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Delta V

  1. Nov 25, 2013 #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,
    [tex]\Delta V = v_{ex} \ln\left(\frac{m_0}{m_1}\right)[/tex]
    and the equation for calculating escape velocity.
    [tex]v_e = \sqrt{\frac{2GM}{r}}[/tex]

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

    is that correct?
     
  2. jcsd
  3. Nov 28, 2013 #2

    BobG

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    That would give you the difference in the required escape velocity. Why would you assume you were travelling 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.

    [tex]\epsilon = -\frac {GM}{2 a}[/tex]

    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.

    [tex]\epsilon = \frac{v^2}{2} - \frac{GM}{r}[/tex]

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

    [tex]\Delta \epsilon = \frac{v2^2}{2}-\frac{v1^2}{2}[/tex]

    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:

    [tex]F = ma[/tex]

    with your delta v being:

    [tex]\Delta v = at[/tex]

    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.
     
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