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January 30, 2015

January 30, 2015

Posted by **mymi** on Tuesday, October 11, 2011 at 9:44am.

- physics -
**tchrwill**, Wednesday, October 12, 2011 at 10:51amExcuse the change in units. 60 year old habits are hard to break.

You left us without a clue as to the weight of the rocket. Propellant fractions of rockets typically range from .85 to .90, i.e., the weight of propellant divided by the fully loaded rocket stage. For this exercise, I will assume .90.

The calculations ignore the energy lost from earth's gravity and the energy gain from the lunar gravity.

The distance to be covered within the 9.5 hr time period is 382,000 km = 237,375 miles. The burnout velocity is therefore 237,375(5280)/9.5(3600) = 36,650 fps.

The rocket is already in an orbit as implied by the problem statement. Assuming the rocket is in a 100 mile high orbit, it already has a circular velocity of 25,618 fps. Therefore, the rocket must only contribute an additional 11,029 fps of velocity. (The escape velocity from this orbit is 36,229 fps.

The delta velocity, dV, derived from the rocket is given by dV = cln(Wo/Wbo where dV = the change in rocket velocity, c = the velocity of the exhaust gases, Wo = the ignition weight and Wbo = the burnout weight.

The rocket weight is 358,533/.9 kg = 398,370 lb. including propellants.

With the 4280 kg payload, the total ignition weight becomes 398,370 + 9437 = 407,807 lb.

The burnout weight is Wbo = 49,274 lb.

The dV contributed by the rocket is dV = cln(407,807/49,274) = 2.1134 making c = 11,029/2,1134 = 5439 fps

These numbers imply a straight line path to the moon, ignoring the more realistic parabolic shape of an escape trajectory.

If you uncover any inconsistencies, feel free to let me know.

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