Delta V and other beginner questions

[Orion Pax]

Greetings exalted ones

Pardon my lack of knowledge but I'm still learning

1. So say that a rocket has 8 km/ s of delta v, and say it's in a vaccum without any gravitational influence, and it's initial velocity is 0 km/ s, after it has burned all it's fuel, what would be its velocity? 8 km/ s?

2. If we have a hypothetical spacecraft with a vtol capability and somehow has enogh delta v to travel outside the solar system, will it be able to get on an escape trajectory out of earth just by going up vertically?

3. Say we have a new type of propulsion that can get us up to 5% the speed of light, and with enough fuel to get us anywhere in the solar system and back, what would be its trajectory to have an intercept with another body? Unlike a hofman transfer here the vessel is traveling at very high speeds and is directly in an escape trajectory out of earth (I know how inefficient that is with current propulsion methods but for the sake of this question...) I mean instead of waiting months or years with the engine on Idle just get to an escape trajectory with high velocity. Correct me if I'm wrong, but this would eliminate the need for the time consuming planetary alignment as we would be able to travel in almost is a straight line.

Say I want to try this in KSP, with infinite fuel, how should I plan the intercept with the target? I managed to get one with the moon when it was dirrectly overhead and it did work, the question is how can I do that with more distant target, which obviously wouldn't be in the same place relative to my successful intercept with the moon?

Thanks

Edited October 28, 2016 by Orion Pax

[p1t1o]

Welcome to the Forums!

There are folks with all sorts of backgrounds, and Im sure someone will elaborate, but here is a start:

 

1. Spot-on. "delta" is science-speak for "change-in" so "delta-V" translates roughly as "change in velocity". Your rocket has the capacity to change its velocity by 8km/s. Note that a rocket which accelerates to 4km/s then decelerates to a (relative) stop also uses 8km/s of dV.

2. Slightly more complex situation. Lets say local gravity is 10m/s^2 ****, then your rocket must use 10m/s of dV per second just to hover against gravity. Ergo, in order to accelerate upwards at 10m/s^s, against a 10m/s^2 gravity field, your rocket must be capable of exerting a 20m/s^2 acceleration in zero-G conditions, ie: have a TWR (thrust-to-weight-ratio) of 2. This is complicated by the fact that the gravity field weakens with distance. The rest of the question is a little unclear (since any vehicle not using aerodynamic lift technically has "VTOL" capability). Google "gravity losses" and "escape velocity". Return here if you have any follow-up questions.

3. Firstly, it makes little sense to categorise propulsion systems by the "speed it can get up to" because the only thing limiting this is the fuel carried (and the speed of light, naturally). However you are correct with the rest of your statement, more direct trajectories are less fuel-efficient but if you have excess dV available you can get there very quickly without having to wait for alignments. 

4. Given infinite fuel, one can simply point directly towards the target and burn. Technically you will find you would need to make corrections as the target moves through space, but the initial condition (infinite fuel - essentially a magic spaceship) is quite fantastic, on an interplanetary trip one would reach speeds high enough that this effect would be minimal (unless you have a very low acceleration, and even then you will attain high velocities requiring only minimal corrections). Matching velocities with the target may be complicated as the target will have some velocity relative to your starting position, but with infinite fuel this is hardly a concern. Planning an ideal trajectory would take some fairly involved maths and data on the properties of the craft, your origin and target (relative velocities, acceleration capabilities and the like). In practice, once you escape Kerbins gravity well, there isnt much to slow you down, so only several km/s dV is needed (once you are out of Kerbins SOI) to get a decent direct-ish intercept, still very much NOT burning all the way, but still very much faster than a Hohmann.

 

Fun Fact: if you have a ship with "infinite fuel" (obviously without infinite mass, lets say it can produce a 1G thrust forever) will, taking into account time dilation, allow you to cross the entire observable universe in only about a century of perceived time ("ship-time") as time slows in your ship relative to the rest of the universe and because constant thrust will allow you to approach an arbitrarily high percentage of the speed of light. Many billions of years will have passed for the rest of us however.

 

 

****Do you understand this notation? Gravity is measured in terms of the acceleration it induces, and acceleration is measured in "metres per second-squared", ie: 10m/s^2 means that your velocity increases by 10m/s, per second.

[Red Iron Crown]

1. Yes, 8km/s assuming all the delta-V was expended in the same direction.

2. Yes, it is fairly trivial to launch straight up to escape velocity, though it's not overly efficient due to gravity losses.

3. Given infinite fuel but finite acceleration the type of trajectory desired is called a brachistochrone curve. The math behind it is non-trivial. 

[Gaarst]

45 minutes ago, Orion Pax said:

1. So say that a rocket has 8 km/ s of delta v, and say it's in a vaccum without any gravitational influence, and it's initial velocity is 0 km/ s, after it has burned all it's fuel, what would be its velocity? 8 km/ s?

Yes.

Quote

2. If we have a hypothetical spacecraft with a vtol capability and somehow has enogh delta v to travel outside the solar system, will it be able to get on an escape trajectory out of earth just by going up vertically?

The Earth's escape velocity at its surface is 11.2km/s. Neglecting atmospheric effects and remembering that gravity is a conservative field this means that a craft going over 11.2km/s on the Earth's surface in any direction (though going downwards might be problematic) will eventually escape Earth's attraction.

Since your VTOL can't magically reach 11.2km/s in an instant, you need to accelerate to this speed in a given time. Going vertically means that you will constantly be fighting gravity face off, and this will make you lose dV depending on the time you spend fighting it: gravity losses are non-conservative since they are not a function of path/position; imagine a craft with 1 million km/s of dV but a surface TWR of 0.99, even though you should have enough dV to escape Earth (and probably the local galaxy cluster at this point) you will never get off the ground. A craft with infinite TWR needs 11.2km/s of dV to escape Earth's gravity from its surface, a craft with <1 TWR needs infinite dV to escape Earth's gravity from its surface, your VTOL will stand somewhere in the middle.

Note that you don't need to actually reach 11.2km/s at any point to escape Earth's gravity. If you go up steadily at say 10m/s, you will eventually reach a point where the escape velocity (which is inversely proportional to the square of your distance to Earth) becomes less than 10m/s and you will have effectively escaped Earth's gravity. This is also the reason why you don't need to break physics to escape a black hole's horizon, just rocket engines. Due to gravity losses I mentioned earlier and considering that you will spend a lot of time going up at a fixed speed (so needing continuous 9.81m/s² of acceleration from your engines at the surface, a bit less going up) this method is very inefficient dV wise, which is why we don't do it in real life: we just accelerate a lot in LEO and follow a ballistic trajectory to wherever we want to end up.

Quote

3. Say we have a new type of propulsion that can get us up to 5% the speed of light, and with enough fuel to get us anywhere in the solar system and back, what would be its trajectory to have an intercept with another body? Unlike a hofman transfer here the vessel is traveling at very high speeds and is directly in an escape trajectory out of earth (I know how inefficient that is with current propulsion methods but for the sake of this question...) I mean instead of waiting months or years with the engine on Idle just get to an escape trajectory with high velocity. Correct me if I'm wrong, but this would eliminate the need for the time consuming planetary alignment as we would be able to travel in almost is a straight line.

Say I want to try this in KSP, with infinite fuel, how should I plan the intercept with the target? I managed to get one with the moon when it was dirrectly overhead and it did work, the question is how can I do that with more distant target, which obviously wouldn't be in the same place relative to my successful intercept with the moon?

5% of the speed of light is a lot. The Kerbal sun's escape velocity at its surface is about 100km/s, here we're talking 15000km/s, that's 150 times as fast. At this point you can just use a ruler to plan your trajectories.

Planning an encounter going at this speed is fairly easy. Draw a straight line between you and where your target will be when you reach it (even though you're going very fast, stuff moves so you can't just shoot point blank at other planets, no matter how fast you go) and you're done. If you do have a craft able to reach 5% of c and stop afterwards, then planetary alignement is just a matter of how long your trip will be (and trying to avoid going through the Sun to reach a planet in conjunction).

To calculate in which direction to go you need some basic maths.
Assume your target is at a constant distance D from your departure point (unless you're aiming for another galaxy, the phase change between your planet and the target will be small in the time you need to reach it, so constant distance).
To travel this distance D at 5% of c you will need a time t = 20*D/c.
In this time the planet will have travelled a small fraction of its orbit, this displacement θ is equal to: θ =  t/T*360° with T the target planet's orbital period.
Therefore, you need to aim at the planet's current position plus an angle θ (relative to its orbital motion).

Spoiler

Body	Distance (max)	Travel time	Orb period	Angle (in °)
Mun	12000	0,8	138984	0,00207218
Minmus	47000	3,13333333	1077304	0,00104706
Moho	18836000	1255,73333	2215741	0,20402385
Eve	23432000	1562,13333	5657961	0,09939411
Duna	34326000	2288,4	17315296	0,04757782
Dres	54439000	3629,26667	47892766	0,02728045
Jool	82333000	5488,86667	104660805	0,01887996
Eeloo	103718000	6914,53333	156991108	0,01585588

Under the spoiler is the calculations done for the Kerbal system's planets. Note that:

• the angles are really small, so it was OK to assume constant distance. But don't assume they are zero! If you do, you'll miss each body by a few dozens of radii (except Jool, which you'll still miss by 3 radii though) going from 25 radii for Eve to 244 for Dres
• the further away a body is, the smaller the angle. This is because the orbital period of a body is proportional to its SMA to the power 3/2, therefore the period "rises" quicker than distance, and the angle is smaller

[Racescort666]

2 hours ago, Red Iron Crown said:

The math behind it is non-trivial. 

This describes so many aspects of my life...

[Hesp]

9 hours ago, Orion Pax said:

3. Say we have a new type of propulsion that can get us up to 5% the speed of light, and with enough fuel to get us anywhere in the solar system and back, what would be its trajectory to have an intercept with another body? Unlike a hofman transfer here the vessel is traveling at very high speeds and is directly in an escape trajectory out of earth (I know how inefficient that is with current propulsion methods but for the sake of this question...) I mean instead of waiting months or years with the engine on Idle just get to an escape trajectory with high velocity. Correct me if I'm wrong, but this would eliminate the need for the time consuming planetary alignment as we would be able to travel in almost is a straight line.

Say I want to try this in KSP, with infinite fuel, how should I plan the intercept with the target? I managed to get one with the moon when it was dirrectly overhead and it did work, the question is how can I do that with more distant target, which obviously wouldn't be in the same place relative to my successful intercept with the moon?

Thanks

7 hours ago, Gaarst said:

5% of the speed of light is a lot. The Kerbal sun's escape velocity at its surface is about 100km/s, here we're talking 15000km/s, that's 150 times as fast. At this point you can just use a ruler to plan your trajectories.

Planning an encounter going at this speed is fairly easy. Draw a straight line between you and where your target will be when you reach it (even though you're going very fast, stuff moves so you can't just shoot point blank at other planets, no matter how fast you go) and you're done. If you do have a craft able to reach 5% of c and stop afterwards, then planetary alignement is just a matter of how long your trip will be (and trying to avoid going through the Sun to reach a planet in conjunction).

 

All of this assuming you have an engine capable of bringing you instantly to 0.05c.

With an arbitrary 10g acceleration it will take 42 hours to get to that speed, and 42 hours more to slow down for intercept, covering WAY more distance than necessary for any object in the Kerbal system

(2.25 billion km to reach 0.05c and stop)

I integrated Gaarst's table to calculate travel times for a straight-line transfer at constant acceleration/deceleration (though a non realistic assumption is initial/final speed = 0 in this case):

Spoiler

Body	Distance (max)	Travel time (s)	Travel time (h)	Max speed (km/s)	Orb period	Angle (deg)
Mun	12000	2212.0	0.6	10.8	138984	5.73
Minmus	47000	4377.7	1.2	21.5	1077304	1.46
Moho	18836000	87637.5	24.3	429.9	2215741	14.24
Eve	23432000	97746.3	27.2	479.4	5657961	6.22
Duna	34326000	118306.1	32.9	580.3	17315296	2.46
Dres	54439000	148987.7	41.4	730.8	47892766	1.12
Jool	82333000	183224.0	50.9	898.7	104660805	0.63
Eeloo	103718000	205647.1	57.1	1008.7	156991108	0.47
Acceleration	1	g

 

Edited October 28, 2016 by Hesp
more math

[Orion Pax]

Wow thanks guys, I really appreciate your detailed answers and I will read the wikipedia pages you've suggested, thanks!