5
$\begingroup$

The gravitational potential of an object around the sun, relative to some point far far away (from the sun), should be smaller zero and proportional to $V_{escape}^2$. Hence the speed one needs should be $V_{mission}=\sqrt{V_{escape1}^2-V_{escape2}^2}$ with $V_{escape1}$ The escape velocity at the starting point relative to the sun.

The list of escape velocities states the Escape Velocity at the Earth Orbit relative to the Suns gravity as 42,1 km/s, at Jupiter (relative to the Sun) 18,5 km/s. So I would expect the lowest possible Delta V for a mission to fly to Jupiter would be the difference, or 37.6 km/s without taking into account leaving LEO or braking inside Jupiters gravity well.
Now, the NASA trajectory browser typically gives Mission delta Vs around 5km/s. So obviously my quick and dirty calculation above is wrong.

What is a working, quick way to calculate the lowest possible mission delta v to reach a point higher in a gravity well, using the data in the list of Escape velocities?

$\endgroup$

1 Answer 1

5
$\begingroup$

See this answer for how to do a simplified Hohmann transfer calculation.

It depends on where you're starting from and where you want to end up. Are you starting from low Earth orbit? From the surface of the Earth? Do you want to end up in Jupiter orbit? What orbit? Or do you just want to fly by?

Your 42.1 km/s is going from a dead stop at Earth's distance from the Sun to escaping the solar system. The Earth is not at a dead stop -- it's going around the Sun at 30 km/s. So if you're coming from the orbit of the Earth (without the Earth being there), to escape the solar system takes 12.1 km/s. Similarly, Jupiter is going around at 13 km/s, so escape from the orbit of Jupiter (without Jupiter being there) is 5.5 km/s. That's still not how you do the calculation though, since Earth and Jupiter are there at departure and arrival. So you need to look at the Hohmann transfer calculation in that other answer. By the way, the fact that the escape speed over the orbital speed in both those cases is about 1.4, or $\sqrt{2}$, is not a coincidence. This is left as an exercise for the reader.

You can generally do much better than a Hohmann transfer if you're willing to take longer and use the inner planets for gravity assists. Juno just had its one Earth flyby. Galileo did one Venus and two Earth flybys to get to Jupiter. Those calculations are much more complicated, and requires software that tracks the relative positions of the planets.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.