Numbers in the appendix of Arthur C. Clarke's 1945 classic paper

MB2012

1. The problem statement, all variables and given/known data

This question relates to the numbers in the appendix of Arthur C. Clarke's 1945 classic paper (Wireless World, Oct. 1945, pp. 305-308).

The appendix (on page 308 of the paper) outlines some basic principles of rocket design, including Tsiolkovsky's fundamental equation of rocket motion which is given as:

V = v log_e (R)

where V is the final velocity of the rocket, v is the exhaust velocity, and R is the ratio of initial mass to final mass (payload plus structure).

Clarke states that: "If we assume v to be 3.3 km/sec. R will be 20 to 1." (For V = 10 km/sec.)

Clarke goes on to state that due to the rocket's finite acceleration, it loses velocity as a result of gravitational retardation and the necessary ratio R_g is increased to

R_g = R * ( (a + g) / (a) )

where a is the rocket's acceleration and g is the acceleration due to gravity.

Clarke goes on to say that (and this is where my problem is): "For an automatically controlled rocket a would be about 5g and so the necessary R would be 37 to 1."

How does the necessary R come to be 37 to 1?


2. Relevant equations

Tsiolkovsky's fundamental equation of rocket motion:

V = v log_e (R)

where V is the final velocity of the rocket, v is the exhaust velocity, and R is the ratio of initial mass to final mass (payload plus structure).

R_g = R * ( (a + g) / (a) )

where a is the rocket's acceleration and g is the acceleration due to gravity.



3. The attempt at a solution

V = v * log_e (R)

10 = 3.3 * log_e (R)

R = e^(10/3.3)

R = 20.7

QED (R is approximately 20 to 1)

R_g = R * ( (a + g) / (a) )

R_g = R * ( (5g + g) / (5g) )

R_g = R * (6g / 5g)

R_g = 1.2 * R

(EQUIVALENTLY: R = 0.83 * R_g)

How does R become 37 to 1?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution