We obtained Tsiolkovsky's Rocket Equation:
$$\Delta v = V_e log \left( \frac{m_1}{m_2} \right) $$
Where V_e is the velocity of the gas exhaust from the rocket engine, m_1 is the initial mass of the rocket (body, payload, engine and fuel) and m_2 is the mass of the rocket after some or all the fuel has been expended.
The exhaust velocity depends on the propellant (small molecules are better), temperature (higher the better) and the design of the nozzle, and whether you are in atmosphere or vacuum. Vacuum allows faster exit of gas. Engines work better in outer space.
(Note, that this is not the same as *thrust* as force. It's to do with efficiency but not lifting ability. Americans built larger more powerful engines with less efficiency, such as the mighty F1 for the Saturn V. Soviet Union built smaller engines with less thrust and lifting power but more efficiency. Attempts to mount 24 of them in one stage of their N-1 Moon rocket proved uncontrollable and doomed their Moon Program.)
Let's plug in some numbers. V_e of American engines range between 2300 to 2800 m/s. Soviet engines seem to do better, going as high as 3600 m/s.
Let's say 80 per cent of the rocket mass is fuel. Let's try a typical American engine, Ve as 2500 m/s. After all the fuel is burnt, the final mass m_2 is 20 percent of m_1.
$$ \Delta v = 2500 log ( m_1 / 0.2 m_1) = 1747 m/s $$
A Soviet engine with Ve of 3200 m/s:
$$\Delta v = 3200 log ( m_1 / 0.2 m_1) = 2237 m/s.$$
Having a higher exhaust velocity is clearly desirable.
What if we want to double the delta v.
$$ \Delta v = V_e log \left(\frac{m_1}{m_2}\right) $$Doubling the delta v, multiply the right side by two.
$$ \Delta v = 2 V_e log \left(\frac{m_1}{m_2}\right) $$Twice logarithm of x is equal to the log of x squared.
$$ \Delta v = V_e log \left(\frac{m_1}{m_2}\right)^2 $$ $$ \Delta v = V_e log \left(\frac{m_1^2}{m_2^2}\right) $$Basically, to double the delta v, leads to SQUARING the right hand side of the equation.
Tripling the delta v CUBES (power of 3) the right hand side! $$ \Delta v = 3 \times V_e log \left(\frac{m_1}{m_2}\right) $$ $$ \Delta v = V_e log \left(\frac{m_1}{m_2}\right)^3 $$
Let's put in some numbers. suppose 80 percent of the initial rocket mass m_1 is fuel . That gives a ratio m_1 : m_2 of 5:1, that is the initial mass is 5 times the final mass
But doubling the delta v is squaring the ratio, so the initial mass needs to be a whopping 25 times the final mass!
Tripling the delta v, is \(5^3\), a ratio of 125.
And that is why space rockets are so darn HUGE. The fuel requirements go up exponentially!
You need more fuel to push the rocket increase delta v. And also you need more fuel to push the fuel that you need to push the rocket. Its pretty vicious.