# Rocket problem

1. Mar 25, 2005

### tony873004

Did I do this right? I'm relying on a formula the teacher gave us in class, without quite understanding it intuitively. So I'm just rewriting the formula to solve for my unknown and plugging in numbers.

A sounding rocket launched from Earth’s surface is to achieve a final speed of 1000 m/s. If the exhaust speed of the spent fuel is 2000 m/s, what fraction of the rocket’s total mass at launch must be fuel? (Assume that the engine burns rapidly enough that you may ignore any effects due to Earth’s gravity during the burn.)
$$$v_f =v_i +v_{ex} \ln \left[ {\frac{m_i }{m_f }} \right]$ $v_f -v_i =v_{ex} \ln \left[ {\frac{m_i }{m_f }} \right]$ $\ln \left[ {\frac{m_i }{m_f }} \right]=\frac{v_f -v_i }{v_{ex} }$ $\ln \left[ {\frac{m_i }{m_f }} \right]=\frac{1000m/s-0m/s}{2000m/s}$ $\ln \left[ {\frac{m_i }{m_f }} \right]=0.5$ $\ln ^{-1}\left[ {0.5} \right]=1.6487$ $\frac{m_i }{m_f }=1.67487$ $\frac{m_f }{m_i }=\frac{1}{1.67487}$ $\frac{m_f }{m_i }=0.60653$$$

2. Mar 25, 2005

Im not sure if thats right since gravity is not taken into account. You need to account for the force on the rocket caused by gravity as it is going up.

Regards,

3. Mar 25, 2005

### tony873004

The instructions said to "Assume that the engine burns rapidly enough that you may ignore any effects due to Earth’s gravity during the burn"

4. Mar 25, 2005

### Janitor

Your calculation looks OK to me as long as you remember that the answer is $$1-\frac{m_f }{m_i }$$.

5. Mar 25, 2005

### tony873004

That makes sense. The fraction represents the ratio of the rocket before and after, not the fuel, so subtracting it from 1 which represents the full rocket gives me the fuel ratio.

Thanks guys.