Any ways to recover kinetic energy ?

[Externet]

Hi.

A space probe/craft traveling at very high speed, in need to slow-down or land on a remote destination, has to waste its energy or is there a method that can recover some of it and store it for later takeoff or whatever energy consuming actions needed ?

[Acme]

Hi.

A space probe/craft traveling at very high speed, in need to slow-down or land on a remote destination, has to waste its energy or is there a method that can recover some of it and store it for later takeoff or whatever energy consuming actions needed ?

After the initial slowdown and at the point where a parachute is often deployed, rotor blades could drive a generator to charge batteries. While possible, the additional weight of a generator and short duration of flight strikes me as impractical.

 

See: >> Will NASA's Next Space Capsule Land Like a Helicopter?

It looks like NASA is getting a little more creative with its landing systems. A team of researchers recently tested a new rotor landing system in the 550-foot fall Vehicle Assembly Building at NASA's Kennedy Space Center. The idea is for spinning blades to take the place of parachutes to enable soft and controlled landings on land instead of the ocean.

...

 

You might also use thermocouples in or behind the heat shield to generate some voltage for batteries, but again the weight of them and short duration of use seem impractical.

[Endy0816]

Reverse gravity assist?

[Acme]

Reverse gravity assist?

I'm not familiar with that. Can you describe how it works?

 

In any case, the deceleration period is short and the mass of a spacecraft relatively small and I suspect that even if you could recover/transform all the kinetic energy it would be a negligible amount.

 

Addendum:

OK. I don't usually attempt these calculations but I was curious. Just so, I chose the Mars Curiosity rover to experiment on. I got my values & equations from several sources & I'm pretty confident they are close enough for an estimate.

 

mass of spacecraft just prior to atmospheric entry:2,401 kg

velocity of spacecraft just prior to atmospheric entry: 5,100 meters per second

KE=0.5 * mv²

=31,225,005,000 joules

approximate time of deceleration: 345 seconds

joule to watts=joules/time in seconds = 90,507,261 watts

 

So if I did that right (am I even in the ball park?), I was quite wrong about no appreciable amount of energy to recover, but now there is the problem of how and where to store 90 mega-watts in approximately 6 minutes.

 

sources:

craft mass & velocity

Kinetic energy

joules to watts

descent time

Edited August 31, 2015 by Acme

[imatfaal]

+1 There really is no substitute for getting your hand dirty and checking with maths; I think I was making the same assumption as you. My second though was that it has to be the same ballpark as speeding something up to the position and speed of the landing craft and that must be very energetic.

 

But : "90 mega-watts in approximately 6 minutes."

 

Would make more sense to say store 30 giga-joules in 6 minutes OR store 90 mega-Watts for 6 minutes. Watts are a rate not a simple quantity - as you correctly calculated. I think this was a typo more than anything else.

 

There is also the Potential energy calculation to take into account.

 

We haven't worked out how to do it with simple aeroplane in our atmosphere - not sure it is feasible to expect them do it on a super stripped down craft in the martian atmosphere.

[John Cuthber]

It's easy. All you need to do is ask the aliens on the other planet to build a really big spring, with a hook at the bottom.

Aim your space ship at the top of the spring and then it will bring you to a gradual, safe, halt. Just as you reach the ground ask someone to clip the hook onto the spring so it doesn't ping you back into space.

 

Even better, when you decide to take off again, simply get back in the ship and undo the hook. the spring will send you on your way..

Obviously, there are practical issues with this.

 

Of course, you can, in principle, take your own spring along with you in case the aliens are uncooperative or on holiday when you land. It would be rather heavy.

[Phi for All]

Just as you reach the ground ask someone* to clip the hook onto the spring so it doesn't ping you back into space.

 

Ooooh, I can finally move "Bungee-jumping" to the Professional Experience section of my resume!

[imatfaal]

 

Ooooh, I can finally move "Bungee-jumping" to the Professional Experience section of my resume!

 

I thought everyone already did...or is that why I don't get any calls to interview

[Acme]

+1 There really is no substitute for getting your hand dirty and checking with maths; I think I was making the same assumption as you. My second though was that it has to be the same ballpark as speeding something up to the position and speed of the landing craft and that must be very energetic.

 

But : "90 mega-watts in approximately 6 minutes."

 

Would make more sense to say store 30 giga-joules in 6 minutes OR store 90 mega-Watts for 6 minutes. Watts are a rate not a simple quantity - as you correctly calculated. I think this was a typo more than anything else.

 

...

Thnx Imat. Yeah, typo was what it was when some sweat dripped into my eye. If we supposed using thermocouples behind the heat shield as I earlier suggested, then we could get a real voltage from their rating and then we could get amps from the watts and voltage and then knowing the amp/hr capacity of storage batteries on board we could talk about storing amp/hrs in 6 minutes. Oui? But of course in all practicality we can't and so we resort to the practicality of the springs earlier invoked. The energy in that system would be rated in boings, or more likely tera-boings.

Edited August 31, 2015 by Acme