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?
Simple image editor tool to draw a simple box/rectangle in an existing image
Is it okay / does it make sense for another player to join a running game of Munchkin?
Giant Toughroad SLR 2 for 200 miles in two days, will it make it?
A social experiment. What is the worst that can happen?
What would you call a finite collection of unordered objects that are not necessarily distinct?
When is separating the total wavefunction into a space part and a spin part possible?
Latex for-and in equation
Why isn't KTEX's runway designation 10/28 instead of 9/27?
Adding empty element to declared container without declaring type of element
Calculating the number of days between 2 dates in Excel
Why does this part of the Space Shuttle launch pad seem to be floating in air?
What is the term when two people sing in harmony, but they aren't singing the same notes?
In Star Trek IV, why did the Bounty go back to a time when whales were already rare?
Can I Retrieve Email Addresses from BCC?
I'm in charge of equipment buying but no one's ever happy with what I choose. How to fix this?
Can I use my Chinese passport to enter China after I acquired another citizenship?
Can a controlled ghast be a leader of a pack of ghouls?
Lifted its hind leg on or lifted its hind leg towards?
Indicating multiple different modes of speech (fantasy language or telepathy)
word describing multiple paths to the same abstract outcome
Freedom of speech and where it applies
Are Warlocks Arcane or Divine?
Superhero words!
Can the electrostatic force be infinite in magnitude?
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 3 hours ago
WaterMoleculeWaterMolecule
63028
$endgroup$
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 3 hours ago
WaterMoleculeWaterMolecule
63028
$endgroup$
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
2 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
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, $|mathbfv|$, 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 3 hours ago
WaterMoleculeWaterMolecule
63028
$endgroup$
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 3 hours ago
WaterMoleculeWaterMolecule
63028
asked 3 hours ago
WaterMoleculeWaterMolecule
63028
asked 3 hours ago
WaterMoleculeWaterMolecule
63028
asked 3 hours ago
WaterMoleculeWaterMolecule
63028
asked 3 hours ago
WaterMoleculeWaterMolecule
63028
63028
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
2 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
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
2 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
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
2 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
2 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
1 hour 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
1 hour 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 53 mins ago
GregGreg
3865
$endgroup$
$begingroup$
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
42 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "508"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f35041%2fwhy-is-delta-v-is-the-most-useful-quantity-for-planning-space-travel%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 53 mins ago
GregGreg
3865
$endgroup$
$begingroup$
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
42 mins 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 53 mins ago
GregGreg
3865
$endgroup$
$begingroup$
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
42 mins 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 53 mins ago
GregGreg
3865
$endgroup$
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 53 mins ago
GregGreg
3865
answered 53 mins ago
GregGreg
3865
answered 53 mins ago
GregGreg
3865
answered 53 mins ago
GregGreg
3865
3865
$begingroup$
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
42 mins ago
add a comment |
$begingroup$
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
42 mins 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
42 mins 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
42 mins ago
add a comment |
Thanks for contributing an answer to Space Exploration Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f35041%2fwhy-is-delta-v-is-the-most-useful-quantity-for-planning-space-travel%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
2 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
1 hour ago