# Rocket speed (Is the equation my choose is the correct one and is my (asumption) and calculation is correct with regard to Hawking's sentences

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.

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.

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.

• @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. – probably_someone Dec 11 '18 at 9:41