Sign in to follow this  
MajorTomtom

Understanding launcher's user's manual for interplanetary travel

Recommended Posts

Hi

I wanted to look at performance data for existing launchers in interplanetary travel. I have read the Ariane 5 user's guide and in the part relative to performance for earth-escape missions, I came across this :

"The Ariane 5 ECA version has a performance of 4550kg towards the following earthescape orbit:

•infinite velocity V∞= 3475 m/s

•declination δ = - 3.8°"

 

What do these numbers mean ? Is the declination the relative inclination between the ecliptic and the trajectory ? And what is V∞ ? I hoped to find charts with mass plotted against orbit inclination and stuff but I only found this... (the other orbits are all LEO or GTO with fixed inclination)

 

Thanks in advance guys,

  • Like 1

Share this post


Link to post
Share on other sites

Could the "infinite velocity" mean the velocity at which the orbit becomes open-ended/hyperbolic, ie: no longer an orbit?

Doesnt mean much to me without more figures.

Share this post


Link to post
Share on other sites
6 hours ago, MajorTomtom said:

Hi

I wanted to look at performance data for existing launchers in interplanetary travel. I have read the Ariane 5 user's guide and in the part relative to performance for earth-escape missions, I came across this :

"The Ariane 5 ECA version has a performance of 4550kg towards the following earthescape orbit:

•infinite velocity V∞= 3475 m/s

•declination δ = - 3.8°"

 

What do these numbers mean ? Is the declination the relative inclination between the ecliptic and the trajectory ? And what is V∞ ? I hoped to find charts with mass plotted against orbit inclination and stuff but I only found this... (the other orbits are all LEO or GTO with fixed inclination)

 

Thanks in advance guys,

8000 * SQRT(2) - 8000 = 3313 m/s. This is the approximate dV required to escape earths orbit from LEO. It differs with LEO chosen. What this means is that a 4550kg payload can be carried from LEO to escape velocity with dV of 3475. Large payloads might not reach that height, smaller payloads can go farther. IOW this is a rocket spec.

The RL10B-2 can run for a rated 220 seconds at 110,000 N/sec of thrust for 24,200,000. It has an ISP of 475 (4700 m/sec exhaust velocity). 23.404 kg of fuel per sec. This is 5148 kg of fuel. The engine weighs 277 kg and the fuel tanks are probably 12% of the fuel weight so its fixed mass is 6042 kg. With no payload is can achieve dV of 8700 m/s second. It can carry a PL of 3814 kg and achieve the same velocity spec as Arian 5 rocket.

If this is the desire you would need a rocket that can carry a LEO payload of 9856 kg (~10 tonnes) into LEO.

Note that in order for the RL10B-2 to achieve a higher payload all that needs to be done is re-rate the engines to a longer burn time and increase the proportion of fuel. The problem with at is a exit trajectory requires a fast burn at a single theta. A burn of 220 seconds covers a theta of 16 degrees. If one extends the burn time then one also has a longer burn window and some of the fuel is used in raising the perigee of the orbit.

The following launch vehicles can launch such a payload.  (*** retired, * in developement) Take your pick.

BFR (Space X)*  -15x
ENergia 5b* - 8x -15x
Saturn 5***  - 14x
LongMarch 9* - 14x
SLS Block 2 w/EUS* - 13x
SLS Block 1b* - 10.5
Energia** - 10x
N1** - 9.5x
SLS1* - 8.5x
Falcon heavy* - 6.4x
New Glenn *- 4.5x
Buran *** - 3.0x
Delta4 heavy - 2.9x
Lng March 5 - 2.5x
Space Shuttle*** - 2.4x
Angara A5 - 2.4x
Proton M - 2.4x

Falcon 9 - 2.3x , less if recycled first stage.
Titan 4*** - 2.2x
Arian 5 ECA - 2.1x
Atlas V - 1.4 to 2.0
and everyones most recent entry into the world of national or commercial space flight.

Given that one can increase the fuel, re-rate the engine and increase the payload its certainly possible to achieve higher rating. The spec they did not provide is final stage thrust.

On a delta 4 heavy one could have a much larger payload but confined to 'kicking' the space craft at perigee several times to achieve maximum efficient exit.

 

 


 

 


 

 

Share this post


Link to post
Share on other sites

https://en.m.wikipedia.org/wiki/Escape_velocity

"

When given a speed V greater than the escape speed {\displaystyle v_{e},} the object will asymptotically approach the hyperbolic excess speed {\displaystyle v_{\infty },} satisfying the equation:[4]

0b61bf844cbd4a7737dc2cffee3a136f2557d9f0"

When it says "infinite velocity" it's referring to velocity at infinity, I think. Escape velocity is the velocity that will carry an object to an infinite distance, and its velocity  will always go towards zero. If "infinite velocity" is zero, then you had no excess energy. If it's greater than zero, then you had excess energy, and your velocity approaches the velocity at infinity.

Essentially, it means it can impart about 3913 m/s to a payload of 4550 kg. The 3913 comes from the above formula, using the infinite velocity provided and the escape velocity from the surface ( I know, not accurate, but it's not much of an issue). Infinite velocity shows up in hyperbolic orbit equations a few times. 

I'm not sure if it relates to the velocity after you leave the sphere of influence. Probably not, but maybe some relationship exists.

  • Like 1

Share this post


Link to post
Share on other sites

So if I understood correctly, ksp-wise it would mean that Ariane 5 ultimately provides a 4550kg payload with a speed that is the equivalent to the speed I would have starting in LEO and adding 3913m/s and a corresponding trajectory ?

What does the declination angle represents then ? Is it the angle between the trajectory and the ecliptic plane or something else ?

Share this post


Link to post
Share on other sites

So how is it supposed to be an earth-escape orbit, I mean sure it leaves Earth SOI, but it isn't near the planets orbit plane (which is I believe around 23° from the equator) ?

Share this post


Link to post
Share on other sites
1 hour ago, MajorTomtom said:

So how is it supposed to be an earth-escape orbit, I mean sure it leaves Earth SOI, but it isn't near the planets orbit plane (which is I believe around 23° from the equator) ?

Remember that as the payload leaves Earth, it has some of Earth's velocity, including direction it's traveling in.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this