0
$\begingroup$

In theory, the orbit around a single mass always has the same amount of total energy (Kinetic + Potential). Therefore, if a satellite was launched from earth, it should always require the same amount of energy (and therefore fuel) to attain orbit, no matter at which altitude this orbit is. Of course this assumes all other things (efficiency, total energy at launch, mass of rocket, etc.) are equal.

Thus my question: are there significant differences between the amount of fuel needed to reach different types of earth orbits? And why is this?

I would imagine that the majority of the fuel burned by the rockets is outside of any significant atmosphere and therefore the efficiency is relatively constant.

$\endgroup$
  • 3
    $\begingroup$ In theory, the orbit around a single mass always has the same amount of total energy (Kinetic + Potential): It's not true. A body of mass $m$ in a circular orbit of radius $r$ around a body of mass $M$ has potential energy equal to $-\frac{GMm}r$ and kinetic energy equal to $\frac 12\frac{GMm}r$. So the total energy is $-\frac 12\frac{GMm}r$, which depends on the radius. Satellites in higher orbits have higher total energy than satellites in lower orbits. $\endgroup$ – Litho Nov 25 '15 at 10:25
  • $\begingroup$ Yes you're right, seems I have completely forgotten about my high school physics. I feel stupid now ;-) $\endgroup$ – fishlein Nov 25 '15 at 10:40
  • $\begingroup$ I guess the overall question could still be interesting though $\endgroup$ – fishlein Nov 25 '15 at 10:41
1
$\begingroup$

First, your assumption is wrong for the simple reason we need nearly 8km/s to bring any payload - and all the fuel to accelerate it any further - to LEO. And that means burning enormous amounts of fuel burned during the launch phase; for every kilogram of fuel in orbit you need several kilograms of fuel burned in the atmosphere. This is known as the Tyranny of Rocket Equation, and this is an unavoidable, fixed initial cost to get anything into space - and bring it up there fast, because on top of all the acceleration you give your payload to bring it to orbital speed, every second you're losing 9.8m/s of that speed to "gravitational drag", fighting Earth's pull to keep from falling.

But once in orbit, all the fun begins with orbital mechanics. And there's a lot of savings and caveats there.

Your primary saving is the Oberth Effect. There were many questions on this site about understanding it, and I encourage you to look them up. The essence is that if your fuel has a high kinetic energy (that is - it's moving fast; the ship is flying fast) - then by burning it, you harness this energy into more acceleration.

Orbital mechanics works in such a way, that a body in elliptical orbit moves faster when it's close to the body it's orbiting - so there's more to be gained by burning in LEO than any farther.

OTOH direction corrections are easier at low speeds - near apoapsis of an orbit.

And then there are gravitational slingshots; using other bodies (planets, moons) to steal a part of their kinetic energy, entirely for free.

I can encourage you to play Kerbal Space Program if you want to explore all these intricacies in an intuitive way, not through dry equations but more like a gravitational theme park :)

But to give a "tl;dr" answer: the orbit around a single mass always has the same amount of total energy (Kinetic + Potential). This is true, but only the Kinetic energy can be usefully transformed into speed by spacecraft engines. Therefore burns performed where the Kinetic component is highest are most efficient.

$\endgroup$
  • $\begingroup$ If I understand the first part correctly, you are saying that in order to bring an object into a high orbit, you need to first bring it into low orbit? $\endgroup$ – fishlein Nov 25 '15 at 9:28
  • $\begingroup$ @fishlein: Not exactly - you can launch directly into high orbit, though performing the burn to bring the apoapsis high enough while still at low orbit altitude (but not speed), and the energy consumption will be about the same. One problem is that until you're safely in any orbit, you have only as much time to perform the acceleration as it takes you to fall back down to Earth. This is not a problem with high-thrust rocket engines, but these are not very efficient; wasteful. If you want to use ion engines or such, you need a lot of time - and that means not falling down = staying in orbit. $\endgroup$ – SF. Nov 25 '15 at 9:45
  • $\begingroup$ Let me amend that: if you launch directly to a higher orbit it's slightly more efficient: your linear speed matters and if before achieving LEO altitude you achieve speed higher than LEO orbital speed (even if the resulting elliptical orbit trajectory crosses Earth surface) you're better off. But take three circular orbits, A,B,C with radii $r_A < r_B < r_C$ you're better off transferring directly $A \rightarrow C$ than doing $A \rightarrow B \rightarrow C$ because orbital speed of A is the highest. $\endgroup$ – SF. Nov 25 '15 at 9:58
  • 1
    $\begingroup$ If you are already in a high orbit and want to leave earth's Sphere Of Influence for another destination, it could save delta V to put on the brakes and drop down and then do your major burn close to earth. See hopsblog-hop.blogspot.com/2013/10/what-about-mr-oberth.html $\endgroup$ – HopDavid Nov 27 '15 at 0:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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