Why is delta-v is the most useful quantity for planning space travel?Could a partial space elevator be practical and useful?Gravity assist for manned travelDo we sufficiently understand mechanics of Lagrange point stationkeeping for EML2 rendezvous and assembly?Is there a simple relation between delta-v and travel time?What's the name of this maneuver for unlimited delta-v?How useful is the Interplanetary Transport Network?Calculating dV to raise apoapsis at an arbitrary point of an orbitDid Rosetta improve on models of non-gravitational effects on comet 67P's orbit?How to best think of the State Transition Matrix, and how to use it to find periodic Halo orbits?Why are most of Jupiter's moons retrograde?
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Why is delta-v is the most useful quantity for planning space travel?
Could a partial space elevator be practical and useful?Gravity assist for manned travelDo we sufficiently understand mechanics of Lagrange point stationkeeping for EML2 rendezvous and assembly?Is there a simple relation between delta-v and travel time?What's the name of this maneuver for unlimited delta-v?How useful is the Interplanetary Transport Network?Calculating dV to raise apoapsis at an arbitrary point of an orbitDid Rosetta improve on models of non-gravitational effects on comet 67P's orbit?How to best think of the State Transition Matrix, and how to use it to find periodic Halo orbits?Why are most of Jupiter's moons retrograde?
$begingroup$
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, $|mathbfv|$, 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 7 hours ago
WaterMoleculeWaterMolecule
65528
$endgroup$
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, $|mathbfv|$, 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 7 hours ago
WaterMoleculeWaterMolecule
65528
$endgroup$
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)
$endgroup$
– Magic Octopus Urn
6 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
5 hours 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, $|mathbfv|$, 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 7 hours ago
WaterMoleculeWaterMolecule
65528
$endgroup$
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, $|mathbfv|$, 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 7 hours ago
WaterMoleculeWaterMolecule
65528
asked 7 hours ago
WaterMoleculeWaterMolecule
65528
asked 7 hours ago
WaterMoleculeWaterMolecule
65528
asked 7 hours ago
WaterMoleculeWaterMolecule
65528
asked 7 hours ago
WaterMoleculeWaterMolecule
65528
65528
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
6 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
5 hours 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
6 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
5 hours 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
6 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
6 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
5 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
5 hours ago
add a comment |
1 Answer
1
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 4 hours ago
GregGreg
5266
$endgroup$
1
$begingroup$
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
4 hours ago
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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active
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$begingroup$
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 4 hours ago
GregGreg
5266
$endgroup$
1
$begingroup$
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
4 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 4 hours ago
GregGreg
5266
$endgroup$
1
$begingroup$
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
4 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 4 hours ago
GregGreg
5266
$endgroup$
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 4 hours ago
GregGreg
5266
answered 4 hours ago
GregGreg
5266
answered 4 hours ago
GregGreg
5266
answered 4 hours ago
GregGreg
5266
5266
1
$begingroup$
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
4 hours ago
add a comment |
1
$begingroup$
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
4 hours ago
1
1
$begingroup$
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
4 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
4 hours ago
add a comment |
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$begingroup$
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
6 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
5 hours ago