# Could an ~12ft rocket launch some sort of probe into space

1. Jul 22, 2012

### Turtle^2

could a large 12ft model rocket be used to launch a probe into space, such as a telescope?

2. Jul 22, 2012

### Danger

Welcome to PF, Turtle.
Does "12'" refer to length or diameter?

3. Jul 22, 2012

### krd

No.

People have tried. There are amateur rocket builders who are working on it. First you need a lot of power to get into space - so you'd need a bigger rocket, the bigger the rocket the more power you need, and the bigger the rocket the harder it is to control it. So, pretty big rockets are usually required to put even relatively small things in space.

They need to be big..........

Last edited by a moderator: May 6, 2017
4. Jul 22, 2012

### Nabeshin

Probably not, you need something slightly bigger.

A 21 foot tall amateur rocket using just solid propellant reached past the official boundary of space, up to 116km though! So 12 feet isn't THAT far off.

Source: http://en.wikipedia.org/wiki/Civilian_Space_eXploration_Team

Note however that if you want to launch something into ORBIT, or on an escape trajectory, the amount of power necessary increases manifold.

5. Jul 22, 2012

### rorix_bw

What happens if you launch it from a ballon? Amateurs get ballons up to 30km+.

Probably 4 ballons with the rocket between them, and use a RC/unit with an arduino or raspberry pi to control levelling and launching.

6. Jul 22, 2012

### Staff: Mentor

You simply get yourself 30 km up in the air and don't need to make it up using the rocket. It makes it easier to get into space, but not to get into orbit. Getting into an orbit requires accelerating from near zero to about 8 km/s for a low Earth orbit.

7. Jul 22, 2012

### rorix_bw

I wonder how much extra height you would get? I guess most of the fuel is used closer to the ground so I imagine it would add more than +30km.

8. Jul 22, 2012

### Aero51

Probably not. As was mentioned earlier amateur rocket enthusiasts compete annually with larger rockets and do not reach orbit. Actually, if you have MATLAB I have a program which does a first iteration calculation of altitude achieved based on some rocket parameters. I could find it but I doubt it's accurate enough to be useful.

9. Jul 22, 2012

### rorix_bw

"Not to orbit" is fine but the question was "how much difference it makes, launching from 30km up" :-)

I am pretty sure that because of thinner air, you will less fuel 30-60km than you use 0-30km

I'm trying to find some data on it. No, I don't expect the answer to be "orbit".

I am worried about stability though. High winds up there. Lower winds at ground level and easier to stabilise yourself once in motion.

10. Jul 22, 2012

### krd

8 km/s where are you getting that figure?.....That's 23 times the speed of sound. The max speed of the space shuttle was 1.3 km/s

11. Jul 22, 2012

### papernuke

Hi krd,
Look here:
http://en.wikipedia.org/wiki/Orbit_velocity

The high velocity is required because that is the velocity needed to be in uniform circular motion around the earth, with gravity
being the centripetal force.

Here's another something similar:
http://en.wikipedia.org/wiki/Escape_velocity
It's the velocity required for a body to escape the pull of gravity.
(example: you shoot a cannon ball straight up, but it normally falls back down. If you shoot it fast enough (escape velocity or above), then it'll never fall back down

12. Jul 22, 2012

### Aero51

If you were to use all your kinetic energy at the surface of the earth in 1 instant (time approaches 0 or power approaches infinity), you would need about 8 km/s as an initial velocity. This is the definition of "escape velocity".

13. Jul 22, 2012

### 256bits

Escape elocity for the earth is a little more than 11 km/s.
As drakith stated 8 km/s is needed for a low earth orbit.

14. Jul 22, 2012

### Staff: Mentor

As you want to launch a telescope, I assume that it should (at least) reach a low earth orbit, otherwise it re-enters quickly and the mission lasts just some minutes. With conventional rockets, there is a physical limit on the dry mass to fuel ratio, given by the rocket equation.
$$m_{total} = m_{dry\,mass} \exp\left(\frac{8km/s}{v_{exhaust}}\right)$$
The best chemical propellants reach an exhaust velocity of about 4km/s, therefore the exponential is about e2 or ~7. However, this does not account for atmospheric drag and the required time of the acceleration - both increase the number of 8km/s, which increases the required ratio exponentially. In addition, it does not take into account that the payload is just a small fraction of the dry mass. Staging can improve this ratio a bit, but it cannot avoid the physical limits of the whole design.

Compare it with real rockets:

The space shuttle could launch about 25 tons into low earth orbit, but required a total mass of 1500 tons -> ratio 1/60. However, that reusable shuttle was not really ideal in terms of "payload to orbit"
The Saturn V was bigger, and with a better ratio: 120 tons payload, 2800 tons total mass -> ratio 1/23
Soyuz-U: 7 tons payload, 313 tons total mass -> ratio 1/45
As you need some systems regardless of size, smaller rockets tend to have a worse ratio. Let's look at commercial rockets:
Falcon 1 (21m height, 1.7m diameter) can launch 0.7 tons with a total mass of 39 tons (1/56), the bigger Falcon 9 with 10,5 tons / 333 tons (1/32) is better. Falcon heavy: 53 tons / 1400 tons (1/26)

15. Jul 22, 2012

### BobMarly

I've heard it stated that the most expensive part of a launch is the first 2 inches!

16. Jul 22, 2012

### surajt88

17. Jul 22, 2012

### BobMarly

That was just a qoute I heard during an interview ten or twenty years ago. If you're looking for more info, dare I say Google It!

18. Jul 22, 2012

### Staff: Mentor

In terms of required fuel... well, yes, the "fuel per distance" ratio decreases with velocity, and the initial velocity is 0 for conventional rocket launches (=all rocket launches up to now). But this does not mean that lifting the rocket by 2 inches would help in any way. Accelerating it to something like 500m/s or more would indeed help.

19. Jul 22, 2012

### krd

I think what that means is getting it 2 inches off the ground without it malfunctioning and exploding is the most expensive part of the launch.

20. Jul 27, 2012

### surajt88

Couldn't get anything by googling the first time around. It also seems logical that the cost of fuel and safety would be very high for the initial part of the launch. What I wanted was data resembling breakup of amount of fuel required as the rocket gets higher or something like that.