Why is delta-v the most useful quantity for planning space travel?












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Many of the questions and answers on this site make use of the concept of delta-v. Is there an easy to understand the reason why delta-v, the magnitude of the change of the velocity, $|mathbf{v}|$, is so useful for understanding orbital mechanics and planning travel?



My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities. Delta-v doesn't seem to be a good proxy for any of these quantities, since it's not squared like the kinetic energy, but it's also not a vector like the linear and angular momenta.










share|improve this question











$endgroup$








  • 3




    $begingroup$
    I'd wager that it's because that quantity is an increasing value, with respect to time, it can never decrease. You cannot lose delta-v over time, you can only increase your delta-v. Also it's agnostic to the body, unlike angular momentum. For a transfer to Mars, you could say "It will take X change in velocity from LEO to LMO." Where-as what you would say for momentum you'll have to say "I need a momentum/energy increase of X from LEO then a momentum/energy decrease of X from Mars approach to LMO". (Note I actually do not know)
    $endgroup$
    – Magic Octopus Urn
    Mar 25 at 18:10








  • 5




    $begingroup$
    Ultimately, we use delta-v to determine the amount of fuel needed to change the trajectory to a desired one using impulse thrusts (sudden change in velocity). But since fuel mass grows exponentially with delta-v, it’s easier to work with delta-v instead of fuel mass directly. It doesn’t matter if you slow down or speed up, the fuel consumed is the same for a given delta-v. Thus, you can accumulate each velocity change over a mission to estimate fuel needed
    $endgroup$
    – Paul
    Mar 25 at 19:23






  • 5




    $begingroup$
    The short version: Mass cancels out.
    $endgroup$
    – chrylis
    Mar 26 at 1:20






  • 5




    $begingroup$
    Spend a few days playing Kerbal Space Program and you should develop a much better understanding of why delta-v is by far the most important factor in spacecraft design and mission planning.
    $endgroup$
    – MooseBoys
    Mar 26 at 5:14






  • 2




    $begingroup$
    In addition to other answers and comments, energy, linear momentum, and angular momentum aren't very useful because they depend on mass and in a spacecraft mass is not constant. A workaround could be using change of momentum per mass unit, which is (unsurprisingly) equal to delta-v.
    $endgroup$
    – Pere
    Mar 26 at 22:21
















29












$begingroup$


Many of the questions and answers on this site make use of the concept of delta-v. Is there an easy to understand the reason why delta-v, the magnitude of the change of the velocity, $|mathbf{v}|$, is so useful for understanding orbital mechanics and planning travel?



My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities. Delta-v doesn't seem to be a good proxy for any of these quantities, since it's not squared like the kinetic energy, but it's also not a vector like the linear and angular momenta.










share|improve this question











$endgroup$








  • 3




    $begingroup$
    I'd wager that it's because that quantity is an increasing value, with respect to time, it can never decrease. You cannot lose delta-v over time, you can only increase your delta-v. Also it's agnostic to the body, unlike angular momentum. For a transfer to Mars, you could say "It will take X change in velocity from LEO to LMO." Where-as what you would say for momentum you'll have to say "I need a momentum/energy increase of X from LEO then a momentum/energy decrease of X from Mars approach to LMO". (Note I actually do not know)
    $endgroup$
    – Magic Octopus Urn
    Mar 25 at 18:10








  • 5




    $begingroup$
    Ultimately, we use delta-v to determine the amount of fuel needed to change the trajectory to a desired one using impulse thrusts (sudden change in velocity). But since fuel mass grows exponentially with delta-v, it’s easier to work with delta-v instead of fuel mass directly. It doesn’t matter if you slow down or speed up, the fuel consumed is the same for a given delta-v. Thus, you can accumulate each velocity change over a mission to estimate fuel needed
    $endgroup$
    – Paul
    Mar 25 at 19:23






  • 5




    $begingroup$
    The short version: Mass cancels out.
    $endgroup$
    – chrylis
    Mar 26 at 1:20






  • 5




    $begingroup$
    Spend a few days playing Kerbal Space Program and you should develop a much better understanding of why delta-v is by far the most important factor in spacecraft design and mission planning.
    $endgroup$
    – MooseBoys
    Mar 26 at 5:14






  • 2




    $begingroup$
    In addition to other answers and comments, energy, linear momentum, and angular momentum aren't very useful because they depend on mass and in a spacecraft mass is not constant. A workaround could be using change of momentum per mass unit, which is (unsurprisingly) equal to delta-v.
    $endgroup$
    – Pere
    Mar 26 at 22:21














29












29








29


3



$begingroup$


Many of the questions and answers on this site make use of the concept of delta-v. Is there an easy to understand the reason why delta-v, the magnitude of the change of the velocity, $|mathbf{v}|$, is so useful for understanding orbital mechanics and planning travel?



My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities. Delta-v doesn't seem to be a good proxy for any of these quantities, since it's not squared like the kinetic energy, but it's also not a vector like the linear and angular momenta.










share|improve this question











$endgroup$




Many of the questions and answers on this site make use of the concept of delta-v. Is there an easy to understand the reason why delta-v, the magnitude of the change of the velocity, $|mathbf{v}|$, is so useful for understanding orbital mechanics and planning travel?



My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities. Delta-v doesn't seem to be a good proxy for any of these quantities, since it's not squared like the kinetic energy, but it's also not a vector like the linear and angular momenta.







orbital-mechanics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 27 at 10:48









Peter Mortensen

21217




21217










asked Mar 25 at 17:25









WaterMoleculeWaterMolecule

760512




760512








  • 3




    $begingroup$
    I'd wager that it's because that quantity is an increasing value, with respect to time, it can never decrease. You cannot lose delta-v over time, you can only increase your delta-v. Also it's agnostic to the body, unlike angular momentum. For a transfer to Mars, you could say "It will take X change in velocity from LEO to LMO." Where-as what you would say for momentum you'll have to say "I need a momentum/energy increase of X from LEO then a momentum/energy decrease of X from Mars approach to LMO". (Note I actually do not know)
    $endgroup$
    – Magic Octopus Urn
    Mar 25 at 18:10








  • 5




    $begingroup$
    Ultimately, we use delta-v to determine the amount of fuel needed to change the trajectory to a desired one using impulse thrusts (sudden change in velocity). But since fuel mass grows exponentially with delta-v, it’s easier to work with delta-v instead of fuel mass directly. It doesn’t matter if you slow down or speed up, the fuel consumed is the same for a given delta-v. Thus, you can accumulate each velocity change over a mission to estimate fuel needed
    $endgroup$
    – Paul
    Mar 25 at 19:23






  • 5




    $begingroup$
    The short version: Mass cancels out.
    $endgroup$
    – chrylis
    Mar 26 at 1:20






  • 5




    $begingroup$
    Spend a few days playing Kerbal Space Program and you should develop a much better understanding of why delta-v is by far the most important factor in spacecraft design and mission planning.
    $endgroup$
    – MooseBoys
    Mar 26 at 5:14






  • 2




    $begingroup$
    In addition to other answers and comments, energy, linear momentum, and angular momentum aren't very useful because they depend on mass and in a spacecraft mass is not constant. A workaround could be using change of momentum per mass unit, which is (unsurprisingly) equal to delta-v.
    $endgroup$
    – Pere
    Mar 26 at 22:21














  • 3




    $begingroup$
    I'd wager that it's because that quantity is an increasing value, with respect to time, it can never decrease. You cannot lose delta-v over time, you can only increase your delta-v. Also it's agnostic to the body, unlike angular momentum. For a transfer to Mars, you could say "It will take X change in velocity from LEO to LMO." Where-as what you would say for momentum you'll have to say "I need a momentum/energy increase of X from LEO then a momentum/energy decrease of X from Mars approach to LMO". (Note I actually do not know)
    $endgroup$
    – Magic Octopus Urn
    Mar 25 at 18:10








  • 5




    $begingroup$
    Ultimately, we use delta-v to determine the amount of fuel needed to change the trajectory to a desired one using impulse thrusts (sudden change in velocity). But since fuel mass grows exponentially with delta-v, it’s easier to work with delta-v instead of fuel mass directly. It doesn’t matter if you slow down or speed up, the fuel consumed is the same for a given delta-v. Thus, you can accumulate each velocity change over a mission to estimate fuel needed
    $endgroup$
    – Paul
    Mar 25 at 19:23






  • 5




    $begingroup$
    The short version: Mass cancels out.
    $endgroup$
    – chrylis
    Mar 26 at 1:20






  • 5




    $begingroup$
    Spend a few days playing Kerbal Space Program and you should develop a much better understanding of why delta-v is by far the most important factor in spacecraft design and mission planning.
    $endgroup$
    – MooseBoys
    Mar 26 at 5:14






  • 2




    $begingroup$
    In addition to other answers and comments, energy, linear momentum, and angular momentum aren't very useful because they depend on mass and in a spacecraft mass is not constant. A workaround could be using change of momentum per mass unit, which is (unsurprisingly) equal to delta-v.
    $endgroup$
    – Pere
    Mar 26 at 22:21








3




3




$begingroup$
I'd wager that it's because that quantity is an increasing value, with respect to time, it can never decrease. You cannot lose delta-v over time, you can only increase your delta-v. Also it's agnostic to the body, unlike angular momentum. For a transfer to Mars, you could say "It will take X change in velocity from LEO to LMO." Where-as what you would say for momentum you'll have to say "I need a momentum/energy increase of X from LEO then a momentum/energy decrease of X from Mars approach to LMO". (Note I actually do not know)
$endgroup$
– Magic Octopus Urn
Mar 25 at 18:10






$begingroup$
I'd wager that it's because that quantity is an increasing value, with respect to time, it can never decrease. You cannot lose delta-v over time, you can only increase your delta-v. Also it's agnostic to the body, unlike angular momentum. For a transfer to Mars, you could say "It will take X change in velocity from LEO to LMO." Where-as what you would say for momentum you'll have to say "I need a momentum/energy increase of X from LEO then a momentum/energy decrease of X from Mars approach to LMO". (Note I actually do not know)
$endgroup$
– Magic Octopus Urn
Mar 25 at 18:10






5




5




$begingroup$
Ultimately, we use delta-v to determine the amount of fuel needed to change the trajectory to a desired one using impulse thrusts (sudden change in velocity). But since fuel mass grows exponentially with delta-v, it’s easier to work with delta-v instead of fuel mass directly. It doesn’t matter if you slow down or speed up, the fuel consumed is the same for a given delta-v. Thus, you can accumulate each velocity change over a mission to estimate fuel needed
$endgroup$
– Paul
Mar 25 at 19:23




$begingroup$
Ultimately, we use delta-v to determine the amount of fuel needed to change the trajectory to a desired one using impulse thrusts (sudden change in velocity). But since fuel mass grows exponentially with delta-v, it’s easier to work with delta-v instead of fuel mass directly. It doesn’t matter if you slow down or speed up, the fuel consumed is the same for a given delta-v. Thus, you can accumulate each velocity change over a mission to estimate fuel needed
$endgroup$
– Paul
Mar 25 at 19:23




5




5




$begingroup$
The short version: Mass cancels out.
$endgroup$
– chrylis
Mar 26 at 1:20




$begingroup$
The short version: Mass cancels out.
$endgroup$
– chrylis
Mar 26 at 1:20




5




5




$begingroup$
Spend a few days playing Kerbal Space Program and you should develop a much better understanding of why delta-v is by far the most important factor in spacecraft design and mission planning.
$endgroup$
– MooseBoys
Mar 26 at 5:14




$begingroup$
Spend a few days playing Kerbal Space Program and you should develop a much better understanding of why delta-v is by far the most important factor in spacecraft design and mission planning.
$endgroup$
– MooseBoys
Mar 26 at 5:14




2




2




$begingroup$
In addition to other answers and comments, energy, linear momentum, and angular momentum aren't very useful because they depend on mass and in a spacecraft mass is not constant. A workaround could be using change of momentum per mass unit, which is (unsurprisingly) equal to delta-v.
$endgroup$
– Pere
Mar 26 at 22:21




$begingroup$
In addition to other answers and comments, energy, linear momentum, and angular momentum aren't very useful because they depend on mass and in a spacecraft mass is not constant. A workaround could be using change of momentum per mass unit, which is (unsurprisingly) equal to delta-v.
$endgroup$
– Pere
Mar 26 at 22:21










3 Answers
3






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Your orbit is uniquely determined by a current position (three coordinates) and velocity (three more quantities to give magnitude and direction). Going places involves changing your orbit. For instance, from a circular orbit about Earth, enter an elliptical transfer orbit to the moon, then circularize your orbit about the moon. Everything you do in space travel involves changing from one orbit to another orbit, and that is done by changing your velocity.



Heavy spaceships have to change their momentum more than light spaceships, but they both have to change their velocities by the same amount. It can be done with a long, slow acceleration, or a short, fast acceleration. Whatever ship you have, and however you do it, the delta-V is the end result that you must achieve.



Your new orbit definitely does depend on your vector delta-V, but pointing your spaceship is basically a freebie. And you don't get any of your fuel back if you accelerate first in one direction and then in the opposite direction. So, as a characteristic of your spacecraft, it really kind of is a scalar quantity, even if direction does matter when you use it.






share|improve this answer









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  • 4




    $begingroup$
    Ahhh... great point. Its agnostic to mass as well. I knew I was missing something. I am glad I didnt answer :).
    $endgroup$
    – Magic Octopus Urn
    Mar 25 at 20:20










  • $begingroup$
    Note that plane-change operations may also be involved depending on where you want to go. And while plane changing does fall under the broad heading of "changing your velocity" its in a perpendicular direction to your orbital plane, as opposed to along your trajectory's path (either forwards or backwards). For instance, the ideal shuttle launch inclination and that of the ISS are quite different, so the easy math of just matching velocities won't be enough delta v.
    $endgroup$
    – Draco18s
    Mar 26 at 3:25










  • $begingroup$
    In considering trajectories going between different spheres of influence, it's not uncommon to refer to "(mass) specific energy", which is just energy per unit mass.
    $endgroup$
    – Russell Borogove
    Mar 26 at 16:10








  • 1




    $begingroup$
    I think what's missing from this answer is why delta-v is simply additive, i.e., why the relevant cost is linear in the delta-v magnitude (rather than say quadratic, which might seem more natural on grounds of vector algebra and energy).
    $endgroup$
    – nanoman
    Mar 26 at 19:19










  • $begingroup$
    @nanoman Because in space you're not pushing against the ground; the faster you're going in your thrust direction, the slower your propellant in the opposite direction.
    $endgroup$
    – wizzwizz4
    Mar 26 at 20:11



















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Delta-v determines the amount of propellant needed.



Suppose a craft with mass $m$ and velocity $mathbf{v}$ burns a small mass $|Delta m|$ of propellant and ejects it at relative velocity $mathbf{u}$, so that the craft mass changes by $Delta m < 0$. This occurs over a time $Delta t$ in a local gravitational field $mathbf{g}$. Then the new craft velocity $mathbf{v} + Deltamathbf{v}$ is given by
$$text{initial momentum} + text{change in momentum due to gravity} = text{final momentum of propellant} + text{final momentum of craft},$$
$$mmathbf{v} + (Delta t)mmathbf{g} = -Delta m(mathbf{v} + mathbf{u}) + (m + Delta m)(mathbf{v} + Deltamathbf{v}).$$
Given that the increments are small, this simplifies to
$$(Delta m)mathbf{u} = mDeltamathbf{v} - (Delta t)mmathbf{g}.$$
Dividing through by $(Delta t)m$ and passing to derivatives, we have
$$frac{dot mmathbf{u}}{m} = dot{mathbf{v}} - mathbf{g}.$$



Taking magnitudes (remembering $dot m < 0$) and integrating over time, we obtain the rocket equation
$$|mathbf{u}|lnfrac{m_0}{m} = int dt,|dot{mathbf{v}} - mathbf{g}|,$$
where $|mathbf{u}|$ is constant since it's a characteristic of the propulsion system. The right-hand side is the general definition of delta-v. We see that it is directly linked to the initial craft mass $m_0$, determining the initial amount of propellant needed.



Now, suppose the propellant is utilized in quick burns, during each of which $mathbf{u}$ is constant in direction and $|dot{mathbf{v}}| gg |mathbf{g}|$, separated by coasting intervals during which $dot{mathbf{v}} = mathbf{g}$ (i.e., $dot m = 0$). Then delta-v simplifies to
$$int dt,|dot{mathbf{v}} - mathbf{g}| = |Deltamathbf{v}_{text{burn 1}}| + |Deltamathbf{v}_{text{burn 2}}| + cdots,$$
hence its name. (In this equation $Deltamathbf{v}$ is not required to be small.)






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  • $begingroup$
    Good answer! I've had a hard time deciding which answer to accept. This one has the mathematical logic that I was looking for, but Greg's has the intuition that I was also looking for.
    $endgroup$
    – WaterMolecule
    Mar 26 at 15:55



















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$begingroup$


My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities.




To put @Greg's answer short: delta-V is a mass-normalized measure to all of the quantities you mention.



(not addressing vector vs. scalar)






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    3 Answers
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    3 Answers
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    66












    $begingroup$

    Your orbit is uniquely determined by a current position (three coordinates) and velocity (three more quantities to give magnitude and direction). Going places involves changing your orbit. For instance, from a circular orbit about Earth, enter an elliptical transfer orbit to the moon, then circularize your orbit about the moon. Everything you do in space travel involves changing from one orbit to another orbit, and that is done by changing your velocity.



    Heavy spaceships have to change their momentum more than light spaceships, but they both have to change their velocities by the same amount. It can be done with a long, slow acceleration, or a short, fast acceleration. Whatever ship you have, and however you do it, the delta-V is the end result that you must achieve.



    Your new orbit definitely does depend on your vector delta-V, but pointing your spaceship is basically a freebie. And you don't get any of your fuel back if you accelerate first in one direction and then in the opposite direction. So, as a characteristic of your spacecraft, it really kind of is a scalar quantity, even if direction does matter when you use it.






    share|improve this answer









    $endgroup$









    • 4




      $begingroup$
      Ahhh... great point. Its agnostic to mass as well. I knew I was missing something. I am glad I didnt answer :).
      $endgroup$
      – Magic Octopus Urn
      Mar 25 at 20:20










    • $begingroup$
      Note that plane-change operations may also be involved depending on where you want to go. And while plane changing does fall under the broad heading of "changing your velocity" its in a perpendicular direction to your orbital plane, as opposed to along your trajectory's path (either forwards or backwards). For instance, the ideal shuttle launch inclination and that of the ISS are quite different, so the easy math of just matching velocities won't be enough delta v.
      $endgroup$
      – Draco18s
      Mar 26 at 3:25










    • $begingroup$
      In considering trajectories going between different spheres of influence, it's not uncommon to refer to "(mass) specific energy", which is just energy per unit mass.
      $endgroup$
      – Russell Borogove
      Mar 26 at 16:10








    • 1




      $begingroup$
      I think what's missing from this answer is why delta-v is simply additive, i.e., why the relevant cost is linear in the delta-v magnitude (rather than say quadratic, which might seem more natural on grounds of vector algebra and energy).
      $endgroup$
      – nanoman
      Mar 26 at 19:19










    • $begingroup$
      @nanoman Because in space you're not pushing against the ground; the faster you're going in your thrust direction, the slower your propellant in the opposite direction.
      $endgroup$
      – wizzwizz4
      Mar 26 at 20:11
















    66












    $begingroup$

    Your orbit is uniquely determined by a current position (three coordinates) and velocity (three more quantities to give magnitude and direction). Going places involves changing your orbit. For instance, from a circular orbit about Earth, enter an elliptical transfer orbit to the moon, then circularize your orbit about the moon. Everything you do in space travel involves changing from one orbit to another orbit, and that is done by changing your velocity.



    Heavy spaceships have to change their momentum more than light spaceships, but they both have to change their velocities by the same amount. It can be done with a long, slow acceleration, or a short, fast acceleration. Whatever ship you have, and however you do it, the delta-V is the end result that you must achieve.



    Your new orbit definitely does depend on your vector delta-V, but pointing your spaceship is basically a freebie. And you don't get any of your fuel back if you accelerate first in one direction and then in the opposite direction. So, as a characteristic of your spacecraft, it really kind of is a scalar quantity, even if direction does matter when you use it.






    share|improve this answer









    $endgroup$









    • 4




      $begingroup$
      Ahhh... great point. Its agnostic to mass as well. I knew I was missing something. I am glad I didnt answer :).
      $endgroup$
      – Magic Octopus Urn
      Mar 25 at 20:20










    • $begingroup$
      Note that plane-change operations may also be involved depending on where you want to go. And while plane changing does fall under the broad heading of "changing your velocity" its in a perpendicular direction to your orbital plane, as opposed to along your trajectory's path (either forwards or backwards). For instance, the ideal shuttle launch inclination and that of the ISS are quite different, so the easy math of just matching velocities won't be enough delta v.
      $endgroup$
      – Draco18s
      Mar 26 at 3:25










    • $begingroup$
      In considering trajectories going between different spheres of influence, it's not uncommon to refer to "(mass) specific energy", which is just energy per unit mass.
      $endgroup$
      – Russell Borogove
      Mar 26 at 16:10








    • 1




      $begingroup$
      I think what's missing from this answer is why delta-v is simply additive, i.e., why the relevant cost is linear in the delta-v magnitude (rather than say quadratic, which might seem more natural on grounds of vector algebra and energy).
      $endgroup$
      – nanoman
      Mar 26 at 19:19










    • $begingroup$
      @nanoman Because in space you're not pushing against the ground; the faster you're going in your thrust direction, the slower your propellant in the opposite direction.
      $endgroup$
      – wizzwizz4
      Mar 26 at 20:11














    66












    66








    66





    $begingroup$

    Your orbit is uniquely determined by a current position (three coordinates) and velocity (three more quantities to give magnitude and direction). Going places involves changing your orbit. For instance, from a circular orbit about Earth, enter an elliptical transfer orbit to the moon, then circularize your orbit about the moon. Everything you do in space travel involves changing from one orbit to another orbit, and that is done by changing your velocity.



    Heavy spaceships have to change their momentum more than light spaceships, but they both have to change their velocities by the same amount. It can be done with a long, slow acceleration, or a short, fast acceleration. Whatever ship you have, and however you do it, the delta-V is the end result that you must achieve.



    Your new orbit definitely does depend on your vector delta-V, but pointing your spaceship is basically a freebie. And you don't get any of your fuel back if you accelerate first in one direction and then in the opposite direction. So, as a characteristic of your spacecraft, it really kind of is a scalar quantity, even if direction does matter when you use it.






    share|improve this answer









    $endgroup$



    Your orbit is uniquely determined by a current position (three coordinates) and velocity (three more quantities to give magnitude and direction). Going places involves changing your orbit. For instance, from a circular orbit about Earth, enter an elliptical transfer orbit to the moon, then circularize your orbit about the moon. Everything you do in space travel involves changing from one orbit to another orbit, and that is done by changing your velocity.



    Heavy spaceships have to change their momentum more than light spaceships, but they both have to change their velocities by the same amount. It can be done with a long, slow acceleration, or a short, fast acceleration. Whatever ship you have, and however you do it, the delta-V is the end result that you must achieve.



    Your new orbit definitely does depend on your vector delta-V, but pointing your spaceship is basically a freebie. And you don't get any of your fuel back if you accelerate first in one direction and then in the opposite direction. So, as a characteristic of your spacecraft, it really kind of is a scalar quantity, even if direction does matter when you use it.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Mar 25 at 20:09









    GregGreg

    87638




    87638








    • 4




      $begingroup$
      Ahhh... great point. Its agnostic to mass as well. I knew I was missing something. I am glad I didnt answer :).
      $endgroup$
      – Magic Octopus Urn
      Mar 25 at 20:20










    • $begingroup$
      Note that plane-change operations may also be involved depending on where you want to go. And while plane changing does fall under the broad heading of "changing your velocity" its in a perpendicular direction to your orbital plane, as opposed to along your trajectory's path (either forwards or backwards). For instance, the ideal shuttle launch inclination and that of the ISS are quite different, so the easy math of just matching velocities won't be enough delta v.
      $endgroup$
      – Draco18s
      Mar 26 at 3:25










    • $begingroup$
      In considering trajectories going between different spheres of influence, it's not uncommon to refer to "(mass) specific energy", which is just energy per unit mass.
      $endgroup$
      – Russell Borogove
      Mar 26 at 16:10








    • 1




      $begingroup$
      I think what's missing from this answer is why delta-v is simply additive, i.e., why the relevant cost is linear in the delta-v magnitude (rather than say quadratic, which might seem more natural on grounds of vector algebra and energy).
      $endgroup$
      – nanoman
      Mar 26 at 19:19










    • $begingroup$
      @nanoman Because in space you're not pushing against the ground; the faster you're going in your thrust direction, the slower your propellant in the opposite direction.
      $endgroup$
      – wizzwizz4
      Mar 26 at 20:11














    • 4




      $begingroup$
      Ahhh... great point. Its agnostic to mass as well. I knew I was missing something. I am glad I didnt answer :).
      $endgroup$
      – Magic Octopus Urn
      Mar 25 at 20:20










    • $begingroup$
      Note that plane-change operations may also be involved depending on where you want to go. And while plane changing does fall under the broad heading of "changing your velocity" its in a perpendicular direction to your orbital plane, as opposed to along your trajectory's path (either forwards or backwards). For instance, the ideal shuttle launch inclination and that of the ISS are quite different, so the easy math of just matching velocities won't be enough delta v.
      $endgroup$
      – Draco18s
      Mar 26 at 3:25










    • $begingroup$
      In considering trajectories going between different spheres of influence, it's not uncommon to refer to "(mass) specific energy", which is just energy per unit mass.
      $endgroup$
      – Russell Borogove
      Mar 26 at 16:10








    • 1




      $begingroup$
      I think what's missing from this answer is why delta-v is simply additive, i.e., why the relevant cost is linear in the delta-v magnitude (rather than say quadratic, which might seem more natural on grounds of vector algebra and energy).
      $endgroup$
      – nanoman
      Mar 26 at 19:19










    • $begingroup$
      @nanoman Because in space you're not pushing against the ground; the faster you're going in your thrust direction, the slower your propellant in the opposite direction.
      $endgroup$
      – wizzwizz4
      Mar 26 at 20:11








    4




    4




    $begingroup$
    Ahhh... great point. Its agnostic to mass as well. I knew I was missing something. I am glad I didnt answer :).
    $endgroup$
    – Magic Octopus Urn
    Mar 25 at 20:20




    $begingroup$
    Ahhh... great point. Its agnostic to mass as well. I knew I was missing something. I am glad I didnt answer :).
    $endgroup$
    – Magic Octopus Urn
    Mar 25 at 20:20












    $begingroup$
    Note that plane-change operations may also be involved depending on where you want to go. And while plane changing does fall under the broad heading of "changing your velocity" its in a perpendicular direction to your orbital plane, as opposed to along your trajectory's path (either forwards or backwards). For instance, the ideal shuttle launch inclination and that of the ISS are quite different, so the easy math of just matching velocities won't be enough delta v.
    $endgroup$
    – Draco18s
    Mar 26 at 3:25




    $begingroup$
    Note that plane-change operations may also be involved depending on where you want to go. And while plane changing does fall under the broad heading of "changing your velocity" its in a perpendicular direction to your orbital plane, as opposed to along your trajectory's path (either forwards or backwards). For instance, the ideal shuttle launch inclination and that of the ISS are quite different, so the easy math of just matching velocities won't be enough delta v.
    $endgroup$
    – Draco18s
    Mar 26 at 3:25












    $begingroup$
    In considering trajectories going between different spheres of influence, it's not uncommon to refer to "(mass) specific energy", which is just energy per unit mass.
    $endgroup$
    – Russell Borogove
    Mar 26 at 16:10






    $begingroup$
    In considering trajectories going between different spheres of influence, it's not uncommon to refer to "(mass) specific energy", which is just energy per unit mass.
    $endgroup$
    – Russell Borogove
    Mar 26 at 16:10






    1




    1




    $begingroup$
    I think what's missing from this answer is why delta-v is simply additive, i.e., why the relevant cost is linear in the delta-v magnitude (rather than say quadratic, which might seem more natural on grounds of vector algebra and energy).
    $endgroup$
    – nanoman
    Mar 26 at 19:19




    $begingroup$
    I think what's missing from this answer is why delta-v is simply additive, i.e., why the relevant cost is linear in the delta-v magnitude (rather than say quadratic, which might seem more natural on grounds of vector algebra and energy).
    $endgroup$
    – nanoman
    Mar 26 at 19:19












    $begingroup$
    @nanoman Because in space you're not pushing against the ground; the faster you're going in your thrust direction, the slower your propellant in the opposite direction.
    $endgroup$
    – wizzwizz4
    Mar 26 at 20:11




    $begingroup$
    @nanoman Because in space you're not pushing against the ground; the faster you're going in your thrust direction, the slower your propellant in the opposite direction.
    $endgroup$
    – wizzwizz4
    Mar 26 at 20:11











    9












    $begingroup$

    Delta-v determines the amount of propellant needed.



    Suppose a craft with mass $m$ and velocity $mathbf{v}$ burns a small mass $|Delta m|$ of propellant and ejects it at relative velocity $mathbf{u}$, so that the craft mass changes by $Delta m < 0$. This occurs over a time $Delta t$ in a local gravitational field $mathbf{g}$. Then the new craft velocity $mathbf{v} + Deltamathbf{v}$ is given by
    $$text{initial momentum} + text{change in momentum due to gravity} = text{final momentum of propellant} + text{final momentum of craft},$$
    $$mmathbf{v} + (Delta t)mmathbf{g} = -Delta m(mathbf{v} + mathbf{u}) + (m + Delta m)(mathbf{v} + Deltamathbf{v}).$$
    Given that the increments are small, this simplifies to
    $$(Delta m)mathbf{u} = mDeltamathbf{v} - (Delta t)mmathbf{g}.$$
    Dividing through by $(Delta t)m$ and passing to derivatives, we have
    $$frac{dot mmathbf{u}}{m} = dot{mathbf{v}} - mathbf{g}.$$



    Taking magnitudes (remembering $dot m < 0$) and integrating over time, we obtain the rocket equation
    $$|mathbf{u}|lnfrac{m_0}{m} = int dt,|dot{mathbf{v}} - mathbf{g}|,$$
    where $|mathbf{u}|$ is constant since it's a characteristic of the propulsion system. The right-hand side is the general definition of delta-v. We see that it is directly linked to the initial craft mass $m_0$, determining the initial amount of propellant needed.



    Now, suppose the propellant is utilized in quick burns, during each of which $mathbf{u}$ is constant in direction and $|dot{mathbf{v}}| gg |mathbf{g}|$, separated by coasting intervals during which $dot{mathbf{v}} = mathbf{g}$ (i.e., $dot m = 0$). Then delta-v simplifies to
    $$int dt,|dot{mathbf{v}} - mathbf{g}| = |Deltamathbf{v}_{text{burn 1}}| + |Deltamathbf{v}_{text{burn 2}}| + cdots,$$
    hence its name. (In this equation $Deltamathbf{v}$ is not required to be small.)






    share|improve this answer








    New contributor




    nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    $endgroup$













    • $begingroup$
      Good answer! I've had a hard time deciding which answer to accept. This one has the mathematical logic that I was looking for, but Greg's has the intuition that I was also looking for.
      $endgroup$
      – WaterMolecule
      Mar 26 at 15:55
















    9












    $begingroup$

    Delta-v determines the amount of propellant needed.



    Suppose a craft with mass $m$ and velocity $mathbf{v}$ burns a small mass $|Delta m|$ of propellant and ejects it at relative velocity $mathbf{u}$, so that the craft mass changes by $Delta m < 0$. This occurs over a time $Delta t$ in a local gravitational field $mathbf{g}$. Then the new craft velocity $mathbf{v} + Deltamathbf{v}$ is given by
    $$text{initial momentum} + text{change in momentum due to gravity} = text{final momentum of propellant} + text{final momentum of craft},$$
    $$mmathbf{v} + (Delta t)mmathbf{g} = -Delta m(mathbf{v} + mathbf{u}) + (m + Delta m)(mathbf{v} + Deltamathbf{v}).$$
    Given that the increments are small, this simplifies to
    $$(Delta m)mathbf{u} = mDeltamathbf{v} - (Delta t)mmathbf{g}.$$
    Dividing through by $(Delta t)m$ and passing to derivatives, we have
    $$frac{dot mmathbf{u}}{m} = dot{mathbf{v}} - mathbf{g}.$$



    Taking magnitudes (remembering $dot m < 0$) and integrating over time, we obtain the rocket equation
    $$|mathbf{u}|lnfrac{m_0}{m} = int dt,|dot{mathbf{v}} - mathbf{g}|,$$
    where $|mathbf{u}|$ is constant since it's a characteristic of the propulsion system. The right-hand side is the general definition of delta-v. We see that it is directly linked to the initial craft mass $m_0$, determining the initial amount of propellant needed.



    Now, suppose the propellant is utilized in quick burns, during each of which $mathbf{u}$ is constant in direction and $|dot{mathbf{v}}| gg |mathbf{g}|$, separated by coasting intervals during which $dot{mathbf{v}} = mathbf{g}$ (i.e., $dot m = 0$). Then delta-v simplifies to
    $$int dt,|dot{mathbf{v}} - mathbf{g}| = |Deltamathbf{v}_{text{burn 1}}| + |Deltamathbf{v}_{text{burn 2}}| + cdots,$$
    hence its name. (In this equation $Deltamathbf{v}$ is not required to be small.)






    share|improve this answer








    New contributor




    nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$













    • $begingroup$
      Good answer! I've had a hard time deciding which answer to accept. This one has the mathematical logic that I was looking for, but Greg's has the intuition that I was also looking for.
      $endgroup$
      – WaterMolecule
      Mar 26 at 15:55














    9












    9








    9





    $begingroup$

    Delta-v determines the amount of propellant needed.



    Suppose a craft with mass $m$ and velocity $mathbf{v}$ burns a small mass $|Delta m|$ of propellant and ejects it at relative velocity $mathbf{u}$, so that the craft mass changes by $Delta m < 0$. This occurs over a time $Delta t$ in a local gravitational field $mathbf{g}$. Then the new craft velocity $mathbf{v} + Deltamathbf{v}$ is given by
    $$text{initial momentum} + text{change in momentum due to gravity} = text{final momentum of propellant} + text{final momentum of craft},$$
    $$mmathbf{v} + (Delta t)mmathbf{g} = -Delta m(mathbf{v} + mathbf{u}) + (m + Delta m)(mathbf{v} + Deltamathbf{v}).$$
    Given that the increments are small, this simplifies to
    $$(Delta m)mathbf{u} = mDeltamathbf{v} - (Delta t)mmathbf{g}.$$
    Dividing through by $(Delta t)m$ and passing to derivatives, we have
    $$frac{dot mmathbf{u}}{m} = dot{mathbf{v}} - mathbf{g}.$$



    Taking magnitudes (remembering $dot m < 0$) and integrating over time, we obtain the rocket equation
    $$|mathbf{u}|lnfrac{m_0}{m} = int dt,|dot{mathbf{v}} - mathbf{g}|,$$
    where $|mathbf{u}|$ is constant since it's a characteristic of the propulsion system. The right-hand side is the general definition of delta-v. We see that it is directly linked to the initial craft mass $m_0$, determining the initial amount of propellant needed.



    Now, suppose the propellant is utilized in quick burns, during each of which $mathbf{u}$ is constant in direction and $|dot{mathbf{v}}| gg |mathbf{g}|$, separated by coasting intervals during which $dot{mathbf{v}} = mathbf{g}$ (i.e., $dot m = 0$). Then delta-v simplifies to
    $$int dt,|dot{mathbf{v}} - mathbf{g}| = |Deltamathbf{v}_{text{burn 1}}| + |Deltamathbf{v}_{text{burn 2}}| + cdots,$$
    hence its name. (In this equation $Deltamathbf{v}$ is not required to be small.)






    share|improve this answer








    New contributor




    nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$



    Delta-v determines the amount of propellant needed.



    Suppose a craft with mass $m$ and velocity $mathbf{v}$ burns a small mass $|Delta m|$ of propellant and ejects it at relative velocity $mathbf{u}$, so that the craft mass changes by $Delta m < 0$. This occurs over a time $Delta t$ in a local gravitational field $mathbf{g}$. Then the new craft velocity $mathbf{v} + Deltamathbf{v}$ is given by
    $$text{initial momentum} + text{change in momentum due to gravity} = text{final momentum of propellant} + text{final momentum of craft},$$
    $$mmathbf{v} + (Delta t)mmathbf{g} = -Delta m(mathbf{v} + mathbf{u}) + (m + Delta m)(mathbf{v} + Deltamathbf{v}).$$
    Given that the increments are small, this simplifies to
    $$(Delta m)mathbf{u} = mDeltamathbf{v} - (Delta t)mmathbf{g}.$$
    Dividing through by $(Delta t)m$ and passing to derivatives, we have
    $$frac{dot mmathbf{u}}{m} = dot{mathbf{v}} - mathbf{g}.$$



    Taking magnitudes (remembering $dot m < 0$) and integrating over time, we obtain the rocket equation
    $$|mathbf{u}|lnfrac{m_0}{m} = int dt,|dot{mathbf{v}} - mathbf{g}|,$$
    where $|mathbf{u}|$ is constant since it's a characteristic of the propulsion system. The right-hand side is the general definition of delta-v. We see that it is directly linked to the initial craft mass $m_0$, determining the initial amount of propellant needed.



    Now, suppose the propellant is utilized in quick burns, during each of which $mathbf{u}$ is constant in direction and $|dot{mathbf{v}}| gg |mathbf{g}|$, separated by coasting intervals during which $dot{mathbf{v}} = mathbf{g}$ (i.e., $dot m = 0$). Then delta-v simplifies to
    $$int dt,|dot{mathbf{v}} - mathbf{g}| = |Deltamathbf{v}_{text{burn 1}}| + |Deltamathbf{v}_{text{burn 2}}| + cdots,$$
    hence its name. (In this equation $Deltamathbf{v}$ is not required to be small.)







    share|improve this answer








    New contributor




    nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    share|improve this answer



    share|improve this answer






    New contributor




    nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    answered Mar 26 at 7:19









    nanomannanoman

    2012




    2012




    New contributor




    nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





    New contributor





    nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.












    • $begingroup$
      Good answer! I've had a hard time deciding which answer to accept. This one has the mathematical logic that I was looking for, but Greg's has the intuition that I was also looking for.
      $endgroup$
      – WaterMolecule
      Mar 26 at 15:55


















    • $begingroup$
      Good answer! I've had a hard time deciding which answer to accept. This one has the mathematical logic that I was looking for, but Greg's has the intuition that I was also looking for.
      $endgroup$
      – WaterMolecule
      Mar 26 at 15:55
















    $begingroup$
    Good answer! I've had a hard time deciding which answer to accept. This one has the mathematical logic that I was looking for, but Greg's has the intuition that I was also looking for.
    $endgroup$
    – WaterMolecule
    Mar 26 at 15:55




    $begingroup$
    Good answer! I've had a hard time deciding which answer to accept. This one has the mathematical logic that I was looking for, but Greg's has the intuition that I was also looking for.
    $endgroup$
    – WaterMolecule
    Mar 26 at 15:55











    3












    $begingroup$


    My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities.




    To put @Greg's answer short: delta-V is a mass-normalized measure to all of the quantities you mention.



    (not addressing vector vs. scalar)






    share|improve this answer









    $endgroup$


















      3












      $begingroup$


      My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities.




      To put @Greg's answer short: delta-V is a mass-normalized measure to all of the quantities you mention.



      (not addressing vector vs. scalar)






      share|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$


        My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities.




        To put @Greg's answer short: delta-V is a mass-normalized measure to all of the quantities you mention.



        (not addressing vector vs. scalar)






        share|improve this answer









        $endgroup$




        My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities.




        To put @Greg's answer short: delta-V is a mass-normalized measure to all of the quantities you mention.



        (not addressing vector vs. scalar)







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 26 at 12:36









        Everyday AstronautEveryday Astronaut

        2,235832




        2,235832






























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