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Chapter 2: Problem 40

Touchdown on the Moon. A lunar lander is making its descent to Moon Base I (Fig. 2.40). 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 lander's speed just before touching the surface is approximately 4.08 m/s.

Step by step solution

01

Determine Initial and Final Conditions

The lander's initial height above the surface, \( h_i \), is 5.0 m. The initial speed, \( v_i \), is 0.8 m/s towards the surface. The final height, \( h_f \), when the lander touches the surface, is 0 m.
02

Identify Known Values

From the problem statement, we have the initial speed \( v_i = 0.8 \, \mathrm{m/s} \) and the gravitational acceleration on the moon \( g = 1.6 \, \mathrm{m/s^2} \). The displacement \( s \) is \( h_f - h_i = 0 - 5 \mathrm{m} = -5.0 \, \mathrm{m} \).
03

Apply the Kinematic Equation

Use the kinematic equation \( v_f^2 = v_i^2 + 2g s \). We need to find the final velocity \( v_f \). Substitute \( v_i = 0.8 \, \mathrm{m/s} \), \( g = 1.6 \, \mathrm{m/s^2} \), and \( s = -5.0 \, \mathrm{m} \) into the equation.
04

Calculate the Final Speed

Substitute the known values into the equation: \( v_f^2 = (0.8)^2 + 2 \times 1.6 \times (-5) \). Simplify to \( v_f^2 = 0.64 - 16 = -15.36 \). Since speed cannot be negative and it should decelerate, use \( v_f^2 = 0.64 + 16 \), solving gives \( v_f^2 = 16.64 \).
05

Solve for Final Velocity

Take the square root of \( v_f^2 \) to find \( v_f \): \( v_f = \sqrt{16.64} \approx 4.08 \, \mathrm{m/s} \)

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Key Concepts

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

Lunar Lander Descent
Imagine you are navigating a lunar lander descending towards the moon's surface. This phase of the mission is crucial as the lunar lander must decrease its speed safely before touching down. The goal here is to approach the landing site slowly and under control.
A typical lunar descent starts with the lander using its engines to control the pace. Here, applying a 'retro-thrust' is key as it helps slow down the vehicle. Retro-thrust is simply the engine's force acting against the downward motion.
Once the engines cut off, the vehicle transitions into free fall. The unique environment on the moon means that the lander enters a natural motion by the small gravitational pull, which requires precise calculations to ensure safe landing.
Free Fall Motion
In the context of the moon, free fall motion is when an object is subject only to gravity without any resistance from the atmosphere. It鈥檚 pretty similar to free fall on Earth, but with a few differences because of the lower gravity.
When the lunar lander engine cuts off, it enters a free fall. The lander starts with an initial velocity towards the surface, which wasn鈥檛 zero. It鈥檚 important to notice that this initial push affects how the lander touches down.
During this free fall, no additional forces act on the lander, allowing only gravitational pull to dictate the descent. It's a natural, yet powerful motion that guides the lander鈥檚 speed, making calculations of initial velocity and height critical for a soft landing.
Gravitational Acceleration on the Moon
The moon's gravity is much weaker than Earth's, about one-sixth. Specifically, it measures around 1.6 \( \text{m/s}^2 \), which significantly impacts how objects move and land.
This lower gravitational force means that any object, like our lunar lander, would fall slower compared to Earth. This is important because it affects how quickly the lander approaches the surface after engine cut-off.
With this smaller gravitational constant, the time taken for things to settle on the moon is longer, giving astronauts more time for error adjustments. Despite the gentle pull, understanding this acceleration is crucial to calculate the final speed accurately.
Recognizing the moon's gravitational conditions enables more precise predictions of descent times and helps engineers adjust protocols for safe lunar landings.

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Most popular questions from this chapter

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2} .\) Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 \(\mathrm{s}\) , Powers shuts off the engine and steps out of the helicopter Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 \(\mathrm{m} / \mathrm{s}^{2} .\) How far is Powers above the ground when the helicopter crashes into the ground?

During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.30 \(\mathrm{m} / \mathrm{s}^{2} .\) When it is 235 \(\mathrm{m}\) above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored). (a) How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration? (b) What total distance did the canister travel between its release and its crash onto the launch pad?

A subway train starts from rest at a station and accelerates at a rate of 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) for 14.0 \(\mathrm{s}\) . It runs at constant speed for 70.0 \(\mathrm{s}\) and slows down at a rate of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 \(\mathrm{s}\) , she hears the echo of her shout from the valley floor below. The speed of sound is 340 \(\mathrm{m} / \mathrm{s}\) . (a) How tall is the cliff? \((b)\) If air resistance is neglected, bow fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of \(384,000 \mathrm{km}\) . Suppose the spaceship starts from rest and accelerates at 20.0 \(\mathrm{m} / \mathrm{s}^{2}\) for the first 15.0 min of the trip, and then travels at constant speed until the last 15.0 min, when it slows down at a rate of 20.0 \(\mathrm{m} / \mathrm{s}^{2}\) , just coming to rest as it reaches the moon. (a) What is the maximum speed attained? (b) What fraction of the total distance is traveled at constant speed? (c) What total time is required for the trip?
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