# Power Required to Accelerate Space Craft to Near Light Speed

1. Nov 16, 2009

### edgepflow

I hope I can explain my question clearly enough for someone to answer.

Suppose we wish to estimate the power required to accelerate a space craft from say 0.9c to 0.9999c in a certain time. By basic definition: Power = Change in Energy / Change in Time.

Due to special relativity, there is a difference in these quantities as measured by the stationary earth and the space craft. The classical change in energy would be m*v^2 /2, while the relativistic energy would include the term for increase in relativistic energy. The time dilation effect may also be directly calculated.

So to figure the power, which energy (classical or relativistic) and which time (earth or spaceship) should be used as seen on (a) earth, or (b) the spaceship. Are these powers equal?

2. Nov 16, 2009

### pervect

Staff Emeritus
If your spaceship is moving at relativistic speeds, you need to use the relativistic energy. As far as power levels go, it depends on what you're interested in. You can calculate either, depending on your interest, and should note which one you are calculating.

Note that rockets are not 100% efficient though. You might be interested in the relativistic rocket equation, which would give you answers to what would be required if the spaceship were a rocket.

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

Last edited by a moderator: May 4, 2017
3. Nov 17, 2009

### yuiop

The time as measured in the Earth frame (t) for a rocket accelerating with constant acceleration as measured in the rocket frame (a) to reach a given velocity (v) is:

$$t = \frac{v}{a\sqrt{1-(v/c)^2}}$$

The time interval measured in the Earth frame to go from 0.9c to 0.9999c is:

$$\Delta t = t_2 - t_1 = \frac{0.9999c}{a\sqrt{1-(0.9999)^2}} - \frac{0.9c}{a\sqrt{1-(0.9)^2}}$$

The energy of the rocket at a given instant in the Earth frame (ignoring loss of mass due to using fuel) is:

$$E = m c^2 \sqrt{1+(at/c)^2}$$

The change in energy in the Earth frame is:

$$\Delta E = E_2 - E_1 = m c^2 \sqrt{1+(at_2/c)^2} - m c^2 \sqrt{1+(at_1/c)^2}$$

where m is the invariant mass.

The average power going from 0.9c to 0.9999c in the Earth frame is then:

$$W = \Delta E / \Delta t$$

Now if we switch to the point of view of an inertial observer moving at 0.9c relative to the Earth frame, the initial velocity of the rocket is zero and the final velocity using the relativistic velocity addition formula is:

$$v_2 = \frac {0.9999c-0.9c}{(1-0.9999*0.9)} = 0.9981c$$

The equations given earlier can be reused to calculate the time interval and the power in the new frame using the new velocity values. It turns out that the change in energy and power in the new frame is less than in the Earth frame, but that makes sense because the final velocity of the rocket is lower according to the observer moving at 0.9c relative to the Earth.

The proper time that elapses according to the rocket observers as they accelerate from 0.9c to 0.9999c is given by:

$$\Delta T = (c/a)*atanh(0.9981)$$

As for what the power is from the point of view of accelerating observers onboard the rockets, I am not sure. What would they consider their velocity and energy state to be at any given time?

4. Nov 19, 2009

### qraal

Nice exposition. As for the power seen in the ship frame, it's just the jet-power which doesn't change for a constant thrust scenario. Exhaust velocity and mass ejected would remain the same in that case at all speeds, in the ship's reference frame.