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If a rocket is going 50,000 kilometers per hour and wanted to go 75,000 kilometers per hour wouldn't the rocket just need an additional thrust equaling 25,000 kilometers per hour? Or is it something different?

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  • $\begingroup$ Something different and more complicated. It becomes easier to accelerate in terms of thrust as the rocket looses mass in the pricess. But energetic balance is the crucial factor. For a given rocket with realistic fuel might be impossible too. Let's wait for a details answer which includes equations. .. $\endgroup$ – Alchimista Dec 17 '17 at 21:18
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In some sense, you're correct. It is just that much thrust. The difficulty is that rockets work by expelling mass, so the mass of the rocket is variable, making the thrust required for a specific acceleration (or delta-v) variable as well.

If the rocket wer mostly non-fuel, then the mass wouldn't change much and you might be able to ignore the difference. But such a rocket can't accelerate much and is much less useful than a rocket that is mostly fuel.

Let's say you want to have a specific configuration arrive at the final velocity (the rocket, plus enough fuel to do other things). For this example, I'll make that a mass of 25000kg. Further, let's assume we carry cryogenic hydrogen and oxygen for fuel. If so, the engine produces $v_e$ of $\sim 4400\text{m/s}$.

Now we have everything to determine how much additional fuel is required.

$$\Delta v = v_e \ln \frac{m_0}{m_f}$$ $$\ln \frac{m_0}{m_f} = \frac{\Delta v}{v_e}$$ $$\frac{m_0}{m_f} = e^{\frac {\Delta v}{v_e}}$$ $$m_0 = m_f e^{\frac {\Delta v}{v_e}}$$ $$m_0 = (25000\text{kg}) e^{\frac {6944 \text{m/s}}{4400 \text{m/s}}}$$ $$m_0 = 121000\text{kg}$$

So for this rocket, with this mass, you'd have to carry and burn 96,000kg of fuel to perform that acceleration. More efficient engines (those with greater $v_e$) would require less fuel.

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If a rocket is going 50,000 kilometers per hour and wanted to go 75,000 kilometers per hour wouldn't the rocket just need an additional thrust equaling 25,000 kilometers per hour?

Yes, but ...

The problem is that the material needed to attain that extra 25,000 kph is just dead weight during the interval over which the rocket gains the initial 50,000 kph velocity. This carrying of dead weight that eventually becomes fuel is what results in the ideal rocket equation.

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