Chapter 9: Q127P (page 256)

During a lunar mission, it is necessary to increase the speed of a spacecraft by $2.2 \frac{m}{s}$ when it is moving at $400 \frac{m}{s}$ relative to the Moon. The speed of the exhaust products from the rocket engine is $1000 \frac{m}{s}$ relative to the spacecraft. What fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase?

Short Answer

Expert verified

The fraction of the initial mass of the spacecraft that must be burned and ejected to accomplish the speed increase is 0.0022.

Step by step solution

Step 1: Given

Increase in the speed, $∆ v = 2.2 \frac{m}{s}$
Speed of the exhaust product, $v_{r e l} = 1000 \frac{m}{s}$

Determining the concept

Using the second equation of the rocket, find the fraction of the initial mass of the spacecraft that must be burned and ejected to accomplish the speed increase.

Formula is as follow:

$∆ v = v_{r e l} I n (\frac{M_{1}}{M_{2}})$

Here, v is velocity and M is mass.

Determine the fraction of the initial mass of the spacecraft that must be burned and ejected to accomplish the speed increase

The equation of rocket can be written as,

$∆ v = v_{r e l} I n (\frac{M_{1}}{M_{2}}) \frac{∆ v}{v_{r e l}} = I n (\frac{M_{1}}{M_{2}}) \frac{M_{1}}{M_{2}} = e^{\frac{∆ v}{v_{r e l}}} \frac{M_{2}}{M_{1}} = e^{- \frac{∆ v}{v_{r e l}}}$

Resolve further as:

$\frac{M_{1} - M_{f}}{M_{2}} = 1 - \frac{M_{2}}{M_{1}} \frac{M_{1} - M_{f}}{M_{2}} = 1 - e^{- \frac{∆ v}{v_{r e l}}}$

Substitute the values and solve as:

$\frac{M_{1} - M_{f}}{M_{2}} = 1 - e^{\frac{2.2 \frac{m}{s}}{1000 \frac{m}{s}}} \frac{M_{1} - M_{f}}{M_{2}} = 0.00219 \frac{M_{1} - M_{f}}{M_{2}} \approx 0.0022$

Hence, the fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase is 0.00219.

Therefore, the fraction of the initial mass of the spacecraft that must be burned and ejected to accomplish the speed increase can be found using the second equation of the rocket.