What is the Initial Mass of Fuel Needed for a Rocket to Reach 0.1c?

[Dopplershift]

The problem states:
Typical chemical fuels yield exhaust speeds of the order of 10³ m/s. Let us imagine we had a fuel that gives v₀ = 3 × 10⁵ m/s. What initial mass of fuel would the rocket need in order to attain a final velocity of 0.1c for a final mass of 1 ton?

I derived the equation in the first part of the problem:
\begin{equation}
v - v_0 = v_e \ln(\frac{m_0}{m})
\end{equation}

Solving for the initial mass, m, yields
\begin{equation}
m_0 = me^{\frac{\Delta v}{v_e)}}
\end{equation}

I plug in that.

0.1c = 3.0*10^7
v₀ = 3.0*10^5
v_ee = 1.0*10^3
v - v₀ = 3.0 *10^7 - 3.0*10^5 = 2.97*10^7
m = 1000 kg
I plug in these numbers and I am getting infinity. What am I doing wrong?

Thanks!

 

[mfb]

The initial velocity should be 0, not 3*10⁵ m/s.

3*10⁵ m/s is the exhaust velocity of the hypothetical fuel, the number for chemical rockets is just given as comparison. The number will be very large, but you should't get infinity anywhere.

Please work with units, that makes it easier (especially for you) to spot mistakes.

 

[Dopplershift]

The initial velocity should be 0, not 3*10⁵ m/s.

3*10⁵ m/s is the exhaust velocity of the hypothetical fuel, the number for chemical rockets is just given as comparison. The number will be very large, but you should't get infinity anywhere.

I'm confused, how is the initial velocity zero? It says that the fuel gives us a v₀ of 3.0*10⁵?

The problem is when I plug in the values of the conditions into my final equation, (equation 2).

 

[scottdave]

Based on your formulas, m0 calculates to (1000 kg)*e^29700, which is approx 3.5 x 10^12901 (not infinity), but too large for your calculator.

 

[Dopplershift]

Based on your formulas, m0 calculates to (1000 kg)*e^29700, which is approx 3.5 x 10^12901 (not infinity), but too large for your calculator.

Thank you!

 

[haruspex]

how is the initial velocity zero? It says that the fuel gives us a v0 of 3.0*10⁵?

If that v₀ is the initial speed of the rocket then the question makes no sense. How can the fuel give it that initial speed, before any fuel is burnt? I believe v₀ is the exhaust speed, and the question is saying that although a typical exhaust speed is only 1000m/s, just suppose we had one with an exhaust speed 300 times as great.

 

[mfb]

If that v₀ is the initial speed of the rocket then the question makes no sense. How can the fuel give it that initial speed, before any fuel is burnt? I believe v₀ is the exhaust speed, and the question is saying that although a typical exhaust speed is only 1000m/s, just suppose we had one with an exhaust speed 300 times as great.

That is how I interpreted the question as well.

 

[scottdave]

Thank you!

In perspective, the mass of the Earth is 5.97 x 10^24 kg.

 

[Ray Vickson]

The problem states:
Typical chemical fuels yield exhaust speeds of the order of 10³ m/s. Let us imagine we had a fuel that gives v₀ = 3 × 10⁵ m/s. What initial mass of fuel would the rocket need in order to attain a final velocity of 0.1c for a final mass of 1 ton?

I derived the equation in the first part of the problem:
\begin{equation}
v - v_0 = v_e \ln(\frac{m_0}{m})
\end{equation}

Solving for the initial mass, m, yields
\begin{equation}
m_0 = me^{\frac{\Delta v}{v_e)}}
\end{equation}

I plug in that.

0.1c = 3.0*10^7
v₀ = 3.0*10^5
v_ee = 1.0*10^3
v - v₀ = 3.0 *10^7 - 3.0*10^5 = 2.97*10^7
m = 1000 kg
I plug in these numbers and I am getting infinity. What am I doing wrong?

Thanks!

Your use of $v_0$ is inconsistent: in the rocket equation, $v_0=$ initial speed of the rocket, That is $v_0 = 0$ nowhere near $3 \times 10^5$ (m/s). The "speed" $3 \times 10^5$ (m/s) is the speed of the exhaust gas relative to the rocket body.

One final quibble: the rocket equation as you wrote it is for non-relativistic rockets, using classical mechanics and Newton's equations of motion. The final speed $0.1 c$ sought in this problem is getting a bit into territory where relativity matters, so in pirnciple one ought to use a relativistic version of the rocket eqution. Of course, it will make very little difference in this particular problem---but it is always a good idea to know what kinds of approximations you are using.

 

[mfb]

This problem is from May, it is unlikely that someone is still working on it.