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FIRST STEPS TOWARD SPACE - Smithsonian Institution Libraries

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28 SMITHSONIAN ANNALS OF FLIGHT<br />

The velocity of the body falling freely from infinity to the moon would be<br />

Oao = 2370 m/sec<br />

The time used to go through the second phase can be calculated approximately by<br />

neglecting the moon's action which is entirely negligible during the total journey.<br />

It would be the same time as that taken by the body during a free fall from the<br />

moon to the point where we had stopped the engine:<br />

Third Phase<br />

t = 48 hr 30 min<br />

One must now decrease the speed by turning the body upside-down as said before,<br />

and by re-starting the motor.<br />

What will be the law of this slowing down?<br />

We would establish it in the same manner as we did for the earth; but the moon's<br />

attraction being much smaller, and as we do not at this stage seek a great precision,<br />

we will deduct from the acceleration due to the propulsor, half the acceleration due<br />

to the moon, and we will assume the motion uniformly slowed down under the action<br />

of this fictitious acceleration. We find that the body has to be turned upside-down<br />

at a distance from the moon's surface equal to<br />

d = 250,000 m approximately<br />

This point is so close to the moon, and the present calculations not being rigorous,<br />

the time necessary to reach the surface could be mistaken for the time necessary to<br />

reach the moon itself.<br />

The time of the slowing down will be<br />

t — 226 sec = 3 min 46 sec<br />

The total time for the whole process is approximately then:<br />

First phase 0 hr 24 min 9 sec<br />

Second phase 48 hr 30 min<br />

Third phase 0 hr 3 min 46 sec<br />

48 hr 58 min approximately<br />

The return trip could be done by reversing the process and in the same time.<br />

It must be pointed out that, by this means, the propulsor is used only 28 min going<br />

and the same time coming back unless the earth's atmosphere is used for the slowing<br />

down process, in which case the 28 min used for the departure, and the time necessary<br />

to orient the body properly, would suffice.<br />

We will now consider the power actually needed to realize these minimum conditions<br />

and the resulting efficiency output of the motor with respect to the theoretical<br />

work given.<br />

If we consider a 1000 kg vehicle out of which 300 kg are consumable; and if the<br />

engine has to work 27 min +3.5 min and to have a sufficient flow margin 35 min

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