I'm doing a science project where I have to explain the physics behind a trip to an exoplanet and I also have to explain about rocket equations while doing so. I know that a trip to an exoplanet is pretty much impossible, but it more has to be the theoretical part of the physics. And my problem is with the rocket equation. So far I have this simple equation:

$$\Delta v=v_e \ln\left(\dfrac{M}{M - m_R}\right)$$

This should give the maximum velocity of a rocket with the total mass $M$ and the fuel mass $m_R$. But I'm kind of unsure what to do about $v_e$. I believe it supposed to be the gravitational pull, but I need the rocket equation for a rocket that's already in space so I don't have to think about getting out of earth's atmosphere and all that. So I'm kind of stuck here. I would also like to know the time that it takes the rocket to get to the planet, which is 16 light years away or $1.514 \times 10^{14}$ kilometres away.

I hope this makes sense. Physics isn't my strong suit, so any help would be appreciated.

** Edit **
Okay, so I think I get it a little more now. Say we have a rocket that needs to transport a load with the mass of 500 kg.

$$M-m_R=500\ kg$$

But we still need to get a speed that's somewhere near the speed of light of possible (This is only theoretical of course). So if we say 1/8 the speed of light which is around $37000000\ m/s$ or $3.7\cdot 10^{7}\ m/s$. Now to get a rocket moving that fast, we would need a proper engine. And according to the link @BowlOfRed provided, the most effective spacecraft propulsion method is a Nuclear photonic rocket which has an exhaust velocity of roughly $2.99\cdot 10^{8}\ m/s$. Can we then just add the load amount and the exhaust velocity value to the equation and find how big the fuel motor has to be? I know that I'm doing something wrong about this, since the equation gives a strange and small number when you find $M$. And the Nuclear photonic rocket method probably isn't a very realistic method, but again, my projekt is mostly theoretical and I'm only trying to find a way that it, in theory, should be possible to travel to an exoplanet.


closed as off-topic by John Rennie, user36790, Kyle Kanos, Gert, ACuriousMind Dec 17 '15 at 16:11

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  • $\begingroup$ Where did you get this equation? Did they explain what $v_e$ is? $\endgroup$ – JiK Dec 17 '15 at 8:39
  • 2
    $\begingroup$ That is the Tsiolkovsky equation and $v_e$ is the exhaust velocity i.e. the speed the exaust gas leaves to rocket motor. $\endgroup$ – John Rennie Dec 17 '15 at 8:40
  • $\begingroup$ @JohnRennie so $v_e$ is calculated by how big the area where the gas leaves is and the speed of it? $\endgroup$ – Phoenix1355 Dec 17 '15 at 8:44
  • $\begingroup$ $v_e$ is just the speed of the exhaust gas. The area of the exhaust nozzle doesn't matter. Or are you asking about the relation between $v_e$ and the volumetric flow rate? $\endgroup$ – John Rennie Dec 17 '15 at 9:02
  • $\begingroup$ $v_e$ for some real (and theoretical) engine technologies mentioned at en.wikipedia.org/wiki/Spacecraft_propulsion#Table_of_methods $\endgroup$ – BowlOfRed Dec 17 '15 at 9:08

Assuming that after reading the comments you understand that $v_e$ is the exit velocity of the fuel, you need to further understand what $\Delta v$ is. It's the change in velocity of the spacecraft. For real missions this is not simply the maximum velocity of the craft.

When you want to visit an exoplanet and return, you need to distribute your $\Delta v$ onto several parts of the trip:

  1. Accelerate to leave Earth
  2. Brake to not fly past the exoplanet
  3. Accelerate to leave exoplanet
  4. Brake to no fly past Earth or cause a crater, which we call lithobraking :-)

Aerobraking in an atmosphere may relax some of these $\Delta v$ requirements, as will swing-bys/gravity-assists along the way. See also this cool $\Delta v$ map of the solar system.

So if your fuel allows for a $\Delta v$ of say 40km/s, your actual travelling speed is going to be considerably lower.

And we haven't talked about staging yet, which also changes things a bit.

Now with nuclear fuel exiting near the speed of light, indeed, fuel mass is quite low. There's a factor of 30.000 compared to the $v_e$ of a chemical reaction (generously taken as 10km/s). If I'm not mistaken, a matter/antimatter rocket is ideal, after solving the problem of getting that nasty kind of unobtainium in measurable quantities.


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