text: Taking into account the rotation of the Earth, the delta-v is up to 0.46km/s less (starting at the equator and going east) or more (if going west).
  Voyager 1
  For Voyager 1, with respect to the Sun:
  = 132,712,440,018 km3⋅s−2 is the standard gravitational parameter of the Sun
  r = 17 billion kilometers
  v = 17.1 km/s
  Hence:
  Thus the hyperbolic excess velocity (the theoretical orbital velocity at infinity) is given by
  However, Voyager 1 does not have enough velocity to leave the Milky Way. The computed speed applies far away from the Sun, but at such a position that the potential energy with respect to the Milky Way as a whole has changed negligibly, and only if there is no strong interaction with celestial bodies other than the Sun.
  Applying thrust
  Assume:
  a is the acceleration due to thrust (the time-rate at which delta-v is spent)
  g is the gravitational field strength
  v is the velocity of the rocket
  Then the time-rate of change of the specific energy of the rocket is : an amount  for the kinetic energy and an amount  for the potential energy.
  The change of the specific energy of the rocket per unit change of delta-v is
  which is |v| times the cosine of the angle between v and a.