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I Rocket equation: is there an optimal thrust?

  1. Jul 9, 2017 #1
    This is my first post. I looked through many threads and could not find an answer to the question below.

    Well known equations in rocketry are:

    Delta Velocity = Ejection Velocity * ln (Final Mass / Initial Mass )


    Net Thrust = Mass Ejection Rate * Ejection Velocity – Current Mass * Acceleration

    The first equation does not refer to Mass Ejection Rate. If the Mass Ejection Rate (MER) is too small the thrust will not be sufficient for the rocket to take off. Increasing the MER one might get:

    Rocket takes off once Thrust > Weight

    Rocket takes off immediately at low acceleration

    Rocket takes off immediately at high acceleration

    Questions: Ignoring drag, is there an optimal thrust that provides maximum delta V for a rocket taking off from the ground?
  2. jcsd
  3. Jul 9, 2017 #2


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    The faster the burn and the higher the thrust, the better... as long as the structure and the cargo can handle the acceleration.

    One way of seeing this is to imagine what happens if the thrust is no greater than the weight of the rocket: it hovers, gaining no altitude, until the fuel is exhausted and it falls to the ground.
  4. Jul 10, 2017 #3


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    For a launch from within an atmosphere, one does not want to get too fast too low. Your savings from spending less time in the gravity well could be offset by losses from spending more energy to air resistance. This is in addition worrying about whether the craft can handle the stresses from air resistance. For instance, see https://en.wikipedia.org/wiki/Max_Q
  5. Jul 10, 2017 #4
    Thank you, Nugatory and jbriggs44.
    Yes, that is why when a Falcon 9 is launched at about 1 minute of flight and 12 Km of altitude it is throttled down temporarily while going through maximum dynamic pressure. Once atmospheric pressure is reduced is is throttled back to full power.
    Still, ignoring drag and material strength, is there an optimal thrust for maximum delta V?
  6. Jul 10, 2017 #5


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    Ignoring drag and material strength @Nugatory has provided the answer in post #2 above. The faster the better.
  7. Jul 10, 2017 #6
    Hi Nuggatory
    Tsiolkowski's equation is considered to be THE rocket equation. The higher the thrust the better is a constraint not implied by the equation.

    Thanks jbriggs444
    After I posted my response above I realized something you said did not quite fit. At constant thrust a rocket cannot hover as its mass is being decreased. Right?
  8. Jul 10, 2017 #7
    Sorry kbriggs444. the hovering statement also came from Nugatory.
  9. Jul 10, 2017 #8


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    If you are launching from within a gravity well, the more time you spend in the well, the more delta-V you lose to the well. Yes, Tsiolkovski gives you a fixed delta V regardless of mass flow rate. But you can budget that delta-V toward hovering in place or toward escaping the gravity well by choosing your mass flow rate.

    Edit: I look at a gravity field as if it were a leak in your tank of "delta V". The more time you spend leaking delta V, the more you'll lose. The closer you are to the planet you're launching from, the stronger the gravity and the faster you're leaking delta V.

    A high mass flow rate means that you burn the fuel lower in the well and you spend less time subject to the stronger gravity there. The faster you burn, the better.
    @Nugatory spoke of a rocket using just enough thrust to hover. Yes, the required thrust would not be constant but would decrease as the fuel left on board depletes.
  10. Jul 10, 2017 #9
    Hi jbrigss444
    You are not implying that the thrust would decrease as fuel on board depletes, are you? It takes some smart technology to make a rocket hover.
  11. Jul 10, 2017 #10


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    If you want to hover, you would need to throttle back. Yes.

    Edit: And yes, nobody hovers on rocket engines because it is such an obvious waste of fuel.
    Last edited: Jul 10, 2017
  12. Jul 10, 2017 #11


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    There's also an optimum direction. Once beyond significant effects from the atmosphere, the closer the thrust is to being perpendicular to the force of gravity, the more efficient. If the thrust is perpendicular to the force of gravity, then all of the thrust goes into increasing velocity.

    update - willem2 notes in his post that thrust in the same direction as velocity is most efficient. In the case of a circular orbit, the velocity is perpendicular to gravity, but in the case of an elliptical orbit, the velocity is not perpendicular to gravity. jbrigs444 notes the Oberth effect, where it's most efficient to apply thrust at the perigee (fastest speed) of an orbit for the greatest change in kinetic energy, and this coincides with the velocity being close to perpendicular to gravity.
    Last edited: Jul 13, 2017
  13. Jul 10, 2017 #12


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    If it were not for the pesky planetary surface getting in the way, one could allow the trajectory to dip below the starting point and make use of the Oberth effect.
  14. Jul 12, 2017 #13
    The thrust is most efficient if it is in the same direction as the velocity. That will increase the kinetic energy the most. The only reason you won't use that is if you will crash into the earth.
    Thrust perpendicular to gravity does not always do this. For example if you thrust sideways in a circular orbit, you will thrust perpendicular to gravity, but you will remain in a circular orbit, only the plane of the orbit will change.
  15. Jul 12, 2017 #14


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    Note, however, that this is only a local optimum. If you want to increase the energy of the craft with a single impulsive burn applied immediately, the optimal direction is indeed parallel to the velocity.

    If you want to increase the energy but you have the option of performing multiple burns and if you are in a gravity well then a global optimum might involve firing anti-parallel [and at apogee] to produce a grazing orbit and then firing parallel at perigee to obtain the maximum energy gain.

    If one is at rest on the surface of a rotating spherical body devoid of atmosphere then the local optimum is the same as the global optimum and we are all in agreement: "fire all your engines at once [horizontally] and explode into space"
    Last edited: Jul 12, 2017
  16. Jul 13, 2017 #15


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    True, but I was thinking of the Oberth effect as posted by jbriggs444. The thrust is most efficient at perigee (fastest speed) of an elliptical orbit, where the velocity is close to being perpendicular to gravity. If the goal is to transition from a smaller circular orbit to a larger circular orbit, then for a Hohmann transfer orbit, the two burns are done at perigee and apogee, when velocity is close to perpendicular to gravity.

    Last edited: Jul 13, 2017
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