.0 km/ s
2) The problem basically uses an elementary form of Kepler's
3

T

where T is in years and a in astronomical units or AU. (1 AU = 1.496 x 10

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

^{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 10^{15 }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:T

^{2}~ a^{3}where T is in years and a in astronomical units or AU. (1 AU = 1.496 x 10

^{8}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:

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)

a1 = {(1AU)

^{3}P1]^{2}}^{1/3}
a1 =
{[0.0034 yrs. ]

^{2}}^{1/3 }
a1 = 3.45 x 10

^{6 }m
The gravitational Potential
V would be:

V = - GMm/ a1 =

- [(6.7 x 10

^{-11}Nm^{2}/ kg^{2}) (6.4 x 10^{23}kg) (1.4 x 10^{15 }kg) / (3.45 x 10^{6 }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}
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***– hence these will be the ones to escape.**

*faster than the mean speed*
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