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Chapter 4: Problem 121

Suppose that a space probe can withstand the stresses of a \(20 g\) acceleration. (a) What is the minimum turning radius of such a craft moving at a speed of one-tenth the speed of light? (b) How long would it take to complete a \(90^{\circ}\) turn at this speed?

Short Answer

Expert verified
(a) The minimum turning radius is approximately \(4.59 \times 10^{12} \ m\). (b) The time to complete a 90-degree turn is about \(2.4 \times 10^5 \ s\).

Step by step solution

01

Identify Given Values and Formulas

We are given that the acceleration the craft can withstand is \( 20g \), where \( g = 9.8 \ m/s^2 \). The speed \( v \) is one-tenth the speed of light, \( c = 3.0 \times 10^8 \ m/s \), which means \( v = 3.0 \times 10^7 \ m/s \). We need to find the minimum turning radius \( r \). The formula for centripetal acceleration \( a_c \) in circular motion is \( a_c = \frac{v^2}{r} \).
02

Calculate Minimum Turning Radius

Given \( a_c = 20g = 20 \times 9.8 \ m/s^2 \), we can substitute in the centripetal acceleration formula: \[ 20 \times 9.8 = \frac{(3.0 \times 10^7)^2}{r} \] Solving for \( r \): \[ r = \frac{(3.0 \times 10^7)^2}{20 \times 9.8} \] Calculate \( r \): \[ r \approx \frac{9.0 \times 10^{14}}{196} \approx 4.59 \times 10^{12} \ m \].
03

Calculate Time to Complete a \(90^{\circ}\) Turn

A \(90^{\circ} \) turn corresponds to a quarter of a full circle, so the arc length \( s \) is \( \frac{\pi}{2} r \). The time \( t \) to travel this arc length at speed \( v \) is given by \( t = \frac{s}{v} \). So, \[ t = \frac{\frac{\pi}{2} \times 4.59 \times 10^{12}}{3.0 \times 10^7} \] Calculate \( t \): \[ t \approx \frac{7.2 \times 10^{12}}{3.0 \times 10^7} \approx 2.4 \times 10^5 \ s \].

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Key Concepts

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

Circular Motion
Circular motion refers to the movement of an object along the circumference of a circle or rotation along a circular path. In physics, it's important because it involves centripetal acceleration, which is the acceleration directed toward the center of the circle.
Centripetal force and acceleration keep the object moving in its path. The formula for centripetal acceleration is given by
  • \( a_c = \frac{v^2}{r} \)
where \( a_c \) is the centripetal acceleration, \( v \) is the tangential speed, and \( r \) is the radius of the circle.
Understanding this formula helps you determine how fast an object can safely travel along a curved path. In the case of our space probe, it can withstand a centripetal acceleration of \( 20g \), or 20 times the acceleration due to gravity on Earth. This means the probe has been engineered to handle significant stresses during its motion in space.
Turning Radius
The turning radius is a critical concept when discussing circular motion. It refers to the minimum radius of the path an object must take to change direction safely without exceeding its limits of acceleration.
A smaller turning radius means sharper turns, which require higher centripetal acceleration.
To calculate the minimum turning radius of our space probe, we use the centripetal acceleration equation:
  • \( r = \frac{v^2}{a_c} \)
Given that the probe travels at one-tenth the speed of light, its speed \( v \) is \( 3.0 \times 10^7 \ m/s \), and it can handle an acceleration of \( 20g \), we find the radius necessary to ensure safe turning as \( r \approx 4.59 \times 10^{12} \ m \).
This enormous radius reflects the interplay between mass, velocity, and force in space travel, illustrating the challenges of maintaining control at extreme speeds.
Speed of Light
The speed of light in a vacuum, which is approximately \( 3.0 \times 10^8 \ m/s \), is a fundamental constant in physics.
Often denoted by \( c \), it's used not only in the study of light but in calculations involving relativistic speeds where objects move close to this velocity.
For our exercise, the probe moves at one-tenth of the speed of light, or \( 3.0 \times 10^7 \ m/s \). This high speed requires careful considerations regarding the radius of turns to prevent disastrous stress on the probe.
  • High-speed motion like this highlights phenomena such as time dilation and length contraction, though these aren't part of this problem scope.
  • The immense speed necessitates precise calculations for turning maneuvers, as impacts from even minor deviations can be catastrophic.
An important part of physics is understanding the consequences of near-light-speed travel, shaping both theoretical investigations and practical designs in aerospace engineering.

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Most popular questions from this chapter

A baseball is hit at Fenway Park in Boston at a point \(0.762 \mathrm{~m}\) above home plate with an initial velocity of \(33.53 \mathrm{~m} / \mathrm{s}\) directed \(55.0^{\circ}\) above the horizontal. The ball is observed to clear the \(11.28\) -m-high wall in left field (known as the "green monster") \(5.00 \mathrm{~s}\) after it is hit, at a point just inside the left-field foulline pole. Find (a) the horizontal distance down the left-field foul line from home plate to the wall; (b) the vertical distance by which the ball clears the wall; (c) the horizontal and vertical displacements of the ball with respect to home plate \(0.500 \mathrm{~s}\) before it clears the wall.

Snow is falling vertically at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\). At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of \(50 \mathrm{~km} / \mathrm{h} ?\)

A small ball rolls horizontally off the edge of a tabletop that is \(1.20 \mathrm{~m}\) high. It strikes the floor at a point \(1.52 \mathrm{~m}\) horizontally from the table edge. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?

A moderate wind accelerates a pebble over a horizontal \(x y\) plane with a constant acceleration \(\vec{a}=\left(5.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At time \(t=0\), the velocity is \((4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}\). What are the (a) magnitude and (b) angle of its velocity when it has been displaced by \(12.0 \mathrm{~m}\) parallel to the \(x\) axis?

A centripetal-acceleration addict rides in uniform circular motion with radius \(r=3.00 \mathrm{~m}\). At one instant his acceleration is \(\vec{a}=\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At that instant, what are the val- ues of (a) \(\vec{v} \cdot \vec{a}\) and (b) \(\vec{r} \times \vec{a}\) ?
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