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Inside a starship at rest on the earth, a ball rolls off the top of a horizontal table and lands a distance \(D\) from the foot of the table. This starship now lands on the unexplored Planet \(X\) . The commander, Captain Curious, rolls the same ball off the same table with the same initial speed as on earth and finds that it lands a distance 2.76\(D\) from the foot of the table. What is the acceleration due to gravity on Planet \(X\) ?

Short Answer

Expert verified
The acceleration due to gravity on Planet X is approximately 1.287 m/s².

Step by step solution

01

Understand the Problem

The problem involves analyzing projectile motion. First, identify that on Earth, the ball lands a distance \(D\) from the table. On Planet \(X\), it lands a distance 2.76\(D\) from the table with the same initial horizontal speed.
02

Equation for Range on Earth

The horizontal range \(R\) of a projectile is given by the formula \(R = v_0 t\), where \(v_0\) is the horizontal velocity and \(t\) is the time of flight. On Earth, \(t = \sqrt{\frac{2h}{g}}\), where \(h\) is the height of the table and \(g\) is the gravitational acceleration (9.8 m/s²). Thus, \(D = v_0 \sqrt{\frac{2h}{g}}\).
03

Equation for Range on Planet X

On Planet \(X\), the range is 2.76 times \(D\). Therefore: \[2.76D = v_0 \sqrt{\frac{2h}{g_X}}\]. This implies \[2.76 v_0 \sqrt{\frac{2h}{g}} = v_0 \sqrt{\frac{2h}{g_X}}.\]
04

Equate and Simplify

Dividing both sides by \(v_0\) and squaring both sides results in \[\frac{2.76^2 2h}{g} = \frac{2h}{g_X}.\] Simplify to get \[g_X = \frac{g}{2.76^2}.\]
05

Calculate the Acceleration on Planet X

Substitute \(g = 9.8\, \text{m/s}^2\) into the equation: \[g_X = \frac{9.8}{2.76^2}.\] Calculate this to find \(g_X\).
06

Final Calculation

Perform the final calculation: \[g_X = \frac{9.8}{7.6176} \approx 1.287 \text{ m/s}^2.\]

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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 the rate at which objects accelerate towards the center of a celestial body, such as Earth or another planet, due to gravity. On Earth, this acceleration is typically denoted as "g" and has a value of approximately 9.8 meters per second squared (m/s²). This means that if you drop an object, its speed increases by 9.8 m/s every second
In projectile motion, gravitational acceleration affects how long an object stays in the air and plays a crucial role in determining the vertical motion. Although horizontal motion is constant, vertical motion is subject to gravity, causing the projectile to follow a curved trajectory.
Gravitational acceleration can vary significantly on other planets due to their different masses and radii. Therefore, understanding how gravity affects motion helps astronauts and scientists anticipate how objects will behave on different planetary bodies.
Horizontal Range
The horizontal range of a projectile is the total horizontal distance an object travels while in motion. It is influenced by two primary factors: initial velocity and time of flight. In our exercise, the initial horizontal speed of the ball remains constant both on Earth and Planet X.
The range on Earth can be represented by the formula:\[ R = v_0 \times t, \] where \( v_0 \) is the initial horizontal velocity, and \( t \) is the time of flight. Since the horizontal speed remains the same, any change in range must be due to changes in time of flight, which is affected by gravitational acceleration.
On Planet X, the range was observed to be 2.76 times greater than on Earth, indicating that the projectile’s time in the air was also substantially increased due to the planet's weaker gravity.
Time of Flight
The time of flight refers to how long a projectile remains in the air. It depends on the initial velocity and the gravitational acceleration of the environment in which the projectile is launched.
For a ball rolling off a table, the time of flight is determined solely by the gravitational acceleration and the height of the table. The equation is expressed as:\[ t = \sqrt{\frac{2h}{g}}, \] where \( h \) is the table height and \( g \) is the gravitational acceleration.
On Planet X, with a longer flight time required for a greater range and weaker gravitational pull, the time of flight increases. The weaker gravitational acceleration on Planet X means that objects spend more time in the air compared to Earth's considerably stronger gravitational pull. This is why the ball was able to travel a much greater horizontal distance of 2.76 times that on Earth.
Comparative Planetary Gravity
Comparing gravitational forces between planets helps us understand how gravity influences phenomena such as projectile motion and everyday experiences. Although Captain Curious performed the same experiment on Earth and Planet X, results differed due to the planets’ differing gravitational fields.
In our scenario, Planet X's gravitational acceleration was found to be significantly weaker than Earth's — approximately 1.287 m/s² compared to Earth’s 9.8 m/s². This weaker acceleration means that objects fall more slowly on Planet X, allowing them to travel farther horizontally during the same fall time.
Comparative gravitational studies are essential for space exploration, ensuring equipment and vehicles are designed to operate under various planetary conditions. Understanding these differences also aids in planning missions and conducting experiments on celestial bodies throughout our solar system.

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Most popular questions from this chapter

A cannon, located 60.0 \(\mathrm{m}\) from the base of a vertical 25.0 -m-tall cliff, shoots a 15 -kg shell at \(43.0^{\circ}\) above the horizontal toward the cliff. (a) What must the minimum muzzle velocity be Express the equation for the horizontal displacement of the cannon. \(x=\left(v_{0} \cos \theta\right) t\)for the shell to clear the top of the cliff? (b) The ground at the top of the cliff is level, with a constant elevation of 25.0 \(\mathrm{m}\) above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?

A rocket designed to place small payloads into orbit is carried to an altitude of 12.0 \(\mathrm{km}\) above sea level by a converted airliner. When the airliner is flying in a straight line at a constant speed of 850 \(\mathrm{km} / \mathrm{h}\) , the rocket is dropped. After the drop, the airliner maintains the same altitude and speed and continues to fly in a straight line. The rocket falls for a brief time, after which its rocket motor turns on. Once its rocket motor is on, the combined effects of thrust and gravity give the rocket a constant acceleration of magnitude 3.00\(g\) directed at an angle of \(30.0^{\circ}\) above the horizontal. For reasons of safety, the rocket should be at least 1.00 \(\mathrm{km}\) in front of the airliner when it climbs the airliner's altitude. Your job is to determine the minimum time that the rocket must fall before its engine starts. You can ignore air resistance. Your answer should include (i) a diagram showing the flight paths of both the rocket and the airliner, labeled at several points with vectors for their velocities and accelerations; (ii) an \(x-t\) graph showing the motions of both the rocket and the airliner; and (iii) a \(y-t\) graph showing the motions of both the rocket and the airliner. In the diagram and the graphs, indicate when the rocket is dropped, when the rocket motor turns on, and when the rocket climbs through the altitude of the airliner.

A water hose is used to fill a large cylindrical storage tank of diameter \(D\) and height 2\(D .\) The hose shoots the water at \(45^{\circ}\) the horizontal from the same level as the base of the tank and is a distance 6\(D\) away (Fig. P3.60). For what range of launch speeds \(\left(v_{0}\right)\) will the water enter the tank? Ignore air resistance, and express your answer in terms of \(D\) and \(g .\)

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Firemen are shooting a stream of water at a burning building using a high- pressure hose that shoots out the water with a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation \(\alpha\) of the hose until the water takes 3.00 s to reach a building 45.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level. (a) Find the angle of elevation \(\alpha\) . (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

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