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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 \(\mathrm{X}\) . The com- mander, 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 \(\mathrm{X}\) ?

Short Answer

Expert verified
The acceleration due to gravity on Planet X is approximately 1.28 m/s².

Step by step solution

01

Understand the Problem

The ball rolls off a table with an initial horizontal velocity and lands at a distance from the table foot. On Earth, this distance is \(D\). On Planet \(X\), it lands 2.76 times farther, and we need to find the gravitational acceleration on Planet \(X\).
02

Identify the Relation on Earth

On Earth, let the initial speed of the ball be \(v_0\) and the table height be \(h\). The horizontal distance \(D\) it lands from the table can be expressed as \(D = v_0 \cdot t\), where \(t\) is the time of flight. The time of flight \(t\) is given by \(t = \sqrt{\frac{2h}{g}}\) where \(g\) is Earth's gravity.
03

Express Variables for Planet X

On Planet \(X\), the horizontal distance is \(2.76D\). Given the same initial speed \(v_0\), the distance equation becomes \(2.76D = v_0 \cdot t'\), where \(t'\) is the time of flight on Planet \(X\).
04

Write Time of Flight for Planet X

The time of flight \(t'\) on Planet \(X\) can be determined using the height \(h\) and the unknown gravitational acceleration \(g_x\) as: \[t' = \sqrt{\frac{2h}{g_x}}\].
05

Set Ratios to Compare

Relating the two horizontal distances, set the ratio from Earth and Planet \(X\) in terms of time: \[\frac{v_0 \cdot t'}{v_0 \cdot t} = 2.76 \quad \Rightarrow \quad \frac{t'}{t} = 2.76\].
06

Substitute Time Expressions and Simplify

Substitute the time expressions:\[\frac{\sqrt{\frac{2h}{g_x}}}{\sqrt{\frac{2h}{g}}} = 2.76\] which simplifies to \[\sqrt{\frac{g}{g_x}} = 2.76\].
07

Solve for Gravitational Acceleration on Planet X

Square both sides and solve for \(g_x\):\[\frac{g}{g_x} = (2.76)^2\]Thus, \(g_x = \frac{g}{(2.76)^2}\) where \(g = 9.8\,\text{m/s}^2\). Calculate \(g_x\) to find the gravity on Planet \(X\).
08

Calculate the Value

Using \(g = 9.8\,\text{m/s}^2\), evaluate \(g_x = \frac{9.8}{7.6176} \approx 1.28\,\text{m/s}^2\).

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Key Concepts

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

Projectile Motion
Projectile motion is a type of motion experienced by an object thrown into the air and influenced by gravity only. This motion has two components: horizontal and vertical motion. These two components are independent of each other. The horizontal motion occurs with constant velocity, while the vertical motion is subject to the acceleration due to gravity. Both the vertical and horizontal motions affect where the projectile will land.

In the given exercise, the ball rolling off the table is an example of projectile motion. Once the ball leaves the table, it becomes a projectile. The horizontal distance it travels depends on its initial horizontal velocity and the time it is in the air. On Earth, this time of flight is determined by the gravitational pull of 9.8 m/s². The ball's motion on another planet with different gravity, such as Planet X in this exercise, will result in a different horizontal distance.
Horizontal Velocity
The concept of horizontal velocity is crucial in understanding projectile motion. Horizontal velocity is the speed of an object moving parallel to the horizontal surface. Importantly, once an object is in motion, no horizontal acceleration affects it under the assumption of minimal air resistance. Therefore, this velocity remains constant throughout its flight.

In the scenario provided, the ball maintains the same initial horizontal velocity both on Earth and Planet X. This consistent speed allows us to calculate distances traveled across different gravitational fields, assuming no other forces are acting on the ball. The ball travels 2.76 times further on Planet X than on Earth because the gravity on Planet X influences the time of flight, not the horizontal velocity.
Time of Flight
The time an object spends in the air is termed 'time of flight.' It's a vital component in calculating projectile motion. The time of flight is influenced by the object's initial vertical velocity and the acceleration due to gravity. For objects released horizontally, like the ball in our exercise, the time of flight only depends on the initial vertical drop and gravitational acceleration.

On Earth, the time of flight is derived from the equation: \(t = \sqrt{\frac{2h}{g}}\). This is because the gravitational force pulls the ball to the ground. When calculating the time of flight on Planet X, the difference in gravitational acceleration changes the time the ball spends in the air. With reduced gravity, the ball remains airborne for a longer period, thus covering a greater horizontal distance.
Planetary Gravity Comparison
Different planets exert different gravitational forces on objects, a factor that significantly influences projectile motion. Earth has a standard gravitational acceleration of approximately 9.8 m/s². However, this value is not universal and varies with different celestial bodies.

In our example, Planet X has weaker gravitational pull compared to Earth. This lessened gravity means that objects will fall more slowly, extending their time of flight when thrown horizontally. By knowing the relative distance the ball travels on both surfaces (i.e., Earth and Planet X), and using the formula \(g_x = \frac{g}{(2.76)^2}\), we affirm that Planet X's gravitational force is approximately 1.28 m/s². This significant difference helps compare and understand planetary masses and sizes, providing insights into similar or more diverse extraterrestrial environments.

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Most popular questions from this chapter

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