How much energy would it require to send Earth to Proxima Centauri?

Question

Assuming 100% efficiency of energy usage, and given the current estimates about the mass of known objects in our solar system, including Earth itself, and assuming no other objects are affected by this (for simplicity), how much energy (in Joules) would be required to transport Earth from its current orbit round the sun to Proxima Centauri?

Notes

1. We need to get there in a reasonable amount of time (e.g., 1 million years)
2. We don't need to account for ecosystem survival on Earth
3. We don't need to worry about over-acceleration breaking Earth up (including water)
4. We don't need to worry about stopping upon arrival at Proxima Centauri

Answer 1

If you want to do it in 1 million years then your basic problem is kinetic energy.

The gravitational potential energy of the Earth around the Sun is $-GM_{\odot}M_{E}/(1au) = -5.3\times 10^{33}$ J.

To get to Proxima Centauri within 1 million years requires a relative velocity of at least 1.27 km/s.

So I'm not sure how exact an answer you need. The main issue is how fast is it moving relative to the Sun.

The relative motion of Proxima Centauri and the Sun, is 22 km/s radially towards the Sun and 24 km/s tangentially. Thus while the radial velocity towards the Sun eliminates the need to do anything but escape the Sun's gravitational well, the tangential component must be matched.

Hence my rough answer would be - enough energy to escape the potential well of the Sun plus enough kinetic energy to track the tangential velocity of Proxima Cen. This amounts to $5.3 \times 10^{33} + 1.7\times10^{33} = 7\times10^{33}\ J$.
That should see a collision take place in about 60,000 years.

To refine this, I think you need to take account of the Earth's 30 km/s motion around the Sun and to what extent you can use this kinetic energy to aid you. This depends on the direction in which you need to propel the Earth. Currently Proxima Cen has an ecliptic latitude of 45 degrees, but this will change rapidly, depending on the proper motion direction. However, at best, this could knock about $2.7\times10^{33}\ J$ off the answer in the most propitious circumstances.