3
$\begingroup$

So, a satellite launch takes a huge amount of fuel to launch a relatively small (compared to total rocket weight at takeoff) satellite into orbit. Obviously there is a lot of energy expended during the launch, but what percentage of the expended energy is actually needed to get the satellite itself up into orbit? I suppose it would just be a ratio of the amount of kinetic energy the satellite has at release in orbit compared to the potential energy of the fuel the rocket burns to get it there.

For a specific common scenario, consider a SpaceX Falcon9 launch of a max-weight satellite to LEO. Research on the SpaceX website indicates that its GTO capacity is 22,800kg. Other research indicates that the Falcon9 carries 136,900kg of RP-1 between its first two stages. However, I'm struggling to find out how to calculate the kinetic energy of such a satellite in LEO and the potential energy in the fuel.

Improve this question
$\endgroup$
3
  • $\begingroup$ The kinetic energy of the payload and the energy in the fuel are not going to match, a lot of the energu ends up in the exhaust and discarded stages. $\endgroup$
    – SE - stop firing the good guys
    Aug 24, 2016 at 20:27
  • $\begingroup$ You can be mass efficient by pushing a little mass fast, or energy efficient by pushing a lot of mass slowly. It's difficult to do both simultaneously. Mass efficiency usually rules rockets. $\endgroup$
    – BowlOfRed
    Aug 24, 2016 at 23:20
  • $\begingroup$ @Hohmannfan, I know they're not going to match, which is the basis of my question. If they matched, it would be 100% efficient. But it isn't, due in part to the reasons you mentioned. What I am trying to find is what % of the potential energy in the fuel was transferred to the satellite in terms of kinetic energy due to velocity and potential energy due to altitude of the orbit. The remainder of the energy in the fuel at takeoff is taken by accelerating fuel, rocket parts, inefficient combustion, atmospheric drag, and other factors too. I'm just wondering if it's 1% efficient, .01%, etc.? $\endgroup$
    – Milwrdfan
    Aug 25, 2016 at 17:03

1 Answer 1

Reset to default
5
$\begingroup$

Let's take your example of a Falcon 9. The actual payload to GTO is given as 8300 kg. In this case GTO is likely to be defined as a 185 x 35890 km orbit.

The rocket has to accelerate the payload to orbital speed and also to lift it to the orbit. That is, the payload is lifted by 185km: $$E_h = m \cdot g \cdot h = 1kg \cdot 9.81 m/s \cdot 185km = 1.8 MJ$$ The orbital speed at this point is calculated as $$\sqrt{GM\left(\frac{2}{a}-\frac{1}{r}\right)}$$ with the semi major axis $a = \frac{1}{2}(185 + 25890 + 2\cdot 6380)km = 24,250 km$ and $r = 185 km$ $$v = 10.25 km/s$$ This is a kinetic energy of 52 MJ/kg. That is, the rocket has to deliver 53.8 MJ per kilogram payload.

Launch mass of a Falcon 9 is 550 t, including 155 t RP-1 and 361 t LOX. RP-1 has an enthalpy of 42.9 MJ/kg, the whole rocket releases 6,640 GJ by burning the propellant. This is used to lift the payload of up to 8300 kg, that is 800 MJ/kg. This makes the overall efficiency 6%.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Unfortunately the rocket has to accelerate and lift not only the payload, but also the remaining fuel in the tanks. The structure and the engines of the rocket also must be accelerated and lifted, but when the payload is set out into the orbit, the rocket has no use anymore. That is why the overall efficiency is so low. $\endgroup$
    – Uwe
    Sep 16, 2016 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.