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

Short Answer

Expert verified
The acceleration due to gravity on Enceladus is approximately \(0.47 \mathrm{m/s^2}\). This means that an object will fall slower on Enceladus than it would on Earth due to a lower acceleration due to gravity.

Step by step solution

01

Determine the Time Squared Ratio

Calculate the square of the time it takes for the rock to reach the ground on Earth and the square of the time it takes for the rock to reach the ground on Enceladus. The ratio of these two values (time on Enceladus squared over time on Earth squared) gives us the ratio of the accelerations due to gravity on Enceladus and Earth. \(Ratio_{time^2} = \frac{time_{Enceladus}^2}{time_{Earth}^2} = \frac{18.6^2}{1.75^2}\)
02

Calculate the Acceleration due to Gravity on Enceladus

To find the acceleration due to gravity on Enceladus, we need to divide the acceleration due to gravity on Earth by the time squared ratio calculated in step 1. \(g_{Enceladus} = \frac{g_{Earth}}{Ratio_{time^2}} = \frac{9.8}{\frac{18.6^2}{1.75^2}}\)

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Key Concepts

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

Acceleration Due to Gravity
When discussing celestial bodies, acceleration due to gravity is a fundamental concept. It's a measure of how quickly an object speeds up as it falls, pulled by the gravity of a planet or satellite. For Earth, this value is approximately 9.8 meters per second squared (\(9.8 \text{ m/s}^2\)). This means that every second, a freely falling object will increase its speed by 9.8 m/s.
Understanding this helps explain why objects fall at different rates on other celestial bodies. On planets or moons with less gravitational pull than Earth, like Enceladus, the acceleration due to gravity is lower, causing objects to fall more slowly.

To calculate how fast or slow an object falls on another planet, you use the formula:\(g = \frac{distance}{time^2}\).Here, \(g\) is the acceleration due to gravity, "distance" is how far an object falls, and "time" is how long it takes to fall.
Gravitational Comparison
Gravitational comparison involves comparing how gravity affects objects on different celestial bodies. By looking at the time it takes for objects to fall, we can infer the differences in gravity.
In our exercise, a rock takes longer to fall on Enceladus compared to Earth. This indicates weaker gravity on Enceladus. By comparing the time squared values for each body, we can derive the corresponding gravitational acceleration.
To perform this comparison:
  • Square the fall time for each celestial body
  • Compare these squared times to understand the difference in gravitational pull
The larger the time squared, the weaker the gravity, making this method a straightforward way to evaluate gravitational differences.
Motion in Gravitational Fields
Motion in gravitational fields explains how objects behave when subjected to gravity. On Earth, you've likely seen how objects accelerate as they fall. The path and velocity are determined by the field's strength.
On Enceladus, for instance, weaker gravity translates to slower acceleration. This prolongs the fall time. Such variations in gravitational fields across solar system bodies lead to interesting differences:
  • Satellites orbit slower or faster depending on the planet's gravity
  • Compound flight patterns of spacecrafts are planned by considering these fields
Understanding motion in these fields helps scientists calculate trajectories and predict movements accurately. This knowledge ensures successful space missions and ongoing exploration of the universe.

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Most popular questions from this chapter

A bird is flying due east. Its distance from a tall building is given by \(x(t)=28.0 \mathrm{~m}+(12.4 \mathrm{~m} / \mathrm{s}) t-\left(0.0450 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} .\) What is the instantaneous velocity of the bird when \(t=8.00 \mathrm{~s} ?\)

An object's velocity is measured to be \(v_{x}(t)=\alpha-\beta t^{2}\), where \(\alpha=4.00 \mathrm{~m} / \mathrm{s}\) and \(\beta=2.00 \mathrm{~m} / \mathrm{s}^{3} .\) At \(t=0\) the object is at \(x=0 .\) (a) Calculate the object's position and acceleration as functions of time. (b) What is the object's maximum positive displacement from the origin?

An antelope moving with constant acceleration covers the distance between two points \(70.0 \mathrm{~m}\) apart in \(6.00 \mathrm{~s}\). Its speed as it passes the second point is \(15.0 \mathrm{~m} / \mathrm{s}\). What are (a) its speed at the first point and (b) its acceleration?

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

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

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