Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 km/h (65 mi/h), and the trip takes 1 h and 50 min. On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only 70 km/h (43 mi/h). How much longer does the trip take?

(a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air?

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 s, she hears the echo of her shout from the valley floor below. The speed of sound is 340 m/s. (a) How tall is the cliff? (b) If we ignore air resistance, how fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free