Doubt with delta-v on maneuvers

[Carlos_media]

If i have a velocity of 2000m/s and the delta-v for the maneuver is 200m/s wich is the final velocity my satellite have to have? more or less?:Dthanks

[Taki117]

If i have a velocity of 2000m/s and the delta-v for the maneuver is 200m/s wich is the final velocity my satellite have to have? more or less?:Dthanks

Depends, are you burning Prograde (Forwards) Retrograde (backwards) Normal or antinormal (Up or down) or Radially? While burning Normal, anti-normal, or radially won't change your orbital velocity any buring Prograde (Forwards) will increase your orbital velocity at periapsis to 2200m/s while burning retrograde (backwards) will reduce your speed at apoapsis to 1800m/s a dV is change in velocity, meaning your velocity changes by about that much.

[MarvinKitFox]

If i have a velocity of 2000m/s and the delta-v for the maneuver is 200m/s wich is the final velocity my satellite have to have? more or less?:Dthanks

Will your final velocity be more or less?The answer to that is: Yes!

it will be more, or less.

unless it remains the same, of course, but that is rather unlikely.

You are vector-adding 2000 and 200. In a 3-dimensional, nonlinear, curved, occasionally cyclic universe. With discontinuities (SOI changes, hard surfaces)

If you burn pure prograde, the answer is 2200. If you burn pure retrograde, the answer is 1800.

But only if you are faaaaaaaaaar from a gravity source. Oberth can play havoc with this equation.

And once your burn is anything other than pure pro- or retrograde, the answer becomes strange.

[EdFred]

And of course it will only be 2200m/s only at that point in space if the eccentricity is anything but 0 assuming you burn at the PE. You can have 2200m/s at the PE and near 0 at the AP. In other words, there is no answer to your question.

[Specialist290]

To sum it up: It may be either more or less, depending on what direction your original velocity is and the direction in which the thrust is applied. In simplified form (assuming a linear vector and instantaneous acceleration), it all comes down to a three-dimensional Cartesian geometry problem.

[Red Iron Crown]

Could even be the same if the thrust is applied in the right direction.

[Kryxal]

Right direction? At right ANGLES, yes (instantaneously, and correcting direction as the velocity vector changes, or it doesn't stay a right angle)...

[Red Iron Crown]

Right direction? At right ANGLES, yes (instantaneously, and correcting direction as the velocity vector changes, or it doesn't stay a right angle)...

It can be done in a straight line in the proper direction (mix of radial or normal and retrograde).

[GoSlash27]

"delta-v" means "change in velocity". It doesn't specify how the change in velocity is applied, so it could be either... or neither.

It's like posing the question "you're in an elevator on the 20th floor and the elevator moves 2 floors"; not enough info given to determine whether it ends up higher, lower, or on the same floor.

Best,

-Slashy

Edited September 9, 2014 by GoSlash27