This is from Stephen Hawking's latest book; Brief Answers to the Big Questions.

The speed at which we can send a rocket is governed by two things, the speed of the exhaust and the fraction of its mass that the rocket loses as it accelerates.

  • $\begingroup$ He wrote several, which one? $\endgroup$
    – Solar Mike
    Commented Dec 10, 2018 at 6:11

2 Answers 2


He refers to Tsiolkovsky's Rocket Equation:

$$ \Delta v=v_e \ln {\frac {m_0}{m_f}} $$


$v_e$ is the exhaust velocity;

${\frac {m_0}{m_f}}$ is the fraction of mass; $m_f$ - the "dry mass"/"final mass" (rocket without fuel) and $m_0$ - "wet mass"/"launch mass" (rocket fully fueled up.)

This equation is one of the most important in rocket science - describing the change of velocity a rocket can achieve. The implications are that the larger the difference between the mass of fuel and the mass of the craft, the larger the velocity achievable (but the $\ln$ results in diminishing returns as mass of fuel is increased) and that engines that impart the exhaust with most velocity provide most performance - linearly, without that pesky $\ln$ - but then... with square root of energy needed; $E={1\over 2} mv^2; v = \sqrt{2E \over m}$. So, increasing power of the engine - chemical energy of fuel, amount of electrical energy imparted by ion drive - results in diminishing returns again.

One of Hawking's last ideas - "Breakthrough Starshot" nicely sidesteps both problems. The propellant is photons, moving at speed of light, and the craft doesn't carry any fuel - the propellant is beamed from a ground-based station through a powerful laser.


"the fraction of its mass that the rocket loses as it accelerates" means that assuming the power delivered is constant, the total mass of the rocket decreases as the fuel is spent, therefore altering the thrust (power) to mass ratio.

  • $\begingroup$ available power and total mass... $\endgroup$
    – Solar Mike
    Commented Dec 11, 2018 at 9:40
  • $\begingroup$ @uhoh can you explain what you understand by the term "rocket" - perhaps you need to post your own question... $\endgroup$
    – Solar Mike
    Commented Dec 11, 2018 at 9:43
  • $\begingroup$ I believe it is all about energy - that's where you get momentum and change in velocity from... $\endgroup$
    – Solar Mike
    Commented Dec 11, 2018 at 9:48
  • $\begingroup$ @uhoh shame you cannot or have not provided a canonical answer then, perhaps it would benefit many... And since you use the term weight - surely that should be mass... $\endgroup$
    – Solar Mike
    Commented Dec 11, 2018 at 9:50
  • 1
    $\begingroup$ @uhoh: You didn't hear of rocket engine power? Saturn V's F-1 producing 160,000,000 horsepower? (okay, okay. I know, we don't divide power by mass.) And even if Solar Mike correctly used TWR, it's only very indirectly connected to total delta-V. $\endgroup$
    – SF.
    Commented Dec 11, 2018 at 9:53

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