Why is delta-v is 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 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
asked 13 hours ago
WaterMoleculeWaterMolecule
66529
$endgroup$
add a comment |
$begingroup$
Many of the questions and answers on this site make use of the concept of delta-v. Is there an easy to understand 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
asked 13 hours ago
WaterMoleculeWaterMolecule
66529
$endgroup$
2
$begingroup$
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)
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– Magic Octopus Urn
12 hours ago
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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
11 hours ago
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The short version: Mass cancels out.
$endgroup$
– chrylis
5 hours ago
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
Many of the questions and answers on this site make use of the concept of delta-v. Is there an easy to understand 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
asked 13 hours ago
WaterMoleculeWaterMolecule
66529
$endgroup$
Many of the questions and answers on this site make use of the concept of delta-v. Is there an easy to understand 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
orbital-mechanics
asked 13 hours ago
WaterMoleculeWaterMolecule
66529
asked 13 hours ago
WaterMoleculeWaterMolecule
66529
asked 13 hours ago
WaterMoleculeWaterMolecule
66529
asked 13 hours ago
WaterMoleculeWaterMolecule
66529
asked 13 hours ago
WaterMoleculeWaterMolecule
66529
66529
2
$begingroup$
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
12 hours ago
$begingroup$
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
11 hours ago
$begingroup$
The short version: Mass cancels out.
$endgroup$
– chrylis
5 hours ago
$begingroup$
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
1 hour ago
add a comment |
2
$begingroup$
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
12 hours ago
$begingroup$
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
11 hours ago
$begingroup$
The short version: Mass cancels out.
$endgroup$
– chrylis
5 hours ago
$begingroup$
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
1 hour ago
2
2
$begingroup$
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
12 hours ago
$begingroup$
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
12 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
The short version: Mass cancels out.
$endgroup$
– chrylis
5 hours ago
$begingroup$
The short version: Mass cancels out.
$endgroup$
– chrylis
5 hours ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
1 Answer
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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.
answered 10 hours ago
GregGreg
59616
$endgroup$
2
$begingroup$
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
10 hours ago
$begingroup$
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
3 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
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.
answered 10 hours ago
GregGreg
59616
$endgroup$
2
$begingroup$
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
10 hours ago
$begingroup$
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
3 hours ago
add a comment |
$begingroup$
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.
answered 10 hours ago
GregGreg
59616
$endgroup$
2
$begingroup$
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
10 hours ago
$begingroup$
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
3 hours ago
add a comment |
$begingroup$
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.
answered 10 hours ago
GregGreg
59616
$endgroup$
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.
answered 10 hours ago
GregGreg
59616
answered 10 hours ago
GregGreg
59616
answered 10 hours ago
GregGreg
59616
answered 10 hours ago
GregGreg
59616
59616
2
$begingroup$
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
10 hours ago
$begingroup$
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
3 hours ago
add a comment |
2
$begingroup$
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
10 hours ago
$begingroup$
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
3 hours ago
2
2
$begingroup$
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
3 hours ago
add a comment |
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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
12 hours ago
$begingroup$
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
11 hours ago
$begingroup$
The short version: Mass cancels out.
$endgroup$
– chrylis
5 hours ago
$begingroup$
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
1 hour ago