Calculating hyperbolic capture

[Pablo49]

So I am trying to find out how to calculate the delta-v required to get into an orbit with a desired periapsis around a planet when coming from a hyperbolic orbit, and my googlin' has failed me thusfar. Does anyone know how to do that?

 

[dgatsoulis]

[math] \Delta V_{capt}=\sqrt{V_{enc}^2+V_{esc}^2}-V_{esc} [/math]where,
[math]\Delta V_{capt}[/math] = capture delta-v
[math]V_{enc}[/math] = encounter velocity
[math]V_{esc}[/math] = local escape velocity (the escape velocity for the altitude you want to get captured).

The local escape velocity is:
[math] V_{esc} = \sqrt{\frac{2GM}{R+alt}}[/math]where,
G = gravitational constant
M = mass of the body
R = radius of the body
alt = periapsis altitude of the capture orbit.

If instead of capture, you want to calculate the circularization burn:
[math] \Delta V_{capt}=\sqrt{V_{enc}^2+V_{esc}^2}-V_{orb} [/math]
where, V_orb is the orbital velocity of the parking orbit. [math]V_{orb} = \frac{V_{esc}}{\sqrt{2}}[/math]
You will find these links useful:
Basics of spaceflight
Example problems

 

[Pablo49]

Awesome, thanks a bunch! I'm surprised I didn't find either of those links while doing my own research. Look like an awesome resource.

 

[Keithth G]

dgatsoulis is correct.

However, the definition of 'encounter velocity' is not always clear. For the avoidance of doubt, it should be taken as the 'hyperbolic excess velocity' - i.e., the velocity of the object well beyond the gravitational influence of the planet. It isn't the velocity of the object on its hyperbolic trajectory at, say, periapsis.

 

[Pablo49]

So I am having some difficulty calculating this. Not sure where I am going wrong. Probably misunderstanding one of the variables.

my orbit stats and such:
planets gravitational parameter: 301.36321 km^3/s^2
planet radius = 320 km
(The planet is Duna from KSP, so that is where those values come from)
semi-major axis: -230.07162km


for hyperbolic excess velocity I have:
[math]\sqrt{\frac{301.36321 \frac{km^3}{s^2}}{230.07162 km}} = 1.14449 \frac{km}{s}[/math]
for local escape velocity: (I am targeting a 500km orbit. the periapsis of my hyperbolic orbit is 890.416km. I tried that value too in case I was misunderstanding this step, but still a weird final delta-v)
[math]\sqrt{\frac{2 * 301.36321 \frac{km^3}{s^2}}{320km+500km}} = .85734 \frac{km}{s}[/math]
then my delta-v:
[math] \sqrt{1.14449^2 \frac{km}{s}+ .85734^2 \frac{km}{s}} - .85734 \frac{km}{s}* \frac{1000m}{km} = 572.656 \frac{m}{s} [/math]
But ~573 m/s delta-v isn't enough to drop me out of a hyperbolic orbit. My velocity at periapsis is ~1335 m/s, and escape at that altitude is ~706 m/s.

 

[boogabooga]

The equations do not care what orbit you are "targeting." If you perform the braking at periapsis, then you need to calculate escape velocity for that altitude (890km)

My calculations show ~639 m/s capture and ~846 m/s to circularize at 890km. You will have to spend extra fuel to Hohmann transfer down to 500km, or do a TCM to get your periapsis down to 500 km before the breaking.

How long does it take your craft to decelerate 639 m/s? You may also have finite burn losses.

 

[dgatsoulis]

A trajectory with 890.416 km periapsis has a different semi-major axis than one with 500 km. So which one are you using for that -203.07162 km semi-major axis?

To clarify, your ~1335 velocity is at 500 km or at 890.416 km altitude? (Also clarify, altitude or distance from the center of the body?)

EDIT: (nvm, point made by boogabooga, I'm also getting the same numbers).

[math] \sqrt{\frac{2 * 301.36321 \frac{km^3}{s^2}}{320km+890.416km}} = .706 \frac{km}{s} [/math]
Capture dV
[math] \sqrt{1.14449^2 + .706^2 } - .706 = .639 \frac{km}{s} [/math]
Circ dV

[math] \sqrt{1.14449^2 + .706^2 } - \frac{.706}{\sqrt{2}} = .846 \frac{km}{s} [/math]

 

[Pablo49]

The equations do not care what orbit you are "targeting." If you perform the braking at periapsis, then you need to calculate escape velocity for that altitude (890km)

My calculations show ~639 m/s capture and ~846 m/s to circularize at 890km. You will have to spend extra fuel to Hohmann transfer down to 500km, or do a TCM to get your periapsis down to 500 km before the breaking.

How long does it take your craft to decelerate 639 m/s? You may also have finite burn losses.

Okay, so it was indeed my misunderstanding of what the equations calculate. Those two velocities would be the correct ones then. (same goes for dgatsoulis' work)

The burn time is 100 seconds for 846 m/s. In practice it takes 147s to circularize. 76s calculated for capture, 114s actual. (31% and 33% over time respectively) Is there a reasonable way to calculate how to compensate for burn time loss? Or is it better to just estimate additional time onto the burn and correct later?

 

[boogabooga]

The burn time is 100 seconds for 846 m/s. In practice it takes 147s to circularize. 76s calculated for capture, 114s actual. (31% and 33% over time respectively) Is there a reasonable way to calculate how to compensate for burn time loss?

You kinda just did. You need to add ~30% to the expected delta-V, apparently.
If you don't already, be sure that you center the burn: start ~75 seconds before periapsis. That should help you cut back on losses.

But come on, Kerbal wouldn't be Kerbal if everything was precise. The fun is in winging it! :lol:

 

[Pablo49]

So I decided to play around a bit with the vehicle I had around Duna for funsies, and I noticed my burntimes were still way off for even small orbit changes, which wasn't ever a problem. It so happens I started to mistake my fuel mass as my vehicle mass right as I started tackling the hyperbola stuff, and was too distracted to notice that error while focusing on the new equations, which threw off my burntime calculations. So I totally would have noticed my original mistakes with the periapsis in my trial and error earlier, but with my mass mistake it was hidden. So now the burns work with trivial error.

So basically Pablo is dumb and blind.

Thanks for your help though, guys. I've been learning how to do as much of the calculations needed to do anything in space by hand. My friends and I have done some mission-control style missions together and are trying to make a club at my university, which has made me want to learn how to do everything ourselves, and this part was stumping me for awhile.

 

[dgatsoulis]

Here is another link that will come in handy:
Equations for elliptical, parabolic and hyperbolic orbits.

I also enjoy calculating this stuff with pen and paper (well... I throw in a spreadsheet from time to time).

If you are familiar with the DGIV, I am working on a challenge that you might find interesting:

A coplanar orbital rendezvous without using any MFDs.

 

[kuddel]

I am working on a challenge that you might find interesting:
A coplanar orbital rendezvous without using any MFDs.

...something like this ?

 

[dgatsoulis]

...something like this ?

Thanks for the link, I hadn't seen that one.

Similar, but with a lua script to keep track of progress and a bigger altitude difference. The user will only have the data from the DGIV antenna to rely on, so in a sense it's easier ('cause of the antenna) and harder (longer burn-times, grav-losses need to be taken into consideration).

Also, each time you try the scenario, the spacecrafts will be placed on different orbital altitudes, so you can't solve it by trial and error.

 

[C3PO]

The user will only have the data from the DGIV antenna to rely on...

Sorry for geting really off topic, but if I had ANY capabilities in coding my first port of call would be a RadarMFD. Something similar to DGIV and Dragonfly but without the visual bells and whistles. Numerical output would do just fine. The main difference would be to add a LVLH frame to the vessel-relative frame those two use.

 

[RGClark]

A trajectory with 890.416 km periapsis has a different semi-major axis than one with 500 km. So which one are you using for that -203.07162 km semi-major axis?
To clarify, your ~1335 velocity is at 500 km or at 890.416 km altitude? (Also clarify, altitude or distance from the center of the body?)
EDIT: (nvm, point made by boogabooga, I'm also getting the same numbers).
[math] \sqrt{\frac{2 * 301.36321 \frac{km^3}{s^2}}{320km+890.416km}} = .706 \frac{km}{s} [/math]Capture dV
[math] \sqrt{1.14449^2 + .706^2 } - .706 = .639 \frac{km}{s} [/math]Circ dV
[math] \sqrt{1.14449^2 + .706^2 } - \frac{.706}{\sqrt{2}} = .846 \frac{km}{s} [/math]

Thanks Dgatsoulis for those equations. I had a similar question. I want to do missions to Jupiter and Saturn. Because of the delta-v requirements, most mission designs use Venus/Earth flybys for gravity assist that result in flight times lasting several years. But recall we discussed the New Horizon mission whose high departure speed allowed fast travel times to the outer Solar System.

So I wanted to use a similar high departure speed for the missions to Jupiter or Saturn. From the January, 2006 launch to passing Jupiter took New Horizons only 13 months:

And it only took 2 years, 5 months to reach Saturn's orbit:

However, that would mean they would have high relative speeds on reaching the planets. What would be the delta-v required to enter orbit in these cases?

Bob Clark

 

[dgatsoulis]

I don't have the heliocentric velocity of N-H, when it passed Saturn's orbital distance.

For Jupiter, the closest approach was at 2.3 million km, with a velocity relative to Jupiter of 21.2 km/s. It would take ~13.8 km/s of delta-v to circularize at that distance.

 

[RGClark]

I don't have the heliocentric velocity of N-H, when it passed Saturn's orbital distance.
For Jupiter, the closest approach was at 2.3 million km, with a velocity relative to Jupiter of 21.2 km/s. It would take ~13.8 km/s of delta-v to circularize at that distance.

Thanks dgatsoulis, Keithth G, and boogabooga for the info. Based on all of your responses I could reduce the required delta-v by going closer in to Jupiter. By Wikipedia, escape velocity for Jupiter at the cloud tops is 59.6 km/s. Then dgatsoulis capture equation gives:

sqrt(21.2^2 + 59.6^2) - 59.6 = 3.66 km/s.

The circularization burn though would be too high, above 21 km/s. We could use various combinations of Jupiter's moon's flyby's and possible aerobraking to circularize the orbit.

That 3.66 km/s number though is from skimming the cloud tops. This might be too strong a radiation environment. If we made the periapsis further out, that would reduce the radiation dose though the required delta-v would increase.
As points of comparison Pioneer 11 came within 43,000 km of Jupiter and Juno came within 4,300 km but only by going above the poles where the radiation is less.
Note also Pioneer 11 only took a year and 8 months to flyby Jupiter. Likely it had a lower departure velocity and lower encounter velocity on reaching Jupiter so would require lower delta-v to be captured.

Bob Clark

 

[dgatsoulis]

That's not entirely correct. You are using the periapsis velocity as the encounter velocity.

Think of it like this: If Jupiter was a spacecraft (there was no gravity well), then the New-Horizons velocity relative to Jupiter wouldn't have been 21.2 km/s. But because of Jupiter's gravity well, the N-H sped up relative to Jupiter as it dropped further in. If it had gone even further, it would have sped up more.

First let's calculate the encounter velocity with Jupiter. As KeiththG pointed out, this is known as the hyperbolic excess velocity, aka V-infinity, aka V encounter. We already know that the velocity relative to Jupiter at the point of closest approach (periapsis - 2.3 million km) was 21.2 km/s. The escape velocity at 2.3*10^9 meters is ~10.5 km/s.

[math] V_p^2 = V_{\infty}^2+V_{esc}^2 \rightarrow \; V_{\infty} = \sqrt{V_p^2-V_{esc}^2} = \sqrt{21.2^2-10.5^2} = 18.42 \\ \\ km/s[/math]
So if Jupiter had negligible mass, the N-H spacecraft, would pass by, with a velocity of 18.42 km/s relative to Jupiter.

Using the same encounter velocity, we get this table:

Distance to Jupiter |Capture dV (km/s) |Circ DV (km/s)
80,000 km (cloud tops) |2.94 |19.42|
422,000 km (Io's distance) |6.15 |13.33|
671,000 km (Europa's distance) |7.34 |13.03|
1,071,000 km (Ganymede's distance) |8.62 |13.12|
1,883,000 km (Callisto's distance) |10.17 |13.57|

Remember that the capture dV, gets the spacecraft's apoapsis at infinity, you never actually return to Jupiter.

 

[RGClark]

...
Remember that the capture dV, gets the spacecraft's apoapsis at infinity, you never actually return to Jupiter.

So what would be the required delta-v for an elliptical orbit under this scenario? For the maximum distance, use the Galileo numbers, which, if I'm reading correctly, had a maximum Jupiter centered semi-major axis of 10,000,000+ km:

Galileo Orbital Elements.

IO TO JUPITER APOAPSIS, ORBIT 0: Orbital Elements valid from J.D. = 2450059.24080 ET (95/12/07 17:45:45 GMT) to J.D. = 2450161.86379 ET (96/03/19 08:42:51 GMT) Epoch of Orbital Elements: J.D. = 2450059.24080 ET (95/12/07 17:45:45 GMT) Classical Orbital Elements Jupiter centered, Earth ecliptic and equinox of B1950 Semi-major Axis 10080400. km Eccentricity 0.970253 Inclination 5.212 degree NodeAngle 58.626 degree Argof Periapsis 89.250 degree TrueAnomaly -65.371 degree
http://planet4589.org/space/jsr/notes/orbital.html

Bob Clark

 

[dgatsoulis]

Here are the equations you need:

[math] V_p = \sqrt{V_{\infty}^2+V_{esc}^2} [/math]where,
[math]V_p[/math] = periapsis vel, [math]V_{\infty}[/math] = encounter vel. [math]V_{esc}[/math] = local escape vel.
[math] V_{esc} = \sqrt{\frac{2\mu_j}{R}} [/math]where,
[math]\mu_j[/math] = Jupiter's grav. parameter, [math]R[/math] = distance from Jupiter's centre
[math] V_{p_{ell}} = \sqrt{\frac{2\mu_jR_a}{R_p(R_a+R_p)}}[/math]where,
[math]V_{p_{ell}}[/math] periapsis velocity of elliptical orbit, [math]R_p[/math] = periapsis distance, [math]R_a[/math] = apoapsis distance

[math] a = \left(\frac{R_a+R_p}{2}\right)\Rightarrow \\ R_a = 2a-R_p[/math]where,
a = semi-major axis

[math] \Delta V = V_p - V_{p_{ell}}[/math]where, ΔV = required delta-v

Here is an example for Io's distance from Jupiter:

Given:
----------------------------------------------------------------
Jupiter's grav. parameter:[math]\mu_j = 126,686,534 \;\; \frac{km^3}{s^2}[/math]
Encounter velocity: [math] V_{\infty} = 18.42 \;\;\frac{km}{s}[/math]
Io's mean distance from Jupiter: [math] R,R_p = 421,700\;\; km [/math]
Semi-major axis of capture orbit: [math] a =10,080,400 \;\; km [/math]-----------------------------------------------------------------


Start by calculating the escape velocity at Io's distance:

[math] V_{esc} = \sqrt{\frac{2\mu_j}{R}}= \sqrt{\frac{2\cdot 126,686,534}{421,700}}=24.512 \;\;\frac{km}{s}[/math]
Now the periapsis velocity of the spacecraft:

[math] V_p = \sqrt{V_{\infty}^2+V_{esc}^2} =\sqrt{18.42^2+24.512^2}=30.662\;\;\frac{km}{s}[/math]
The apoapsis of the capture orbit:

[math]R_a = 2a-R_p = 2\cdot 10,080,400 - 421,700=19,739,100\;\;km[/math]
The periapsis vel. of the capture orbit:

[math] V_{p_{ell}} = \sqrt{\frac{2\cdot 126,686,534 \cdot 19,739,100}{421,700\cdot (19,739,100+421,700)}}=24.254\;\;\frac{km}{s} [/math]
And finally the delta-v for the capture-to-elliptical-orbit:

[math] \Delta V = V_p - V_{p_{ell}}=30.662-24.254 = 6.408 \;\;\frac{km}{s} [/math]
You can also calculate the period of the capture orbit with:

[math]T = 2\cdot \pi \sqrt{\frac{a^3}{\mu_j }} \;\; [seconds][/math]
and convert to days by dividing with 86400.

[math]T = 2\cdot \pi \sqrt{\frac{10,080,400^3}{126,686,534}} = 17,866,162.9 \;\;seconds = 206.78 \;\; days[/math]
Source: Basics of Spaceflight: Orbital Mechanics