# B Amount of energy to get a spaceship to 99% C

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1. Mar 12, 2016

### Kaura

So recently a friend and I were pondering how much energy would be required to get a 20,000kg spaceship to 99% the speed of light. This arose from us discussing the raw efficiency of antimatter and how much antimatter would be required to fuel a spaceship to 99% C or 296794533.42 m/s. At first this seem simple since we thought we could use KE = 1/2mv^2 but then we realized we would have to take the change in relativistic mass into account. If I am not mistaken this is a differential equation as the ship's mass at any given point would be a function of its velocity. I am not experienced with problems of this sort. If anyone is willing to work out this problem for the hell of it I would much appreciate it. Also please tell me if I am overlooking something.

2. Mar 12, 2016

### PeroK

The relativistic KE is given by:

$(\gamma - 1)mc^2$

Where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$

3. Mar 12, 2016

### Orodruin

Staff Emeritus
... and m is the invariant mass. The OP should take note that relativistic mass is a largely deprecated term.

Also, to OP, this is clearly not an A level thread. The thread tags are intended for indicating what level of explanation you are comfortable with. Using the A tag indicates that you have an understanding of the subject at the level of a physics graduate student and expect answers directed at that level. This is obviously not the case here. I am going to relabel the thread accordingly.

4. Mar 12, 2016

### Ibix

It's worth noting that the formula given above assumes that the energy source is not mounted on the rocket. If the energy source is mounted on a rocket, the number becomes truly depressing, because not only do you have to accelerate the ship but you have to accelerate the fuel, which means more fuel, which means more fuel to accelerate more fuel.

For a rocket which carries its own fuel, Wikipedia gives the formula: $$\Delta v=c \tanh\left(\frac{I_{sp}}{c}\ln\frac{m_0}{m_1}\right)$$ where $m_1$ is the mass of your ship, $m_0$ is the mass of your ship with fuel, $c$ is the speed of light, and $I_{sp}$ is the specific thrust. There's a table on Wikipedia with some values for that last one - in the case of a 100% efficient matter-antimatter rocket, $I_{sp}/c$ is 1. It's smaller (usually a lot smaller) in the case of a more plausible engine.

Note that you're going to need fuel to slow down again. Simply squaring the number you get out of the formula above will give you a ballpark figure for the total mass needed to accelerate and decelerate, but you can work it out longhand if you want.

In case you haven't come across it before, $\tanh$ is the hyperbolic tangent. A lot of scientific calculators have button marked HYP. Press it and then the TAN button to get the tangent, or HYP, then INV, then TAN to get the inverse hyperbolic tangent.

Last edited: Mar 12, 2016
5. Mar 12, 2016

### Kaura

Thanks for the correction I will do this in the future

6. Mar 13, 2016

### Ibix

I was being lazy last night, but curiosity got the better of me. Note that:$$\tanh^{-1}x=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$$ if $x<1$ and is real. That means that the Wikipedia expression can be re-written as $$m_0 = m_1\left(\frac{1+\Delta v/c}{1-\Delta v/c}\right)^{1/(2I_{sp}/c)}$$You asked about accelerating to 0.99c. With a 100% efficient total conversion rocket, you need 14 times the "dry" mass of your ship to make that acceleration, so around 200 times the mass to accelerate and stop again. Wikipedia quotes $I_{sp}/c=0.119$ for a H→He fusion rocket; that gives you about four and a half billion times the dry mass to accelerate, or on the order of 1019 times if you want to stop again.

Told you it was depressing.

7. Mar 13, 2016

### Kaura

Haha and a while ago my friend used Newtonian Physics and was convinced that it would only take 0.2 grams of antimatter
Thanks for working it out legitimately
I will have to tell him the bad news

8. Mar 13, 2016

### Ibix

Yeah. The non-linearity of relativistic equations can really come back to bite you at high velocities. Incidentally, using PeroK's approach (so a solar sail pushed by some kind of super laser based on Mercury, for example) you still need six times the mass of the ship in matter/antimatter to accelerate and the same to decelerate. That's before you get in to inefficiencies in the system and the question of how to build a super laser at the destination before you get there.

9. Mar 13, 2016

### Ibix

Also your friend's Newtonian calculation has gone wrong somewhere. The Newtonian kinetic energy of something travelling at 0.99c would be something like $0.49mc^2$. So the mass of matter and antimatter needed would be 0.49 times the mass of the ship.

The calculation isn't valid anyway, but that's not 0.2g unless your ship is the size of a thimble.

10. Mar 13, 2016

### Kaura

Yeah I thought something was fishy when he told me he calculated it

11. Mar 14, 2016

### PAllen

You really need those sci-fi inertia-less drives. Just eliminate all inertia without eliminating the matter .. if only ...