NASA formula, can someone decipher

[Pekis]

(04-20-2020 02:35 AM)twdeckard Wrote:  I'll throw in a few discordant guesses.

The e sin E - peaking thru is Keplers equation.

Wdot (first derivative of W, is the rate of change of vehicle or propellant weight)
Wb is the baseline weight of the vehicle at launch or at insertion.
G is gravity to turn the weights into masses.
The exponential decay is the propellant mass fraction
I say the lower case t is time

For a given change in velocity how much time elapses? a guess.

Todd

Hello,

It seems it's the case, according to http://www.braeunig.us/space/propuls.htm , derivating from the Tsiolkovsky's rocket equation:

For many spacecraft maneuvers it is necessary to calculate the duration of an engine burn required to achieve a specific change in Velocity:

t = m0 / q * (1 - exp( -DeltaV / C)

Isp = F / ( q * g0) (specific impulse of a rocket)
C = Isp * g0 [=> F / q] (effective exhaust gas Velocity)
F is thrust, q is the rate of mass flow, and g0 is standard gravity (9.80665 m/s2) [on Earth]

If q is an alias for WDOT
If g0 is an alias for g
If F is an alias for TH * g (?)
If m0 is an alias for WB

Then [Delta]t is indeed WB / WDOT * (1 - exp( -WDOT * DeltaV / (g * TH))


PROBLEM 1.4

A 5,000 kg spacecraft is in Earth orbit traveling at a velocity of 7,790 m/s.
Its engine is burned to accelerate it to a velocity of 12,000 m/s placing it
on an escape trajectory. The engine expels mass at a rate of 10 kg/s and an
effective velocity of 3,000 m/s. Calculate the duration of the burn.

SOLUTION,

Given: M = 5,000 kg
q = 10 kg/s
C = 3,000 m/s
DeltaV = 12,000 - 7,790 = 4,210 m/s

Equation (1.21),

t = M / q × [ 1 - 1 / e(DeltaV / C) ]
t = 5,000 / 10 × [ 1 - 1 / e(4,210 / 3,000) ]
t = 377 s