/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A lunar lander is making its des... [FREE SOLUTION] | 91Ó°ÊÓ

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A lunar lander is making its descent to Moon Base I (Fig. E2.40). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is \(5.0 \mathrm{~m}\) above the surface and has a downward speed of \(0.8 \mathrm{~m} / \mathrm{s}\). With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is \(1.6 \mathrm{~m} / \mathrm{s}^{2}\).

Short Answer

Expert verified
The speed of the lunar lander just before it touches the surface is \(4.08 \mathrm{~m/s}\).

Step by step solution

01

Identify the relevant parameters

The initial velocity (\(v_i\)) of the lander when the engine is cut is \(0.8 \mathrm{~m/s}\). The lander then falls a distance of \(5.0 \mathrm{~m}\) under the influence of moon's gravity of \(1.6 \mathrm{~m/s^{2}}\) (acceleration, \(a\)). The final velocity (\(v_f\)) at the end of this fall, i.e., just before it hits the ground, is what we are asked to find.
02

Apply the second equation of motion

The second equation of motion is \(v_f^2 = v_i^2 + 2ad\). Substituting the known values into this equation, we get \(v_f^2 = (0.8 \mathrm{~m/s})^2 + 2 * 1.6 \mathrm{~m/s^2} * 5.0 \mathrm{~m}\).
03

Solve for the final speed

Upon performing the computation, you’ll find \(v_f^2 = 0.64 \mathrm{~m^2/s^2} + 16 \mathrm{~m^2/s^2} = 16.64 \mathrm{~m^2/s^2}\). Now, we apply the square root on both sides, since velocity is not squared in normal practice. The square root of \(16.64\) will give us the final velocity \(v_f\) as \(4.08 \mathrm{~m/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Free Fall
Free fall is a fascinating concept in physics. Imagine an object that is falling solely under the influence of gravity, without any other forces acting on it. In the case of the lunar lander in the problem, once the engines stop, it enters this state of free fall. Here it is influenced only by the Moon's gravitational force which has an acceleration of \(1.6 \, \mathrm{m/s^2}\). This is different from Earth's gravity, which is much stronger at \(9.8 \, \mathrm{m/s^2}\).

When something is in free fall, its only acceleration is due to gravity. There is no interference from other forces like air resistance, as there would be on Earth. This makes calculations on the Moon both exciting and a tad bit simpler!

It's important to note that the velocity of an object in free fall increases steadily as it falls, as it's being pulled faster and faster towards the gravitational body. This increase in speed under lunar gravity is neatly explained using the equations of motion.
Equations of Motion Simplified
The equations of motion are handy tools that help us predict how objects move. They relate velocities, accelerations, distances, and times in a neat formulaic way.

In this problem, we used the equation: \[v_f^2 = v_i^2 + 2ad\] where:
  • \(v_f\) is the final velocity, which we want to find out.
  • \(v_i\) is the initial velocity (\(0.8 \, \mathrm{m/s}\) when the engines stop).
  • \(a\) is the acceleration due to moon's gravity (\(1.6 \, \mathrm{m/s^2}\)).
  • \(d\) is the distance the lander falls (\(5.0 \, \mathrm{m}\)).
This equation comes from combining other basic motion equations, distilling them into a form perfect for objects under constant acceleration, like our lunar lander's free fall. Using this specific equation allowed us to calculate the final speed effortlessly. It's a clear and systematic way to solve for the unknown while knowing initial conditions and acceleration.
Lunar Gravity: A Different World
The Moon's gravity is significantly weaker than Earth's. It is only about \(1/6\) of Earth's gravitational pull. This difference is essential not only for calculations but also for how we consider landing spacecraft or even imagining how walking would be on the Moon.

Because the Moon has less gravity, objects will fall slower than they would on Earth. This means a descending spacecraft like our lunar lander can take advantage of this weaker pull, requiring less thrust to land softly. The lower gravity affects all aspects of physics on the lunar surface, from how high you can jump to how objects settle.

For astronauts, this means both challenges and advantages, navigating an environment with different physical laws. Understanding lunar gravity is crucial for planning lunar missions, influencing engineering designs and mission protocols. Calculating movement under lunar gravity, as we saw, involves familiar formulas but applied in novel scenarios.

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Most popular questions from this chapter

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