Chapter 6: Problem 708
Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will be \(-m s^{-2}\) \(\left(\mathrm{g}=9.8 \mathrm{~ms}^{2}\right)\) (A) \(19.6\) (B) \(9.8\) (C) \(4.9\) (D) \(2.45\)
Short Answer
Expert verified
The new acceleration due to gravity is \(19.6 \text{ m/s²}\), which corresponds to the choice (A).
Step by step solution
01
Write the formula for acceleration due to gravity
The formula for acceleration due to gravity is:
\[g = \frac{GM}{r^2}\]
Where G is the gravitational constant, M is the mass of the Earth, and r is the radius of the Earth.
02
Write the mass-density relationship formula
The mass-density relationship is given by:
\[M = \rho V\]
Where M is the mass, ρ is the density, and V is the volume.
03
Substitute the mass-density relationship in the gravity formula
Replace M in the gravity formula using the mass-density relationship:
\[g = \frac{G \cdot (\rho V)}{r^2}\]
04
Rewrite the formula in terms of radius
Since the radius of the Earth is constant, we can rewrite the volume in terms of radius (using the formula for the volume of a sphere, \(V = \frac{4}{3} \pi r^3\)):
\[g = \frac{G \cdot (\rho \cdot \frac{4}{3} \pi r^3)}{r^2} =\frac{4}{3}G\pi r \rho\]
05
Calculate acceleration due to gravity when Earth's density is doubled
Now, we double the density (2ρ) and find the new acceleration due to gravity:
\[g' = \frac{4}{3}G\pi r (2 \rho) \]
06
Calculate the ratio of the new gravity to the initial gravity
To find the new gravity, observe the ratio between the new gravity and the initial gravity:
\[\frac{g'}{g} = \frac{\frac{4}{3}G\pi r (2 \rho)}{\frac{4}{3}G\pi r \rho} = 2\]
07
Calculate the new acceleration due to gravity
Since the ratio of the new gravity to the initial gravity is 2, the new acceleration due to gravity (g') is twice the initial gravity (9.8 m/s²):
\[g' = 2 \cdot 9.8 = 19.6 \text{ m/s²}\]
Answer: The new acceleration due to gravity is 19.6 m/s², which corresponds to the choice (A).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Density
Density is a fundamental concept that refers to how much mass is contained within a given volume. It's expressed mathematically as \( \rho = \frac{m}{V} \), where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume.
Imagine density as how tightly matter is packed. When something is dense, it means it has a lot of matter in a small space, like a stone. Comparatively, a feather is less dense because its mass is distributed in a larger volume.
In our exercise, when the density of the Earth is doubled, the mass per unit volume increases. However, because the radius remains unchanged, it affects the Earth's overall mass, influencing gravitational pull (or acceleration due to gravity) as per the formula \( \rho V = M \). Understanding density helps in grasping how mass can change without altering size directly.
Imagine density as how tightly matter is packed. When something is dense, it means it has a lot of matter in a small space, like a stone. Comparatively, a feather is less dense because its mass is distributed in a larger volume.
In our exercise, when the density of the Earth is doubled, the mass per unit volume increases. However, because the radius remains unchanged, it affects the Earth's overall mass, influencing gravitational pull (or acceleration due to gravity) as per the formula \( \rho V = M \). Understanding density helps in grasping how mass can change without altering size directly.
Gravitational Constant
The gravitational constant, often represented by \( G \), is a key factor in the law of universal gravitation. Its value is approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \).
This constant helps us calculate the force of gravity between two objects. It's a fixed number and does not change, regardless of the situation. Its role in our exercise is to remain constant as we manipulate other variables, like mass and distance, to observe how gravity (\( g \)) behaves.
By keeping \( G \) constant, we can focus on how changes in mass or distance affect gravitational forces. Think of \( G \) as a bridge, connecting two masses through the gravitational force, allowing us to predict the gravitational interactions accurately.
This constant helps us calculate the force of gravity between two objects. It's a fixed number and does not change, regardless of the situation. Its role in our exercise is to remain constant as we manipulate other variables, like mass and distance, to observe how gravity (\( g \)) behaves.
By keeping \( G \) constant, we can focus on how changes in mass or distance affect gravitational forces. Think of \( G \) as a bridge, connecting two masses through the gravitational force, allowing us to predict the gravitational interactions accurately.
Mass-Density Relationship
The mass-density relationship is a simple yet crucial formula: \( M = \rho V \), where \( M \) stands for mass, \( \rho \) for density, and \( V \) for volume.
This formula tells us that an object's mass is directly proportional to its density and volume. If either the density or volume increases, so does the mass. This principle was used in the exercise to double Earth's hypothetical mass by doubling its density, keeping the volume constant (since the radius remains the same).
Understanding this relationship helps when predicting how changes in density impact mass. It's like imagining balloons: if a balloon fills with denser air, its total mass increases compared to filling it with regular air. Used in calculations, it helps predict resulting gravitational changes like those explored in the exercise.
This formula tells us that an object's mass is directly proportional to its density and volume. If either the density or volume increases, so does the mass. This principle was used in the exercise to double Earth's hypothetical mass by doubling its density, keeping the volume constant (since the radius remains the same).
Understanding this relationship helps when predicting how changes in density impact mass. It's like imagining balloons: if a balloon fills with denser air, its total mass increases compared to filling it with regular air. Used in calculations, it helps predict resulting gravitational changes like those explored in the exercise.
Sphere Volume Formula
The sphere volume formula \( V = \frac{4}{3} \pi r^3 \) provides the volume of a spherical object using its radius \( r \).
A sphere’s volume depends on the cube of its radius. This means a small change in radius can lead to a significant change in volume. For constant density, this would also affect mass. But in our exercise, the radius remains unchanged, which means the volume stays constant as well.
When we double the density of Earth while keeping its radius constant, the volume formula helps emphasize that it's the density that influences mass changes. Imagine Earth as a solid ball; the volume doesn't expand or shrink. Any change in gravity must come from mass adjustments via density changes, highlighting how interconnected these concepts are.
A sphere’s volume depends on the cube of its radius. This means a small change in radius can lead to a significant change in volume. For constant density, this would also affect mass. But in our exercise, the radius remains unchanged, which means the volume stays constant as well.
When we double the density of Earth while keeping its radius constant, the volume formula helps emphasize that it's the density that influences mass changes. Imagine Earth as a solid ball; the volume doesn't expand or shrink. Any change in gravity must come from mass adjustments via density changes, highlighting how interconnected these concepts are.
