### Author Topic: Basic Rocket Science Q & A  (Read 277112 times)

#### kevin-rf

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##### Re: Basic Rocket Science Q & A
« Reply #840 on: 10/11/2013 10:42 PM »
um physics and calc.

rise time = Initial V / acceleration

In this case acceleration would be a constant 9.81 meters/second

Height = acceleration * time^2 / 2

Gets a bit more complicated for rockets that go to very high altitudes

Going straight up at an orbital velocity of 8 km/s using a constant G of 9.81 (it will drop off a little) you would get a height ~3260 km.
« Last Edit: 10/11/2013 10:49 PM by kevin-rf »
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#### deltaV

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##### Re: Basic Rocket Science Q & A
« Reply #841 on: 10/11/2013 10:48 PM »
How do you calculate the maximum altitude if you know the speed for a suborbital rocket? You would go straight up and omit all propellant and rocket weight kind of things.

If your max altitude is relatively small (at most a few hundred km) you can get a good approximation using little-g as described by kevin. Equation (4) in http://physics.bu.edu/~redner/211-sp06/class01/equations.html makes things easier since it doesn't involve time.

For higher altitude apogees or more accurate calculations you can use conservation of energy. The specific orbital energy GM/r + v^2/2 is invariant during free fall, where G is the universal gravitational constant, M is the mass of Earth, r is the distance from the center of the Earth and v is the speed. You know r and v at launch and v is zero at apogee (when firing straight up) so you can solve for radius at apogee.

Quote
How high could you go with an orbital speed rocket?

If you shoot a rocket capable of LEO (and no more) straight up it'll reach apogee at a distance from the center of the Earth equal to twice the radius of the Earth. (This is an approximation ignoring complications such as the rotation of the Earth and gravity drag.)

Derivation: the vis-via equation (http://en.wikipedia.org/wiki/Vis-viva_equation) is: v^2 = GM(2/r - 1/a) where v is the speed of a satellite at some point in time, r is the satellite's distance to the center of the Earth at that time, and a is the semi-major axis of its orbit. We'll use this equation a couple of times. Note that a rocket launched straight up and one launched downrange (for LEO) end up with the same v and r (ignoring rotation of Earth and assuming impulsive transfer) so by the vis-via equation the semi-major axis of the resulting orbit is independent of the direction of launch. It's obvious from the definition of semi-major axis that a low earth orbit has semi-major axis equal to the radius of the Earth (plus a small altitude that we can ignore). Since we launch straight up at apogee speed equals zero so we can solve the vis-viva equation for the apogee altitude:
0 = v_app^2 = G M (2/r_app - 1/r_LEO)   ==>   r_app = 2 r_LEO.    Q.E.D.

The vis-via equation can also be used to answer questions such as how much harder (in terms of initial speed) it is to escape Earth's gravity than to orbit it.
« Last Edit: 10/11/2013 10:52 PM by deltaV »

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##### Re: Basic Rocket Science Q & A
« Reply #842 on: 10/12/2013 10:24 AM »
Thanks  for answers! The purpose of my question was to figure if you have an orbital tourist vehicle and instead of going to orbit you'd like to do as high suborbital trip as possible and see the Earth as a globe.

#### M129K

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##### Re: Basic Rocket Science Q & A
« Reply #843 on: 10/12/2013 10:32 AM »
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
You mean gravity losses? That's the integral of  g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.
If I understand it correctly, t is total time? If so, thank you! Seems very accurate.
Nope, h(t) and rho(t) are functions of t. Thus, you have to integrate from t=0 to T  the function over dt.
Thanks... I guess I'll have to open my math books again.

#### baldusi

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##### Re: Basic Rocket Science Q & A
« Reply #844 on: 10/13/2013 12:24 AM »
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
You mean gravity losses? That's the integral of  g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.
If I understand it correctly, t is total time? If so, thank you! Seems very accurate.
Nope, h(t) and rho(t) are functions of t. Thus, you have to integrate from t=0 to T  the function over dt.
Thanks... I guess I'll have to open my math books again.
Throttle and rho are your variables for a launch trajectory (if done in 2D, i.e. no plane changes).
g: 9.8*(R^2)/[(R+h)^2]
R: 6371 (in km, the average radius of the Earth)
« Last Edit: 10/14/2013 04:40 PM by baldusi »

#### deltaV

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##### Re: Basic Rocket Science Q & A
« Reply #845 on: 10/13/2013 03:46 PM »
Throttle and rho are your variables for a launch trajectory (if done in 2D, i.e. no plane changes).
g: 9.8*sqrt(R)/sqrt(R+h)
R: 6371 (in km, the average diameter of the Earth)

AIUI the symbol "g" is usually used for the standard acceleration due to gravity near the surface (about 9.8 m/s/s), not the acceleration anywhere else. More importantly the acceleration at altitude is (assuming spherical Earth): (9.8 m/s/s)*(R/(R+h))^2 --- i.e. square, not square-root. Gravity is proportional to distance raised to the -2 power (i.e. inverse square law): http://en.wikipedia.org/wiki/Newtonian_gravity .

Edit: also that's the radius of Earth, not the diameter.
« Last Edit: 10/13/2013 11:55 PM by deltaV »

#### baldusi

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##### Re: Basic Rocket Science Q & A
« Reply #846 on: 10/14/2013 04:25 AM »
Throttle and rho are your variables for a launch trajectory (if done in 2D, i.e. no plane changes).
g: 9.8*sqrt(R)/sqrt(R+h)
R: 6371 (in km, the average diameter of the Earth)

AIUI the symbol "g" is usually used for the standard acceleration due to gravity near the surface (about 9.8 m/s/s), not the acceleration anywhere else. More importantly the acceleration at altitude is (assuming spherical Earth): (9.8 m/s/s)*(R/(R+h))^2 --- i.e. square, not square-root. Gravity is proportional to distance raised to the -2 power (i.e. inverse square law): http://en.wikipedia.org/wiki/Newtonian_gravity .

Edit: also that's the radius of Earth, not the diameter.
Correct in all numerical issues. I simply can't do math on an iPad. So difficult to type that I confuse my own writing. Will correct my post. I thought that Earth surface gravity was g0, since G is the gravitational constant, and I'm use to single letters for functions.
« Last Edit: 10/14/2013 04:33 AM by baldusi »

#### deltaV

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##### Re: Basic Rocket Science Q & A
« Reply #847 on: 10/14/2013 02:07 PM »
Your corrected radius of the Earth (3185 km) is wrong. You had the right radius the first time, you just labeled it wrong.

#### aero

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##### Re: Basic Rocket Science Q & A
« Reply #848 on: 12/07/2013 05:58 AM »
If I had a suitable canon and a totally frictionless shell, no drag of any sort on the shell, say it behaved as a point mass then I shot the shell straight up with an initial velocity of 1.908 km/s (delta-V), the shell would come to a stop at 200 km altitude. Is this 1.908 delta-V the same thing that is called "gravity drag?"
« Last Edit: 12/08/2013 06:00 PM by Chris Bergin »
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#### sdsds

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##### Re: Basic Rocket Science Q & A
« Reply #849 on: 12/07/2013 07:41 AM »
If I had a suitable canon and a totally frictionless shell, no drag of any sort on the shell, say it behaved as a point mass then I shot the shell straight up with an initial velocity of 1.908 km/s (delta-V), the shell would come to a stop at 200 km altitude. Is this 1.908 delta-V the same thing that is called "gravity drag?"

My longer answer: I would not call that "gravity drag" because gravity drag happens over the interval of a propulsive maneuver, and your canon provides its shell with an impulsive (i.e. instantaneous) maneuver. Gravity drag happens when, for example, a vehicle hovers at a fixed altitude. In that case, all of the thrust is lost to gravity drag. In general, gravity drag is delta-v spent fighting gravity.

Note also your shell is going to get back again as it falls all the velocity it lost on ascent. Gravity drag is lost forever.
« Last Edit: 12/07/2013 07:43 AM by sdsds »
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#### Hop_David

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##### Re: Basic Rocket Science Q & A
« Reply #850 on: 12/07/2013 12:48 PM »
If I had a suitable canon and a totally frictionless shell, no drag of any sort on the shell, say it behaved as a point mass then I shot the shell straight up with an initial velocity of 1.908 km/s (delta-V), the shell would come to a stop at 200 km altitude. Is this 1.908 delta-V the same thing that is called "gravity drag?"

Given those conditions, it would be better to send the shell along a horizontal path. Accelerating it to 7.96 km/s would result in an orbit with an apogee at 200 km and a perigee at 0 km.

At apogee the circularization burn is .06 km/s. So total delta v from surface to a circular 200 km altitude orbit would be 8.02 km/s.

At 0 km altitude, orbit speed is 7.9 km/s. Going from a 0 km altitude orbit to a 200 km altitude takes .12 km/s.

------

But earth isn't frictionless. Before accelerating to those velocities we must first climb above the atmosphere. During vertical ascent the rocket's thrust has a vertical component. Let's say vertical acceleration is 19.6 meters/sec^2. Subtract gravity's 9.8 meters/sec^2 from that and your net acceleration is 9.8 meters/sec^2. This would be a thrust to weight ratio of 2.

I like to round 9.8 to 10. Accelerating vertically for 3 minutes would incur a gravity loss of about 180 seconds * 10 meter/sec^2. -- about 1.8 km/s.

It's easy to see longer vertical ascents result in higher gravity loss. To make ascent brief, a lower stage should have a high thrust to weight ratio. Which encourages lower ISP propellants and also encourages more rocket engines.

#### aero

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##### Re: Basic Rocket Science Q & A
« Reply #851 on: 12/07/2013 04:36 PM »
Quote
I like to round 9.8 to 10. Accelerating vertically for 3 minutes would incur a gravity loss of about 180 seconds * 10 meter/sec^2. -- about 1.8 km/s.

Yes, but you'd only reach 162 km in altitude. To reach 200 km you need to accelerate for 200 seconds which is 2 km/s delta-V, using 10 m/s^2 acceleration.

I think the answer boils down to this: Yes, some of the gravity loss goes toward raising the vehicle to orbital altitude, or staging altitude for stage 1, and some of the gravity loss results from holding the vehicle up against the pull of gravity as it accelerates to final horizontal velocity where centrifugal force counterbalances gravity. The more time this takes, the greater this gravity loss. We lump all of this loss together and name it gravity drag.

But no matter how gravity drag is sliced up, the US and payload at 200 km altitude has a gravitational potential energy sufficient to impart 1.9 km/s velocity if it should somehow fall straight down. Looking at conservation of energy, that same delta V is required to raise it up, no matter the path.

My point and the reason for this question in the first place is: Doesn't physics require a minimum gravity drag of 1.9 km/s delta-V, no matter what rocket or canon you use? It follows that people who bandy about a gravity drag of less than 1.9 km/s for their rocket are wrong or more charitably, are being overly optimistic.

Edit add - But I don't understand your example of shooting the shell horizontally. Raising the perigee I got, but getting around the need to add the potential energy of altitude went right over my head. And LEO orbital velocity does not seem to be a fixed number even for a fixed 200 km altitude. I mean, different sources give different values, with 7.8 km/s being common. I guess I'll have to calculate it from first principles to get my own value.
« Last Edit: 12/07/2013 04:52 PM by aero »
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#### aero

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##### Re: Basic Rocket Science Q & A
« Reply #852 on: 12/07/2013 07:26 PM »
I did that, calculate orbital velocity and got the value of 7783.09281 m/s at 200 km altitude. I used this:

centrifugal acceleration equals the acceleration of gravity in orbit.
centrifugal acceleration equals v^2/r, where r = re +h, re = 6371000 meters, h=200000 meters, and v is tangential velocity.
gravitational acceleration at altitude h  is equal to g0 *(re/(re + h))^2, g0 = 9.80665 m/s^2

equating and simplifying/rearranging, v=sqrt (g0*re^2/(re + h)) = 7783.1 m/s, or 7.7831 km/s at 200 km altitude. Note. At the surface, h=0 and this simplifies to v=sqrt(g0*re) = 7904.313199 = 7.9043 km/s

So now I have a handy dandy little formula to tell me orbital velocity around Earth at my altitude of choice without asking or looking it up.

« Last Edit: 12/07/2013 07:42 PM by aero »
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#### Hop_David

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##### Re: Basic Rocket Science Q & A
« Reply #853 on: 12/07/2013 10:07 PM »
My point and the reason for this question in the first place is: Doesn't physics require a minimum gravity drag of 1.9 km/s delta-V, no matter what rocket or canon you use? It follows that people who bandy about a gravity drag of less than 1.9 km/s for their rocket are wrong or more charitably, are being overly optimistic.

Edit add - But I don't understand your example of shooting the shell horizontally. Raising the perigee I got, but getting around the need to add the potential energy of altitude went right over my head.

Orbital energy is a combination of kinetic and potential energy.

Potential energy is -GMm/r. In earth's case the difference in potential energy between 6378 and 6578 km from center is about 1900 joules per kilogram.

To go from a 0 km circular orbit (7.9 km/s) to a 0 x 200 elliptical orbit takes an acceleration of .06 km/s. Kinetic energy is 1/2 mv2. The difference between 1/2 (7.9 km/s)2 and 1/2 (7.96 km/s)2 is 480 joules per kilogram. A .6 circularization burn at apogee takes velocity from 7.72 to 7.78 km/s, another 470 joules per kilogram.

The difference in kinetic energy between a 0 km orbit (7.9 km/s) and a 200 km orbit (7.78 km/s) is 950 joules per kg. Spending 950 joules to get a kilogram from an orbit where it's kinetic energy differs by 950 joules, the total energy difference is 1900 joules per kilogram.

If you're already going fast, a little increase in speed gives you a lot of extra energy. Also known as the Oberth benefit.
« Last Edit: 12/07/2013 10:08 PM by Hop_David »

#### aero

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##### Re: Basic Rocket Science Q & A
« Reply #854 on: 12/08/2013 02:02 AM »
Yes, I know about the Oberth benefit. It has to do with work done by the force of rocket thrust. Since work equals force times distance, applying that force while moving at high speed means the force will act through a longer distance hence doing a lot more work and adding a lot more energy than when that same force (engine burn) is applied at a slow speed. And the speed of a vehicle in orbit is highest at periapsis so that is the place to apply the force for maximum work hence orbital energy added.

What is truly counter intuitive is the result of adding 950 J kinetic energy to reap 1900 J of potential energy. That potential energy would convert to 1.91 km/s if it could respond to the gravity field and fall, (your 1900 J). Atmosphere does not enter because of course you could repeat the same steps while raising the 200 km orbit to a 400 km orbit with very little difference in the delta-V’s and arrive at a very similar result.

So, back to finding the minimum possible gravity drag for a launch vehicle, does your example above mean that gravity drag must always exceed 0.48 km/s? Even though it is not a very achievable minimum, it is good to know that gravity drag numbers larger than that are subject to reduction by “What,” more powerful boosters and wiser choice of launch trajectory profile. Traded against atmospheric drag of course.

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#### Hop_David

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##### Re: Basic Rocket Science Q & A
« Reply #855 on: 12/08/2013 03:17 AM »
What is truly counter intuitive is the result of adding 950 J kinetic energy to reap 1900 J of potential energy.

I made an arithmetic error. My spreadsheet uses units of kilograms, kilometers and seconds. In converting kg * km2/sec2 to joules I multiplied by 1,000 when I should have multiplied by 1,000,000.

But since I made the same error in potential as well as kinetic energies, the two sides of the ledger still came out.

That potential energy would convert to 1.91 km/s if it could respond to the gravity field and fall, (your 1900 J).

As it's falling potential converts to kinetic energy. and kinetic energy  is 1/2 mv2

1900000 joules = 1/2 kg v2
3800000 joules =  kg v2
sqrt(3800000 joules/kg) = v
v = 1949 meters/sec

So your 1.91 km/s sounds in the right ball park.

Atmosphere does not enter because of course you could repeat the same steps while raising the 200 km orbit to a 400 km orbit with very little difference in the delta-V’s and arrive at a very similar result.

Periapsis and apoapsis altitudes can be the entered in the pink cells (F38 and F39) of my spreadsheet. Periapsis and apoapsis circularization burns can be seen at cells J38 and J39. Entering 200 and 400 into periapsis and apoapsis altitudes gives me circularization burns of .0576 and .0581. So I'd agree with you, the burns are nearly the same.

So, back to finding the minimum possible gravity drag for a launch vehicle, does your example above mean that gravity drag must always exceed 0.48 km/s? Even though it is not a very achievable minimum, it is good to know that gravity drag numbers larger than that are subject to reduction by “What,” more powerful boosters and wiser choice of launch trajectory profile. Traded against atmospheric drag of course.

We're attaching different meanings to the term. What I call gravity drag (or gravity loss) is loss incurred while doing an acceleration with a vertical component.

On worlds with an atmosphere, a vertical ascent must be made to get above the atmosphere.

Even on our airless moon, a departing rocket's thrust must have a vertical component or the rocket will fall to the surface. Unless the rocket is on a nearly frictionless track that allows horizontal acceleration over a long distance. In this case, gravity loss would be virtually zero.
« Last Edit: 12/08/2013 03:20 AM by Hop_David »

#### aero

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##### Re: Basic Rocket Science Q & A
« Reply #856 on: 12/08/2013 07:06 AM »
Quote
We're attaching different meanings to the term.

Are you sure? My position was that when the vertical component of displacement was considered, ~1.9 km/s delta V is needed to reach 200 km. Of course, to do that with a real rocket takes time and gravity will act during that time causing additional losses but my canon example eliminated that time to accelerate. During the 210 second freefalling coast upward, the canon shell would need to accelerate horizontally to orbital velocity but that could be done by most boosters, if only horizontal acceleration were needed. (I'd have to check the available fuel on the booster.)

You countered that by using my canon, the altitude could be gained more cheaply by firing horizontally at orbital velocity so that centrifugal acceleration countered gravity then using orbital mechanics to gain the altitude.

Quote
What I call gravity drag (or gravity loss) is loss incurred while doing an acceleration with a vertical component.

On worlds with an atmosphere, a vertical ascent must be made to get above the atmosphere.

I agree, and since no one has that magic canon the rocket's vertical thrust component must be used for support. When considering real rockets, calculating a minimum gravity loss is complicated by the cosine factor causing the sum of the vertical and horizontal components of the thrust to be greater than the total thrust of the engine. But their must be a minimum gravity drag for a particular engine because their certainly can be more or less.

Maybe I should have asked for a minimum gravity drag trajectory. I bet it has something to do with the vertical and horizontal component of engine thrust during ascent to orbit. And perhaps atmospheric drag although that seems to be much less than gravity drag so that if minimizing gravity drag comes at the expense of a little aero drag, it is still a benefit overall. So, is there a simple control rule like, always maintain the vehicle vertical acceleration positive, or maybe, point the engine down at 45 degrees and blast away? I'm looking for a way that I can know that my rocket simulation is doing the best that it can do. Minimizing gravity drag looks like an area of improvement.

Quote
... moon, ... snip
yes, ok .

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#### Hop_David

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##### Re: Basic Rocket Science Q & A
« Reply #857 on: 12/08/2013 04:00 PM »
Quote
We're attaching different meanings to the term.

Are you sure? My position was that when the vertical component of displacement was considered, ~1.9 km/s delta V is needed to reach 200 km.

My position is that it can be done with less if you have horizontal velocity. The most extreme example would be a 7.9 km/s circular orbit at 0 km altitude. (if the shell is frictionless, as you stipulated, this would be possible). From this orbit it only takes .06 km/s to achieve a 200 km vertical displacement.

In the real world we can't enjoy the Oberth savings conferred by high speeds from earth's surface. Orbital speeds in the troposphere aren't an option. But neither do we go straight up, turn a hard right angle and then  do the burn for horizontal velocity. The rocket typically starts vertically, then gradually tips towards the horizontal. During vertical ascent it gains horizontal velocity. This is a path somewhere between your vertical cannon shot and my horizontal departure.

You want to drop your shell 200 kilometers and call the final speed your delta V for gravity loss. It just isn't so.

Exploiting the Oberth effect can get you the energy needed with less than 1.9 km/s delta V.

On the other hand, if thrust to weight ratio is bad, you can consume much more than 1.9 km/s. If T/W ≤ 1, no amount of propellant will get you off the ground.

I will say it again. Potential energy difference between altitudes and gravity loss are two different animals.

Maybe I should have asked for a minimum gravity drag trajectory.

Optimum ascent path is complicated and the subject of much study. Above my pay grade, sorry to admit.

You might try Googling "Apollo ascent profile" or "shuttle ascent profile". Looking at various ascent profiles might give you something to copy.

#### ClaytonBirchenough

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##### Re: Basic Rocket Science Q & A
« Reply #858 on: 12/22/2013 08:35 PM »
Alright, I'm having some trouble calculating the ISP of certain rocket engines.

As I understand, specific impulse can be calculated by dividing the total impulse of a rocket motor by its weight. I'm particularly interested in calculating the specific impulse of a rocket motor built by Richard Nakka.

It's properties are:

Motor mass: 2.735 kg
Total impulse: 2008 N * s
ISP: 126 sec

So if I am correct (which I'm obviously not):

2008 N * s / (2.735 kg * 9.8 m/s^2) = 74.9 sec

What did I do wrong?

(Screenshot via Richard Nakka's website.)
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#### deltaV

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##### Re: Basic Rocket Science Q & A
« Reply #859 on: 12/22/2013 09:38 PM »
Alright, I'm having some trouble calculating the ISP of certain rocket engines.

As I understand, specific impulse can be calculated by dividing the total impulse of a rocket motor by its weight.

You need to divide the total impulse by the weight of the propellant, not the weight of the entire motor.

2008 N * s / (1.618 kg * 9.8 m/s^2) = 127 s.
« Last Edit: 12/22/2013 09:39 PM by deltaV »