^{2}, I

_{sp}is the specific impulse of the stage in seconds, and W

_{i}is the initial mass of the stage plus its fuel and whatever else it is carrying. dV is the increase in velocity provided by the stage.

Somebody with more skill at rocket-science than me has provided an even more ferocious condemnation of Richard Hoagland's faulty mathematics in attempting to calculate the velocity addition of Stages 2,3 and 4 of Explorer 1.

He points out that, using Hoagland's notation, Konstantin Tsiolkovsky's rocket equation is normally written as:

dV = g x I_{sp}x ln(W_{i}/W_{f})

where W_{f}is the final mass at burnout (but before discarding the burnt-out stage)

He then points out that it's not permissible to aggregate the stages and evaluate the equation once. You are obliged to take it stage by stage, and this is his result:

STAGE 2

----------

W_{i}= 1020+280+80

= 1380

W_{f}= 1380 - 530 (weight of the burned fuel)

= 850

dV = 32.2 x 220 x ln(1380/850)

= 7084 x 0.482

= 3,414 ft/sec

STAGE 3

------------

W_{i}= 280+80

= 360

W_{f}= 360 - 140 (weight of the burned fuel)

= 220

dV = 32.2 x 235 x ln(360/220)

= 7567 x 0.492

= 3,723 ft/sec

STAGE 4

-----------

W_{i}= 80

W_{f}= 80 - 48.5 (weight of the burned fuel)

= 31.5

dV = 32.2 x 235 x ln(80/31.5)

= 7567 x 0.932

= 7,052 ft/sec

TOTAL delta-V

---------------------

3414 + 3723 + 7052 = 14,189 ft/sec

An additional 600 ft/sec represents 4.2% over-performance.

There is no need to resort to an anti-gravity field to account for this. Hoagland's entire thesis is therefore falsified. Cheers.

There's another, possibly simpler, way of looking at this. Since the parameters of the elliptical orbit are precisely known, the orbit velocity can be calculated from a reliable equation. Deduct the burnout velocity of the first stage -- also known, but with less precision -- and what's left is the dV of the solid upper stages. Calculations done this way were offered to the "Dark Mission" blog, but blocked. Hoagland and Bara accept nothing that challenges their ridiculous and misguided ideas.

Richard Hoagland writes that the solid upper stages of the Juno 1 rocket that launched Explorer 1 only contributed 3,520 ft/sec of the total orbital injection velocity. Well, let's see now...

DATA:

Planned orbit 220 x 1,000 miles (352 x 1,600 kM)

Actual orbit 223 x 1,592 miles (357 x 2,547 kM)

Radius of Earth 6,375 kM

Gravitational constant, µ, of Earth 398,660 kM^{3}/s^{2}

Cut-off velocity of Jupiter-C rocket (Juno's liquid first stage) 9,020 mph = 13,229 ft/sec [1]

CALCULATION:

semi-major axis of actual orbit, L_{smaj}, (357+6375+6375+2547)/2 = 7827 kM

distance from center of Earth to orbit point, R, 6375+357 = 6732 kM

velocity at orbit injection, V_{orb}= √(µ(2/R - 1/L_{smaj})) [2]

2/R - 1/L_{smaj}= 0.000169

V_{orb}= √67.493 = 8.215 kM/sec = 27,111 ft/sec (actually only 2.5% greater than planned)

less the velocity achieved by the first stage: 13,229 ft/sec

Ladies and gentlemen, get out your calculators! In about 3 seonds you will falsify Hoagland's entire theory.

[1] http://www.astronautix.com/lvs/jupiter.htm

[2] http://en.wikipedia.org/wiki/Orbital_velocity

There's a small discrepancy between the two calculations, to be sure -- due to uncertainty about the velocity at first-stage burnout, perhaps.

Other factors that would be considered in a really accurate calculation:

* Correction to the value of g as the rocket ascends away from Earth -- a small positive vertical increment

* Boost given by rotation of Earth -- separately calculated as 1100 ft/sec horizontal

* Gravity drag -- perhaps negative 1000 ft/sec vertical

However, the discrepancy is nothing to compare with the difference between 14,000 and 3,520.

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