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Hawking wrote,

Exhaust speed of chemical rockets is 3 kilometer/second. By dropping 30 percent of their mass, they can achieve speed of about 0.5 kilometer per second and then slow down again.

So, I would like to know does "0.5" mean delta v (change of velocity)? I think it is about Tsiolkovsky equation:

$$\Delta v = v_E \ln{ \left(\frac{m_i}{m_f} \right) } $$

So my calculation is that he says (jettisoning 30 % of mass), so m(1)= 100 and m(f)=70. ln (100/70)* 3 km/s = 1.07.

So, it is not equal to what Hawking says (0.5 km/s). But if we use log (base 10) instead of natural logarithm, we get (0.46 km/s) which is nearly equal to Hawking's amount. But the real equatin uses natural logarithm. So, I would like to know what's wrong.

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You've used the equation correctly, including the natural logarithm. Have another look at the quote:

Exhaust speed of chemical rockets is 3 km/s. By jettisoning 30 % of their mass, they can achieve speed of about 0.5 km per second and then slow down again.

So that's using 30% of the mass to accelerate to 0.5 km/s, rotate and point the engine in the other direction, and then decelerate back down to 0.

If you didn't turn around, you'd be going 1.0 km/s like you calculated.

The logarithmic term

$$v_e \ln{ \left(\frac{m_i}{m_f} \right) } = v_e \ln{ \left(\frac{1.0}{0.7} \right) } \approx 0.36 \ v_e $$

applies to the whole trip. You can take the square root of the term inside to estimate the mass used to do the first half and get to 0.5 km/s

$$\left(\frac{1}{0.7} \right)^{1/2} \approx \left(\frac{1}{0.84} \right)$$

So you would use 16% of your original mass to speed up to 0.5 km/s, and then

$$\left(\frac{0.84}{0.70} \right)$$

14% of your original mass to slow down again. That makes sense because the rocket is now lighter.

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The original calculation, giving $\Delta v=1.07$ km/s, is correct. The crucial part is that the rocket is assumed to speed up and then slow down again. The rocket uses 0.5 km/s of $\Delta v$ to reach a speed of 0.5 km/s, and then uses another 0.5 km/s of $\Delta v$ to slow back down to being at rest.

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    $\begingroup$ @Harry Because that's what the operator of the rocket decided to do with their $\Delta v$. In this case, slowing down involves pointing the exhaust in the opposite direction. $\endgroup$ – probably_someone Dec 11 '18 at 9:41

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