Gravity used in rocket equation

[jasonjason]

Hi

Can anyone explain why earth gravity is used in rocket equation calculations when landing on mars, and not use mars gravity instead?

[93143]

Gravity isn't used at all in rocket equation calculations.

deltaV = v_exh*ln(M0/M1)

The only place it could show up is in the conversion of Isp into v_exh, which involves multiplying by Earth gravity - because the definition of Isp as lbf·s/lbm assumes Earth gravity equivalence between lbf and lbm (or kgf and kg; it doesn't actually matter).  So you're removing Earth gravity from the value before using it in the equation.

[cpooley]

Another way to put it is g is the needed conversion from the absolute unit of force to the force expressed in units of weight.  So g*Newtons = kg force.  Or 1 kg exerts 9.82 Newtons per kg of mass due to acceleration of Earth gravity. 

The US English units equivalent is poundal, pound force.

[jasonjason]

If your landing on Mars and you want to determine the Delta V why does Ve not use 3.711 as opposed to 9.8 since mars is doing the pulling?

[jasonjason]

As an example

In http://www.ssdl.gatech.edu/papers/conferencePapers/IEEE-2009-1219.pdf  page 3 

it has  an example of a propulsive landing on Mars and use this to calculate Mp propellant mass

m = T /Isp* g0  and I would think Mars gravity would be used but most examples show earth, why is this?

[QuantumG]

ISP has nothing to with gravity.. it's just easier to talk about numbers like "325" than what the actual exhaust velocity of the rocket is.

Some Europeans, most notably Russians, don't even use ISP.

It's just a convention.

[NovaSilisko]

it has  an example of a propulsive landing on Mars and use this to calculate Mp propellant mass

m = T /Isp* g0  and I would think Mars gravity would be used but most examples show earth, why is this?

If you use mars gravity to calculate mass, you'll run into problems. Mass doesn't change under gravity. Weight changes.

[jasonjason]

Thanks

I can see why mass won't change, but if you were on mars you would not plug 9.8 in for gravity you would use 3.711 so why is it not used when you are in the gravity field of mars? Why is 9.8 still used?

[Proponent]

As 93143 said above, what matters in the rocket equation is exhaust velocity:  Δv = v_exh ln(M₀/M₁).  Engineers working on Earth have found it convenient to define specific impulse as I_sp = v_exh / g_Earth, so they sometimes write the rocket equation as Δv = g_Earth I_sp ln(M₀/M₁).

If there were rocket engineers on Mars, they would find I_sp values about three times higher than those found by their counterparts on Earth, because g_Mars is about 1/3 g_Earth.  But then when they plugged those values into the rocket equation written for Martian gravity, they would get the same Δv values as their terrestrial colleagues.  What matters is the value of g in the labs where propulsion system performance is measured, not the value where the propulsion system is used.

Let's put it another way.  Never mind Mars, what about deep space, where there is no gravity.  Do you want to use g=0?  Of course not.

[jasonjason]

Thank you

Now I get it.

[jongoff]

As 93143 said above, what matters in the rocket equation is exhaust velocity:  Δv = v_exh ln(M₀/M₁).  Engineers working on Earth have found it convenient to define specific impulse as I_sp = v_exh / g_Earth, so they sometimes write the rocket equation as Δv = g_Earth I_sp ln(M₀/M₁).

One reason I still prefer Isp to v_exh is that "v_exh" isn't really the speed of the exhaust products leaving the nozzle. It's the "effective exhaust velocity" which factors in things like backpressure losses. If you could actually measure the average velocity of particles in the exhaust, you'd find that it only equaled v_exh when the nozzle was perfectly optimally expanded (ie where the outlet pressure equaled the ambient pressure). If you're overeexpanded, the actual exhaust velocity you'd measure would be higher than v_exh, and if you're underexpanded, the actual exhaust velocity you'd measure would be lower than the v_exh. Since v_exh isn't something you can actually measure per se, I prefer Isp. It at least is straightforward--the thrust at that altitude divided by the 1g weight-flow of propellant to get that thrust level. Since most rocket test stands are on earth, near sea-level, and since you can simulate most relevant back pressures reasonably well with the appropriate stand, that seems more practical.

Sorry for the anti-metric-snob pedantic rant, but sometimes dead monarch units really do make more physical sense.

~Jon

[Proponent]

But that's not really a units thing, is it?  As long as we quote effective exhaust velocity rather than actual exhaust velocity (which is itself still an abstraction, since velocity isn't uniform across the nozzle exit), we're talking about the relevant quantity.

The place that I find Isp more intuitively useful than effective exhaust velocity is in eye-balling launch-vehicle designs.  And that's because one of the key parameters, the initial thrust-to-weight ratio, generally lies within a pretty narrow range.  If R is the thrust-to-weight ratio and t is the time of powered flight, then the rocket equation becomes Δv = - g I_sp ln(1 - R t / I_sp).

PS  But sometimes it's just plain fun to use inchic units.  I don't know why, but sometimes I get a bit of a thrill out of converting pounds-mass to slugs or pounds-force to poundals.  Maybe sometime when I'm bored I'll try to work out a consistent set of English units for electromagnetic quantities.

[cpooley]

As in the second post, both ways to express Isp are valid.  Put it this way:  Impulse is the product of force and time.  It can be in N-sec or kg-sec or pound-sec.  So Isp is N-s/kg of propellant, Kg-s/kg, pound-sec/pound
The 1st one gives a number equal to exhaust velocity in m/sec and the other 2 the lower number.

So 300 isp in the 2nd two ways is equivalent to 2946 for metric and 9660 in English units.  Meters/sec or feet/sec.   N-sec/kg  or poundals-sec/pound.

[Proponent]

Or pound-seconds per slug!  I wonder whether Tsiolkovsky ever did rocket calculations in traditional Russian units.

[pippin]

Can't be any worse than imperial American units