tsiolkovsky's rocket equation question

enfield

How would the equation look if instead of knowing the effective exhaust velocity we knew the force the exhaust was exerting on the rocket.

The equation is:

[tex] \Delta V = v_e * ln(\dfrac{m_0}{m_1}) [/tex]

would [tex] \Delta V [/tex] still be proportional to the log of the initial mass over the final mass?

http://en.wikipedia.org/wiki/Tsiolko...ocket_equation

K^2

[itex]\Delta v[/itex] does not depend on thrust. Only on I_sp of the propellant. As long as I_sp is some constant, the [itex]\Delta v[/itex] will always be proportional to natural log of the mass ratio.

enfield

ah k. i followed the link to specific impulse and it helped me understand.

http://en.wikipedia.org/wiki/Specific_impulse

Specific impulse (usually abbreviated Isp) is a way to describe the efficiency of rocket and jet engines. It represents the force with respect to the amount of propellant used per unit time.[1] If the "amount" of propellant is given in terms of mass (such as in kilograms), then specific impulse has units of velocity. If it is given in terms of weight (such as in kiloponds), then specific impulse has units of time. The conversion constant between the two versions of specific impulse is g.[2] The higher the specific impulse, the lower the propellant flow rate required for a given thrust, and in the case of a rocket the less propellant is needed for a given delta-v per the Tsiolkovsky rocket equation.

and it says the effective exhaust velocity is also the thrust divided by the rate of the flow of mass from the rocket. that makes sense. (and yeah, this is with the specific impulse being constant).

cjl

Quote by enfield

ah k. i followed the link to specific impulse and it helped me understand.

http://en.wikipedia.org/wiki/Specific_impulse

and it says the effective exhaust velocity is also the thrust divided by the rate of the flow of mass from the rocket. that makes sense. (and yeah, this is with the specific impulse being constant).

Actually, your parenthetical note at the end there is somewhat irrelevant - specific impulse is directly proportional to effective exhaust velocity. The two terms are just two different representations of the same value (specifically, effective exhaust velocity = I_sp*g₀)

K^2

Depending on definition. For impulse per weight, v = I_sp*g. For impulse per mass, v = I_sp. Both conventions are used, mostly, depending on application. For rocket taking off from Earth's surface, I_sp per weight is a more directly useful quantity. For rocket accelerating in deep space, you just want the exhaust velocity, so I_sp per mass.

cjl

Quote by K^2

Depending on definition. For impulse per weight, v = I_sp*g. For impulse per mass, v = I_sp. Both conventions are used, mostly, depending on application. For rocket taking off from Earth's surface, I_sp per weight is a more directly useful quantity. For rocket accelerating in deep space, you just want the exhaust velocity, so I_sp per mass.

Nope. I_sp*g₀ = V_{e, effective}. It isn't a matter of per weight or per mass, and g₀ is strictly a conversion factor in this case. That's simply the definition of specific impulse and effective exhaust velocity. No matter where the rocket is, an I_sp of 300 seconds is exactly the same as an effective exhaust velocity of 2940 meters per second.

(Note that I use g₀ rather than g - this is because no matter where you are in the solar system (or elsewhere), g₀ = 9.8 m/s², and since it is a conversion factor rather than a variable, it is independent of the local gravity field)

K^2

Quote by cjl

Nope. I_sp*g₀ = V_{e, effective}. It isn't a matter of per weight or per mass, and g₀ is strictly a conversion factor in this case.

There are two alternative definitions of I_sp.

1) I_sp = dp/dw = (dp/dm)/g
2) I_sp = dp/dm = v_e

Both are used in the literature and you differentiate by the units. First definition gives you units of inverse seconds. Second definition gives you units of m/s and is identical to exhaust velocity for a conventional rocket.

cjl

I've never seen the second one called I_sp - everywhere I've seen it used, it was called effective exhaust velocity. If it is called I_sp anywhere, it is at least a somewhat nonstandard usage. Also, just because I'm in a somewhat nitpicky mood at the moment, it's not necessarily identical to exhaust velocity. It's identical to exhaust velocity if and only if the nozzle is perfectly expanded (and thus the pressure thrust is zero). Otherwise, there will be a difference between effective exhaust velocity and actual exhaust velocity.

D H

Quote by cjl

Nope. I_sp*g₀ = V_{e, effective}.

Not necessarily. Americans tend to specify I_sp in seconds, Europeans in units of meters/second. American engineers tend to use customary units, where there's a problem with the pound: Is it a unit of mass or a unit of force? This problem doesn't exist in SI units, and since "specific" typically means per mass, European engineers tend to specify I_sp in units of meters/second.

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K^2 Recognitions: Science Advisor	[itex]\Delta v[/itex] does not depend on thrust. Only on I_sp of the propellant. As long as I_sp is some constant, the [itex]\Delta v[/itex] will always be proportional to natural log of the mass ratio.