How Do You Calculate Delta V in a 1-D Gravitational Field?

  • Thread starter Thread starter Ashiataka
  • Start date Start date
  • Tags Tags
    Delta Delta v
AI Thread Summary
To calculate delta V in a 1-D gravitational field, the Tsiolkovsky equation and escape velocity formula are essential. The delta V needed to transition between two radii can be derived from the difference in escape velocities at those points. It's important to consider that delta V either adds or subtracts energy from the orbit, impacting the total energy and size of the orbit. The change in energy between orbits can be calculated using specific energy formulas, emphasizing that kinetic energy is crucial for achieving the desired delta V. Ultimately, the Tsiolkovsky equation is vital for assessing fuel requirements for the necessary delta V.
Ashiataka
Messages
21
Reaction score
1
I want to calculate the delta V needed to move objects around in a 1-d gravitational field. The relevant equations as far as I can see are the Tsiolkovsky equation,
\Delta V = v_{ex} \ln\left(\frac{m_0}{m_1}\right)
and the equation for calculating escape velocity.
v_e = \sqrt{\frac{2GM}{r}}

Now I'm guessing that if I want to find the delta v to get from r to R I would use
v_e = \sqrt{2GM}\left(\frac 1R - \frac 1r\right)

is that correct?
 
Astronomy news on Phys.org
Ashiataka said:
Now I'm guessing that if I want to find the delta v to get from r to R I would use
v_e = \sqrt{2GM}\left(\frac 1R - \frac 1r\right)

is that correct?

That would give you the difference in the required escape velocity. Why would you assume you were traveling at escape velocity at both places?

Your delta V either adds energy or subtracts energy from the orbit. The total energy determines the size of the orbit.

\epsilon = -\frac {GM}{2 a}

where a is the semi-major axis and epsilon is the specific energy per unit of mass. Determine the difference in energy between the two different orbits. Since kinetic energy is the only type of energy that can be added, use a modified version of the formula for orbital energy to determine how much energy needs to be added.

\epsilon = \frac{v^2}{2} - \frac{GM}{r}

Assuming the delta v is instantaneous, meaning r remains unchanged, the change in energy between the two orbits will be:

\Delta \epsilon = \frac{v2^2}{2}-\frac{v1^2}{2}

All closed orbits have a negative specific energy. An specific energy of zero is a parabolic orbit - your minimum escape trajectory.

Assuming you have enough fuel, any size delta v is possible. You just keep firing your thruster until the required delta v is obtained. The acceleration is:

F = ma

with your delta v being:

\Delta v = at

Of course, for really large delta v's, such as at launch, assuming r remains unchanged seems like a bad assumption. It doesn't matter, since your kinetic energy is being converted to potential energy the entire time and your velocity really isn't changing as fast as you think it is. It's the change in energy that matters and that's the only relevance of the change in velocity - velocity is kinetic energy.

The Tsiolkovsky equation is used to determine whether or not you have enough fuel to perform the required delta v. Obviously, that's a pretty important consideration in the real world.
 
This thread is dedicated to the beauty and awesomeness of our Universe. If you feel like it, please share video clips and photos (or nice animations) of space and objects in space in this thread. Your posts, clips and photos may by all means include scientific information; that does not make it less beautiful to me (n.b. the posts must of course comply with the PF guidelines, i.e. regarding science, only mainstream science is allowed, fringe/pseudoscience is not allowed). n.b. I start this...
Asteroid, Data - 1.2% risk of an impact on December 22, 2032. The estimated diameter is 55 m and an impact would likely release an energy of 8 megatons of TNT equivalent, although these numbers have a large uncertainty - it could also be 1 or 100 megatons. Currently the object has level 3 on the Torino scale, the second-highest ever (after Apophis) and only the third object to exceed level 1. Most likely it will miss, and if it hits then most likely it'll hit an ocean and be harmless, but...
Back
Top