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Chapter 2: Problem 71

A certain volcano on earth can eject rocks vertically to a maximum height \(H\). (a) How high (in terms of \(H\)) would these rocks go if a volcano on Mars ejected them with the same initial velocity? The acceleration due to gravity on Mars is 3.71 m/s\(^2\); ignore air resistance on both planets. (b) If the rocks are in the air for a time \(T\) on earth, for how long (in terms of \(T\)) would they be in the air on Mars?

Short Answer

Expert verified
(a) Rocks on Mars would reach 2.64 times the height on Earth. (b) Rocks on Mars would be in the air for 2.64 times as long as on Earth.

Step by step solution

01

Write the formula for maximum height

To find the maximum height \( H \) a rock reaches when ejected vertically, the formula is: \[ H = \frac{v^2}{2g} \]where \( v \) is the initial velocity and \( g \) is the acceleration due to gravity.
02

Compare gravitational accelerations

On Earth, the acceleration due to gravity is \( g_e = 9.81 \text{ m/s}^2 \), whereas on Mars, it is \( g_m = 3.71 \text{ m/s}^2 \). Use these values to find the height on Mars.
03

Calculate maximum height on Mars

Since the initial velocity \( v \) is the same for both planets, the maximum height on Mars, \( H_m \), can be written as:\[ H_m = \frac{v^2}{2g_m} \]Given \( H_e = \frac{v^2}{2g_e}, \) find the ratio of the heights:\[ \frac{H_m}{H_e} = \frac{g_e}{g_m} = \frac{9.81}{3.71} \approx 2.64 \]Thus, \( H_m = 2.64 \times H_e \).
04

Write the formula for total time in air

The total time \( T \) a rock is in the air can be given by: \[ T = \frac{2v}{g} \]where \( g \) is the acceleration due to gravity.
05

Calculate time in air on Mars

Using the same initial velocity \( v \) as before, the time in air on Mars, \( T_m \), would be:\[ T_m = \frac{2v}{g_m} \]The ratio of times between Mars and Earth is:\[ \frac{T_m}{T_e} = \frac{g_e}{g_m} = \frac{9.81}{3.71} \approx 2.64 \]So, \( T_m = 2.64 \times T_e \).

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is a measure of the intensity of gravity at a particular location. Essentially, it tells us how strongly a planet pulls objects towards its surface. On Earth, this acceleration is approximately 9.81 m/s虏, meaning that in one second, an object increases its velocity by 9.81 m/s if dropped.

This is a crucial factor when calculating how high or how long an object will travel in vertical motion. The formula for maximum height an object can reach is given by \[ H = \frac{v^2}{2g} \]where:
  • \( v \) is the initial velocity
  • \( g \) is the gravitational acceleration
Thus, gravitational acceleration directly impacts the height reached by any projectile. The lesser the gravitational pull, the higher the object will go with the same initial velocity. This principle is helpful when analyzing motions on different celestial bodies, such as Mars.
Mars vs Earth Gravity
When discussing gravity, it鈥檚 fascinating to consider how it varies across planets. Mars鈥� gravitational acceleration is only 3.71 m/s虏, noticeably weaker than Earth's 9.81 m/s虏. This difference substantially impacts projectile motion.

For instance, if a rock were launched with the same initial speed on both planets, it would reach a far greater height on Mars. The formula \[ \frac{H_m}{H_e} = \frac{g_e}{g_m}\]shows us that the height on Mars is roughly 2.64 times that on Earth.

Similarly, because of the reduced gravity on Mars, the time a projectile remains airborne is longer when compared to Earth. Using a comparable formula for time \[ \frac{T_m}{T_e} = \frac{g_e}{g_m}\]we find the duration is again about 2.64 times longer. This showcases how even our solar system neighbors exhibit remarkably different physics from Earth due to variations in gravitational forces.
Vertical Motion
Vertical motion under gravity is a type of projectile motion where the object moves straight up and then comes back down. The forces acting on it throughout are solely due to gravity, and this makes the concepts of maximum height and total time in the air quite simple to grasp using formulas.

In our example of rocks being ejected from a volcano, as soon as a rock is launched, it decelerates due to gravity until it stops at the topmost height. At this point, gravity pulls it back to the ground.
  • The maximum height reached depends on the initial launch speed and gravitational force.
  • The formula for the time it stays airborne, \[ T = \frac{2v}{g} \], indicates that longer durations occur when gravity is weaker.
Understanding these core concepts helps clarify how objects behave under different gravitational fields, such as those present on Mars versus Earth. This approach not only aids in solving textbook problems but also extends to real-life scenarios and planning space missions.

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Most popular questions from this chapter

A small object moves along the \(x\) -axis with acceleration \(a_{x}(t)=-\left(0.0320 \mathrm{~m} / \mathrm{s}^{3}\right)(15.0 \mathrm{~s}-t) .\) At \(t=0\) the object is at \(x=-14.0 \mathrm{~m}\) and has velocity \(v_{0 x}=8.00 \mathrm{~m} / \mathrm{s} .\) What is the \(x\) -coordinate of the object when \(t=10.0 \mathrm{~s} ?\)

A turtle crawls along a straight line, which we will call the \(x\)-axis with the positive direction to the right. The equation for the turtle's position as a function of time is \(x(t) =\) 50.0 cm + (2.00 cm/s)\(t -\) (0.0625 cm/s\(^2)t^2\). (a) Find the turtle's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 cm from its starting point? What is the velocity (magnitude and direction) of the turtle at each of those times? (e) Sketch graphs of \(x\) versus \(t, v_x\) versus \(t\), and \(a_x\) versus \(t\), for the time interval \(t =\) 0 to \(t =\) 40 s.

A 15-kg rock is dropped from rest on the earth and reaches the ground in 1.75 s. When it is dropped from the same height on Saturn's satellite Enceladus, the rock reaches the ground in 18.6 s. What is the acceleration due to gravity on Enceladus?

A student is running at her top speed of 5.0 m/s to catch a bus, which is stopped at the bus stop. When the student is still 40.0 m from the bus, it starts to pull away, moving with a constant acceleration of 0.170 m/s\(^2\). (a) For how much time and what distance does the student have to run at 5.0 m/s before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x-t\) graph for both the student and the bus. Take \(x =\) 0 at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is 3.5 m/s, will she catch the bus? (f) What is the \(minimum\) speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

A ball is thrown straight up from the ground with speed \(v_0\). At the same instant, a second ball is dropped from rest from a height \(H\), directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of \(H\) in terms of \(v_0\) and \(g\) such that at the instant when the balls collide, the first ball is at the highest point of its motion.
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