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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?

Short Answer

Expert verified
Gravity on Enceladus is approximately 0.0868 m/sĀ².

Step by step solution

01

Understand the Problem

We need to find the acceleration due to gravity on Enceladus. We know the time it takes for the rock to reach the ground on Earth and Enceladus, and that the initial velocity is zero as it is dropped. We can use these times to find the gravitational accelerations.
02

Determine Earth's Gravitational Acceleration

We know gravity on Earth is constant as \( g = 9.81 \, \text{m/s}^2 \). We will compare this with Enceladus to find the latter's gravitational acceleration.
03

Apply Free Fall Equation for Earth

For Earth, use the free fall formula \( d = \frac{1}{2} g t^2 \) where \( d \) is the distance fallen, \( g = 9.81 \, \text{m/s}^2 \), and \( t = 1.75 \, \text{s} \).Calculate the distance \( d = \frac{1}{2} \times 9.81 \times (1.75)^2 = 15.0319 \, \text{m} \).
04

Use Distance for Enceladus

Use the same distance \( d = 15.0319 \, \text{m} \) for Enceladus. Apply the same formula with time \( t = 18.6 \, \text{s} \) to calculate the gravitational acceleration \( g_e \) for Enceladus: \( d = \frac{1}{2} g_e t^2 \).
05

Solve for the Gravitational Acceleration on Enceladus

Rearrange the formula and solve for \( g_e \):\[ g_e = \frac{2d}{t^2} = \frac{2 \times 15.0319}{(18.6)^2} \approx 0.0868 \, \text{m/s}^2 \].

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Key Concepts

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

Free Fall Equation
The free fall equation is a fundamental part of physics that describes the motion of objects falling solely under the influence of gravity. This situation implies no air resistance or other forces are acting on the object, such as a rock dropped from a height. The equation used to calculate the distance an object falls is:
  • \( d = \frac{1}{2}gt^2 \)
This formula considers:
  • \(d\): the distance fallen in meters,
  • \(g\): the gravitational acceleration in meters per second squared \(m/s^2\),
  • \(t\): the time in seconds \(s\).
Using this equation, we can determine how far an object will fall in a certain amount of time given the gravitational force acting on it. The rock dropped in the exercise is a perfect example of a free fall scenario where the equation helps find important parameters such as distance.
Gravitational Acceleration
Gravitational acceleration refers to the acceleration of an object caused by the force of gravity from a massive body like a planet or a moon. On Earth, this value is standard and constant at approximately \( 9.81 \, \text{m/s}^2 \). This means that any object falling freely near the surface of the Earth accelerates at this rate.
  • Any changes in this value are due to other factors such as air resistance, which isn't considered in basic free fall calculations.
  • When calculating gravitational acceleration on another celestial body, like Enceladus, it is vital to notice how it differs from Earth. If a rock takes longer to fall, this indicates a weaker gravitational pull.
Understanding gravitational acceleration is crucial in various applications, including space exploration, where knowing this factor helps in landing and maneuvering on other planets or moons.
Saturn's Satellite Enceladus
Saturn's satellite Enceladus is an intriguing celestial body, known for its icy surface and geysers that suggest a subsurface ocean. Although not as gravitationally potent as Earth, Enceladus offers a unique environment for studying gravitational forces. In our exercise:
  • A rock took 18.6 seconds to fall from the same height it took 1.75 seconds on Earth.
  • This significant difference highlights the weaker gravitational field Jones there.
  • The calculated gravitational acceleration on Enceladus is about \( 0.0868 \, \text{m/s}^2 \), much less than Earth's \( 9.81 \, \text{m/s}^2 \).
These measurements are crucial for missions that aim to land instruments or probes on Enceladus, as they guide design and fuel calculations for landings.
Distance Calculation in Physics
Distance calculation in physics involves determining how far an object travels over a period, often within the dynamics of forces and motion. Using the free fall equation, we can deduce how far the object has "fallen" under gravity.
  • It's important because it defines factors like time, velocity, and acceleration in motion problems.
  • In the example provided, the distance the rock fell when dropped is a constant that applies to both Earth and Enceladus, enabling us to directly compare gravitational effects across different environments.
  • By maintaining the same falling distance when comparing two bodies, we isolate and manipulate only the variables of time and gravitational acceleration, simplifying complex calculations into understandable physics.

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Most popular questions from this chapter

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. (a) At what time after being ejected is the boulder moving at 20.0 m/s upward? (b) At what time is it moving at 20.0 m/s downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch \(a_y-t, v_y-t\), and \(y-t\) graphs for the motion.

You are standing at rest at a bus stop. A bus moving at a constant speed of \(5.00 \mathrm{~m} / \mathrm{~s}\) passes you. When the rear of the bus is \(12.0 \mathrm{~m}\) past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of \(0.960 \mathrm{~m} / \mathrm{~s}^{2}\). How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

An object is moving along the \(x\)-axis. At \(t =\) 0 it has velocity \(v_{0x}\) = 20.0 m/s. Starting at time \(t =\) 0 it has acceleration \(a_x = -Ct\), where \(C\) has units of m/s\(^3\). (a) What is the value of \(C\) if the object stops in 8.00 s after \(t =\) 0? (b) For the value of \(C\) calculated in part (a), how far does the object travel during the 8.00 s?

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 \(t=10.0 \mathrm{~s} ?\)

The rocket-driven sled \(\textit{Sonic Wind No. 2,}\) used for investigating the physiological effects of large accelerations, runs on a straight, level track 1070 m (3500 ft) long. Starting from rest, it can reach a speed of 224 m/s(500 mi/h) in 0.900 s. (a) Compute the acceleration in m/s\(^2\), assuming that it is constant. (b) What is the ratio of this acceleration to that of a freely falling body (\(g\))? (c) What distance is covered in 0.900 s? (d) A magazine article states that at the end of a certain run, the speed of the sled decreased from 283 m/s (632 mi/h) to zero in 1.40 s and that during this time the magnitude of the acceleration was greater than 40\(g\). Are these figures consistent?

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