Question

During a lunar mission, it is necessary to make a midcourse correction of \(22.6 \mathrm{~m} / \mathrm{s}\) in the speed of the spacecraft, which is moving at \(388 \mathrm{~m} / \mathrm{s}\). The exhaust speed of the rocket engine is \(1230 \mathrm{~m} / \mathrm{s}\). What fraction of the initial mass of the spacecraft must be discarded as exhaust?

Short answer

After computing, we find that the required fraction of the mass to be discarded is approximately 0.018 or 1.8 %.

Step-by-step solution

Write down the known values

From the given problem, we know the change in velocity (\(\Delta v = 22.6 \, \mathrm{m/s}\)), the velocity of the spacecraft (\(v = 388 \, \mathrm{m/s}\)), and the exhaust speed of the engine (\(v_e = 1230 \, \mathrm{m/s}\)).

Write down the rocket equation

The rocket equation is \(\Delta v = v_e \ln(\frac{m_i}{m_f})\). Here, we need to solve for \(m_f\), which is the final mass of the spacecraft after the midcourse correction.

Use the rocket equation to solve for \(m_f\)

First, arrange the rocket equation to represent \(m_f\), giving \(\frac{m_i}{m_f} = e^{\frac{\Delta v}{v_e}}\). Then, we find that \(m_f = \frac{m_i}{e^{\frac{\Delta v}{v_e}}}\).

Compute the ratio

We then should find the ratio \(m_r\) of the discarded mass to the initial mass which gives us \(m_r = \frac{m_i - m_f}{m_i} = \frac{m_i}{m_i} - \frac{m_f}{m_i} = 1 - \frac{m_f}{m_i} = 1 - e^{-\frac{\Delta v}{v_e}}\).

Substitute the known values into the formula to get the answer

Substituting the known values gives \(m_r = 1 - e^{-\frac{22.6}{1230}}\).

Key concepts

Midcourse Correction

In space missions, precise navigation is key. Sometimes, even tiny errors can add up over long distances. That's where a midcourse correction comes in. A midcourse correction is an adjustment made to a spacecraft's trajectory during its journey. This modification helps ensure the spacecraft stays on the correct path towards its intended destination, such as the Moon in a lunar mission.

Think of a midcourse correction as steering your car to keep it in the lane on a long highway drive. In space, this "steering" uses a change in speed and direction to adjust the route, which is essential due to the vast distances involved. The need for a correction arises from initial launch imprecision or external forces like gravitational pulls that might have altered the craft's course slightly. In our exercise the spacecraft needs a velocity change of 22.6 meters per second to maintain its right course.

Exhaust Speed

Exhaust speed in space science refers to the speed at which gases are expelled from the rocket engine. This is crucial for propulsion as it dictates how effectively the rocket can change its speed. In simpler terms, the higher the exhaust speed, the more efficiently a rocket can execute maneuvers such as midcourse corrections. The exhaust speed acts like the push you need when sledding down a hill; the stronger the push (or jet), the faster you can move.

In the context of our exercise, we are given an exhaust speed of 1230 meters per second. This value indicates the speed at which the thurst can alter the spacecraft's velocity. A high exhaust speed is beneficial because it means that less propellant is needed to achieve significant velocity changes, making the spacecraft more fuel-efficient over long journeys.

Mass Ratio

The mass ratio is a key concept in rocket science, particularly in scenarios involving the rocket equation. It describes the relationship between the initial mass of the spacecraft, including fuel, and the final mass once the fuel has been consumed (or in this case, once course correction has been completed).

When a spacecraft uses a portion of its fuel, part of its initial mass (which includes everything in the spacecraft at the start of the journey) is expelled as exhaust to generate thrust. The mass ratio \(m_r\) aids us in understanding how much mass needs to be shed to achieve a given change in velocity (in this exercise, 22.6 m/s).

In the exercise, this is calculated through the formula: \ m_r = 1 - e^{-\frac{\Delta v}{v_e}} \ where \(\Delta v\) is the desired change in speed (22.6 m/s) and \(v_e\) is the exhaust speed (1230 m/s). This calculation helps in determining how much of the spacecraft's total mass must be discarded as exhaust to achieve the midcourse correction.