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It is a consequence of Newton's law of gravitation that near the surface of any planet, the distance \(D\) fallen by a rock in time \(t\) is given by \(D=c t^{2}\). That is, distance fallen is proportional to the square of the time, no matter what planet one may be on. But the value of \(c\) depends on the mass of the planet. For Earth, if time is measured in seconds and distance in feet, the value of \(c\) is 16 . a. Suppose a rock is falling near the surface of a planet. What is the comparison in distance fallen from 2 seconds to 6 seconds into the drop? (Hint: This question may be rephrased as follows: "If time increases by a factor of 3 , by what factor will distance increase?") b. For objects falling near the surface of Mars, if time is measured in seconds and distance in feet, the value of \(c\) is \(6.4\). If a rock is dropped from 70 feet above the surface of Mars, how long will it take for the rock to strike the ground? c. On Venus, a rock dropped from 70 feet above the surface takes \(2.2\) seconds to strike the ground. What is the value of \(c\) for Venus?

Step by step solution

01

Calculate Distance on Earth at 2 and 6 Seconds

For part a, use the formula \( D = ct^2 \) where \( c = 16 \) for Earth. Calculate the distance at \( t = 2 \) seconds as \( D = 16 \times 2^2 = 64 \) feet. Similarly, calculate the distance at \( t = 6 \) seconds as \( D = 16 \times 6^2 = 576 \) feet.

02

Determine Factor Increase in Distance

To find the factor by which the distance increases as time increases from 2 seconds to 6 seconds, divide the distance fallen at 6 seconds by that at 2 seconds. This gives \( \frac{576}{64} = 9 \). The distance increases by a factor of 9.

03

Set Up Equation for Mars

For part b, use the formula \( D = 6.4t^2 \) for Mars, with \( D = 70 \) feet. Set up the equation \( 70 = 6.4t^2 \) to find the time \( t \) it takes for the rock to strike the ground.

04

Solve for Time on Mars

Solve the equation \( 70 = 6.4t^2 \) to find \( t \). Dividing both sides by 6.4 gives \( t^2 = \frac{70}{6.4} \). Calculating this gives \( t^2 = 10.9375 \), and taking the square root gives \( t \approx 3.31 \) seconds.

05

Use Known Variables for Venus

For part c, use the formula \( D = ct^2 \) for Venus, with \( D = 70 \) feet and \( t = 2.2 \) seconds. Substitute these values into the equation to find \( c \): \( 70 = c \times 2.2^2 \).

06

Solve for c on Venus

Calculate \( 2.2^2 = 4.84 \), then rearrange to find \( c \) as \( c = \frac{70}{4.84} \). Performing the division gives \( c \approx 14.46 \).

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Key Concepts

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

Gravitational Acceleration

Gravitational acceleration is a fundamental concept that explains how gravity pulls objects toward the center of a planet. Near the surface of a planet, this uniform acceleration can be depicted with a constant, often represented as "c." This constant varies depending on the mass of the planet and determines how fast an object will accelerate when dropped.

• On Earth, this gravitational acceleration is approximately 32 feet per second squared, but often divided by 2 for simplicity, yielding "c = 16" when using the formula for distance fallen over time.
• Mars, with a smaller mass compared to Earth, has a lower gravitational acceleration constant of "c = 6.4," making objects fall more slowly there.

Understanding these constants helps us predict how different planetary masses affect the acceleration of falling objects.

Motion Equations

Motion equations are used to analyze and predict the movement of objects under various forces, including gravity. These equations simplify the understanding of how objects travel over time and the effect of gravitational acceleration. One key motion equation in relation to falling objects near a planet's surface is:

• Distance fallen, given by the formula \( D = ct^2 \), where \( D \) is the distance, \( c \) is the gravitational acceleration constant, and \( t \) is time.

This quadratic relationship shows that the distance an object falls is proportional to the square of the time it has been falling.
This equation helps students to comprehend the dynamics of motion by evaluating how the distance changes over time and by calculating the effects of different gravitational forces from varied planetary masses.

Planetary Motion

Planetary motion refers to the way planets and objects on or near them move, influenced by gravitational forces. This concept is not just about how planets orbit the sun, but also how gravity affects objects near the surfaces of these celestial bodies.

• Newton's law of gravitation is foundational here, indicating that every mass attracts every other mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.
• This principle affects both large-scale body movements like planets and smaller-scale ones such as a rock falling on a planet's surface.

Understanding planetary motion within these contexts allows us to predict and analyze behaviors of objects in different gravitational scenarios, helping scientists in fields like astrophysics and space exploration.

Falling Object Dynamics

Falling object dynamics explores how and why objects fall, particularly focusing on the factors that influence the speed and distance of a fall. These dynamics are determined largely by the gravitational force exerted by the body the object is falling towards.

• For example, on planets with greater mass, like Earth, falling objects are subjected to greater gravitational force, thus accelerating faster compared to those that fall on Mars or Venus.
• The factors affecting these dynamics include the initial height from which an object is dropped, the gravitational constant "c" specific to the planet, and external factors like air resistance, although the latter is often ignored in basic calculations.

Exploring how these variables interact offers insights into not only fundamental physics but also practical applications like space travel and landing strategies on other planets.