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A \(15-k g\) rock is dropped from rest on the carth and reaches the ground in 1.75 s. When it is dropped from the same height on Saturn's satellite Enceladus, it 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 0.0866 m/s².

Step by step solution

01

Find the initial conditions

The rock is dropped from rest, meaning its initial velocity is 0 m/s. We are given two scenarios: one on Earth and one on Enceladus.
02

Write the equation of motion for free fall

The equation for the distance fallen under uniform acceleration due to gravity is: \[ d = v_i \cdot t + \frac{1}{2} g t^2 \] where \( v_i \) is the initial velocity (0 in this case), \( t \) is the time, and \( g \) is the acceleration due to gravity.
03

Calculate the distance fallen on Earth

On Earth, we know \( g = 9.8 \ m/s^2 \) and the time is \( t = 1.75 \ s \). Substitute these into the equation: \[ d = 0 \cdot 1.75 + \frac{1}{2} \cdot 9.8 \cdot (1.75)^2 \] \[ d = 0 + \frac{1}{2} \cdot 9.8 \cdot 3.0625 \] \[ d = 4.9 \cdot 3.0625 \approx 14.985625 \ m \] Therefore, the rock falls approximately 14.99 meters on Earth.
04

Use this distance for Enceladus

Now, using the same distance (\( d = 14.99 \ m \)) for Enceladus, and the new time \( t = 18.6 \ s \), we will solve for \( g \), Enceladus's gravity. We use: \[ 14.99 = \frac{1}{2} g (18.6)^2 \] \[ 14.99 = \frac{1}{2} g \cdot 345.96 \] \[ g = \frac{14.99 \cdot 2}{345.96} \approx 0.0866 \ m/s^2 \].
05

Verify the units and magnitude

Check the final value of \( g \) on Enceladus to ensure it is in the correct units of \( m/s^2 \) and appears reasonable given the conditions.

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Key Concepts

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

Free Fall
Free fall describes the motion of an object under the influence of gravitational force alone. When an object is "dropped," it experiences free fall, meaning no other forces, like air resistance, are acting on it. This state allows us to examine the effects of gravity without interference.
For any object in free fall, its initial velocity is usually zero (if dropped from rest). During free fall, the only acceleration acting on the object is due to gravity, denoted by \( g \). This makes understanding gravitational acceleration crucial. In free fall calculations, if air resistance is insignificant, objects at different weights fall at the same rate, accelerating at \( g \).
  • Free fall simplifies many physics problems since only gravity is in play.
  • On Earth, objects accelerate at approximately \( 9.8 \, m/s^2 \) due to gravity.
Understanding free fall is essential as it allows us to apply motion equations to predict outcomes accurately in different environments.
Motion Equations
Motion equations help us predict the future position, velocity, and acceleration of objects in motion when forces like gravity act on them. For a rock in free fall, the basic motion equation is:
\[ d = v_i \cdot t + \frac{1}{2} g t^2 \]
where:
  • \( d \) is the distance fallen.
  • \( v_i \) is the initial velocity.
  • \( t \) is the time taken to fall.
  • \( g \) is the acceleration due to gravity.
Substituting known values into this equation allows us to determine unknown variables like gravitational acceleration on different planetary bodies.
For instance, with Earth, we assume \( g = 9.8 \ m/s^2 \). The distance a rock falls can be precisely calculated using this motion equation.
Motion equations are fundamental for problems involving various initial conditions, such as non-zero initial velocities or other forces acting on the falling object.
Planetary Gravity
Planetary gravity refers to the gravitational attraction exerted by a planet or moon. This force affects objects in free fall, along with many other dynamics.
The acceleration due to gravity (\( g \)) on different celestial bodies varies depending on their mass and size. For example, Earth's \( g \) is about \( 9.8 \, m/s^2 \), whereas on the moon and other satellites like Enceladus, it is significantly less.
  • On Enceladus, we calculated \( g \approx 0.0866 \, m/s^2 \), indicating weaker gravity compared to Earth.
  • The strength of gravity influences how fast objects will fall towards the planetary surface.
  • Lesser gravity results in slower falls, explored in scenarios where objects take much longer to reach the ground compared to Earth.
Understanding planetary gravity helps scientists and engineers predict and plan for conditions in space missions, effectively accommodating the physical laws that govern different planets or moons.

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Most popular questions from this chapter

Two sunt drivers drive directly toward each other. At time \(t=0\) the two cars are a distance \(D\) apart, car 1 is at rest, and \(\operatorname{car} 2\) is moving to the left with speed \(v_{0}\) . Car 1 begins to move at \(t=0\) , speeding up with a constant acceleration \(a_{x}\) Car 2 continues to move with a constant velocity. (a) At what time do the two cars collide? \((b)\) Find the speed of car 1 just before it collides with \(\operatorname{car} 2 .\) (c) Sketch \(x-t\) and \(v_{x}-t\) graphs for car 1 and \(\operatorname{car} 2 .\) For each of the two graphs, draw the curves for both cars on the same set of axes.

Raindrops. If the effects of the air acting on falling raindrops are ignored, then we can treat raindrops as freely falling objects. (a) Rain clouds are typically a few hundred meters above the ground. Estimate the speed with which raindrops would strike the ground if they were freely falling objects. Give your estimate in \(\mathrm{m} / \mathrm{s}, \mathrm{km} / \mathrm{h},\) and \(\mathrm{mi} / \mathrm{h}\) . (b) Estimate (from your own personal observations of rain) the speed with which raindrops actually strike the ground. (c) Based on your answers to parts (a) and (b), is it a good approximation to neglect the effects of the air on falling raindrops? Explain.

During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.30 \(\mathrm{m} / \mathrm{s}^{2} .\) When it is 235 \(\mathrm{m}\) above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored). (a) How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration? (b) What total distance did the canister travel between its release and its crash onto the launch pad?

A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) . Air resistance may be ignored, so the water balloon is in free fall after it leaves the thrower's hand. (a) What is its speed after falling for 2.00 s? (b) How far does it fall in 2.00 \(\mathrm{s?}\) (c) What is the magnitude of its velocity after falling 10.0 \(\mathrm{m} ?\) (d) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion.

A car's velocity as a function of time is given by \(v_{x}(t)=\) \(\alpha+\beta t^{2},\) where \(\alpha=3.00 \mathrm{m} / \mathrm{s}\) and \(\beta=0.100 \mathrm{m} / \mathrm{s}^{3}\) , (a) Calculate the average acceleration for the time interval \(t=0\) to \(t=5.00 \mathrm{s}\) (b) Calculate the instantaneous acceleration for \(t=0\) and \(t=\) 5.00 \(\mathrm{s}\) . (c) Draw accurate \(v_{x}-t\) and \(a_{x}-t\) graphs for the car's motion between \(t=0\) and \(t=5.00 \mathrm{s} .\)

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