Delta-v described as m^2 / s^2

[WyDavies]

I was reading an article on the NASA web site (which I am now unable to find - I was on a different computer at the time) and the author was talking about the energy that would be required for another mission to Pluto. He expressed the delta-v as 130 km^2 / s^2.

I've never seen delta-V expressed like that and it doesn't make much sense in my head. Was this a typo or is it a different way of describing delta-V?

[Bill Phil]

That's velocity squared. Probably a multiple for the kinetic energy required. Or just a typo.

[p1t1o]

m2/s2 is J/kg, perhaps it was statement of kinetic energy, rather than delta-v?

[Bill Phil]

16 minutes ago, p1t1o said:

m2/s2 is J/kg, perhaps it was statement of kinetic energy, rather than delta-v?

There's a 2, or a 1/2, in there somewhere. Might be in the 130km^2 / s^2.

[cryogen]

That's characteristic energy, C₃. It's not a delta-v figure per se; it's equal to v_∞², i.e. the asymptotic limit ("at infinity") of the Earth-relative velocity as you leave the Earth's SOI. Squared. So if you're leaving Earth with a C₃ of 130 km²/s², you're departing with an Earth-relative heliocentric velocity of more than 11 km/s.

It has a wikipedia page:

https://en.wikipedia.org/wiki/Characteristic_energy

It's often used to measure launcher performance (on the x-axis):

https://www.nasa.gov/sites/default/files/files/NAC-July2014-Hill-Creech-Final.pdf

Edited July 9, 2016 by cryogen

[Bill Phil]

Yeah, that's exactly what it is. Pretty strange, but it appears to be a way of presenting specific orbital energy. What's interesting is the performance graph.

In this case it's MJ/kg.