[syhprum (OP)]

If I accelerate a body to near light speed and send it off into space so that it loops around a massive body and then retraces its path back to me and is then brought to rest according to GR it is now somewhat younger than other bodies that have not experienced this.
what is the energy requirment for this ie how many Joules are required to make 1 KG one second less old.

[CliffordK]

What you are doing with relativity is not aging one item, while aging another item.

Is it better to just put your treasure in the deep freeze?

Typically much of what you do with space is "coasting".  The expensive part is the initial acceleration.  So, once you get up to speed, you could technically maintain that speed/time differential forever.

However, you will have troubles reversing your object as you may need either a black hole, or to pass by a series of stars.  And there are not any black holes that we know of anywhere close to Earth.  Hmm, can you even do a "slingshot maneuver" at the speed of light, as the method for doing a slingshot is to accelerate towards the object, and reach the highest speed at the closest point near the object, then decelerate as one departs.  The problem is that if you are already at lightspeed, you aren't supposed to be able to accelerate to a faster speed, then you loose some speed as you depart from the gravity pull of the object.  Anyway, I believe you need the black hole to get adequate inward gravity pull (centripetal force) to reverse the direction of your near lightspeed object.  And, if done right, it will be a weightless maneuver, so you won't squash your astronauts.  Just make sure you stay on the right side of the event horizon.

Earlier, I tried to do the calculations of how much energy it took to reach the speed of light by looking at momentum and kinetic energy (with classical mechanics).

Momentum:
     p=mv  (momentum (kg*m/s) = mass x velocity).
And Kinetic Energy:
     E_k = (1/2) mv² = p²/2m


Substitute in c for v, and one gets:

p=mc
E_k = (1/2) mc²

With relativity, you add the Lorenz factor
p = γm_ov

where


And, as you approach the speed of light, you multiply in a factor approaching infinity, and both your momentum and energy go towards infinity.

But, even with classical mechanics, one finds:

E_k = (1/2) mc²

With the famous: E=mc²

One finds that to accelerate 1kg of matter to near the speed of light (classically), it takes the equivalent amount of energy as converting (1/2) kg of matter into pure energy.

Now, a couple of other equations.
Say you want to keep your acceleration under 10G's, or say 100m/s².
To reach the speed of light, it would take:
299,792,458m/s / 100m/s² = 2,997,924 seconds / (60*60*24) = 35 days of just over 10G acceleration.

Say you made a track in a loop around the moon, you'd quickly go beyond your specified 10G's centrifugal acceleration long before you get to the speed of light.  Likewise, if you could install a track around Jupiter (which would be hard due to it being a gas giant), you would still quickly get to your 10G's centrifugal acceleration.

So, you really have to use an onboard energy source.  I.E.  a lot of antimatter, and consuming most of your ship.

Also, I really wouldn't recommend aerobraking for the same reason it takes about 35 days to get to the speed of light, and the heat and friction coefficients (would you really want to dump the equivalent of half your mass in pure energy in a matter of seconds?) .

[syhprum (OP)]

Thanks for the calculation it confirmed my worst fears, it would seem that time travell not only to say dangerous is a very expencive business if only to slip back one second.

[CliffordK]

My calculations were to reach the speed of light, or a high fraction of the speed of light.

Current research seems to indicate that one can "time travel" at lower speeds too.

The Hafele and Keating experiment concluded that two loops around the world on normal jets, Eastbound lost 59 nanoseconds, and two loops around the world westbound gained 273 nanoseconds.

nano = 10^-9 = 1/1,000,000,000.

So...  according to the theory, if you were a commercial airline pilot, and manged to fly around the world 10,000,000 times or so, always in the same direction, one should either gain or loose about a second.

Ok, so it would take a bit of flying.

But, the International Space station was supposed to have completed 75296 orbits in a little over 10 years.  So...  hmmm, over a period of about 1,000 years, it should complete on the order of 7 million orbits, and loose about a second.  Now, if NASA doesn't choose to crash it back to Earth in the next few years!!!

Is it going Eastbound (and loosing time), so I would have to count 5x as long? 
Can I simply count loops around the Earth, or do I have to also consider the speed?