/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 The mythical planet Mongo has tw... [FREE SOLUTION] | 91Ó°ÊÓ

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The mythical planet Mongo has twice the mass and twice the radius of Earth. Compute the acceleration due to gravity at its surface. We know from Problem \(3.40\) that in general $$ g_{R}=\frac{G M}{R^{2}} $$ Then for the Earth at its surface $$ g_{E}=\frac{G M_{E}}{R_{E}^{2}}=g=9.81 \mathrm{~m} / \mathrm{s}^{2} $$ where \(R_{E}\) is the Earth's radius and \(M_{E}\) is its mass. For Mongo $$ g_{M}=\frac{G M_{M}}{R_{M}^{2}}=\frac{G\left(2 M_{E}\right)}{\left(2 R_{E}\right)^{2}} $$ and or $$ \begin{array}{c} g_{M}=\frac{1}{2} \frac{G M_{E}}{R_{E}^{2}}=\frac{1}{2}\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) \\ g_{M}=4.91 \mathrm{~m} / \mathrm{s}^{2} \end{array} $$

Short Answer

Expert verified
The acceleration due to gravity on Mongo's surface is \( 4.91 \, \text{m/s}^2 \).

Step by step solution

01

Understand the given formula for gravity

The formula to find the acceleration due to gravity at any planet's surface is given by \( g = \frac{G M}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.
02

Relate given values to Earth's properties

We know from the problem's statement that Earth's gravity is \( g_E = 9.81 \, \text{m/s}^2 \), and it's calculated as \( g_E = \frac{G M_E}{R_E^2} \). This will serve as a reference for calculating Mongo's gravity.
03

Analyze changes in mass and radius for Mongo

For the planet Mongo, both the mass and radius are twice that of Earth's. So, \( M_M = 2M_E \) and \( R_M = 2R_E \). This is crucial for substituting into the gravity formula for Mongo.
04

Substitute values into Mongo's gravity formula

Substitute the mass and radius of Mongo into the gravity formula: \( g_M = \frac{G \times (2 M_E)}{(2 R_E)^2} \).
05

Simplify the expression

Simplify the expression: \( g_M = \frac{2 G M_E}{4 R_E^2} = \frac{1}{2} \frac{G M_E}{R_E^2} \).
06

Calculate the gravity on Mongo

Since \( \frac{G M_E}{R_E^2} = g_E = 9.81 \, \text{m/s}^2 \), we find \( g_M = \frac{1}{2} \times 9.81 \, \text{m/s}^2 = 4.91 \, \text{m/s}^2 \).

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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 we talk about gravity on the surface of a planet, we specifically refer to the acceleration objects experience when they fall. This is known as the acceleration due to gravity, commonly denoted as \( g \). On Earth, we usually consider \( g \) as \( 9.81 \, \mathrm{m/s}^2 \). However, this value can vary based on a planet's mass and radius. On larger planets, gravitational acceleration can be higher, but that's not always the case if the radius also increases.

Gravity isn't just a force pulling everything down to a planet's surface. It's a crucial factor that affects everything from how we move to how atmospheric layers are structured. Understanding gravity's role helps us appreciate the unique conditions present on different worlds, such as Mongo in this example.
Gravitational Formula
The gravitational formula is an expression that helps us calculate the acceleration due to gravity on any celestial body. It is given by the equation \( g = \frac{G M}{R^2} \), where:
  • \( G \) is the universal gravitational constant, approximately \( 6.674 \times 10^{-11} \, \mathrm{Nm^2/kg^2} \).
  • \( M \) is the mass of the planet.
  • \( R \) is the radius of the planet from its center to its surface.
This equation shows that gravity is directly proportional to the mass of the planet and inversely proportional to the square of its radius. Thus, if you increase a planet's mass, gravity will increase, but if you increase its radius, gravity will decrease.

Using this formula, we can calculate the different gravity experienced on any planet by simply plugging in the values of mass and radius specific to that planet.
Mass and Radius Relation
The mass and radius of a planet are fundamental properties that determine the gravity experienced on the planet's surface. For the planet Mongo, we have twice the mass and twice the radius compared to Earth. This results in a unique scenario where both properties change simultaneously.

When we adjust these properties:
  • Mass increases by a factor (2), doubling it compared to Earth.
  • Radius also increases by the same factor (2), altering how gravity is distributed over the planet's surface.
Given this scenario, the radius change affects gravity more dramatically due to its square relationship in the gravitational formula. This means that while the mass increases gravity, the increase in radius, which is squared, reduces the overall acceleration. In the end, these changes balance out and provide a new gravitational acceleration.
Step-by-Step Calculation
Creating a step-by-step calculation is essential to understanding how all the parts of the problem fit together. Here's how we calculated the acceleration due to gravity on Mongo:1. **Understand the given formula:** Recognize that the gravity on the surface is calculated by \( g = \frac{G M}{R^2} \).2. **Relate given values to Earth's properties:** Use Earth's known gravity \( g_E = 9.81 \, \text{m/s}^2 \) as a reference point to understand calculations for Mongo.3. **Analyze changes in mass and radius for Mongo:** For Mongo, both the mass \( (M_M = 2M_E) \) and radius \( (R_M = 2R_E) \) are double those of Earth's.4. **Substitute values into Mongo's gravity formula:** Plug these values into Mongo's formula: \( g_M = \frac{G \times (2 M_E)}{(2 R_E)^2} \).5. **Simplify the expression:** Simplify to find \( g_M = \frac{1}{2} \frac{G M_E}{R_E^2} \).6. **Calculate the gravity on Mongo:** Compute the final value as \( g_M = \frac{1}{2} \times 9.81 \, \text{m/s}^2 = 4.91 \, \text{m/s}^2 \).Each step captures not only the mechanical process of substitution and simplification but also highlights the underlying principles of how mass and radius affect gravitational acceleration.

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Most popular questions from this chapter

A \(600-\mathrm{N}\) object is to be given an acceleration of \(0.70 \mathrm{~m} / \mathrm{s}^{2} .\) How large an unbalanced force must act upon it? Notice that the weight, not the mass, of the object is given. Assuming the weight was measured on the Earth, use \(F_{W}=m g\) to find $$ m=\frac{F_{W}}{g}=\frac{600 \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=61.2 \mathrm{~kg} $$ Now that we know the mass of the object \((61.2 \mathrm{~kg})\) and the desired acceleration \(\left(0.70 \mathrm{~m} / \mathrm{s}^{2}\right)\), the force is $$ F=m a=(61.2 \mathrm{~kg})\left(0.70 \mathrm{~m} / \mathrm{s}^{2}\right)=42.8 \mathrm{~N} \text { or } 43 \mathrm{~N} $$

A \(5.0-\mathrm{kg}\) block rests on a \(30^{\circ}\) incline. The coefficient of static friction between the block and the incline is \(0.20\). How large a horizontal force must push on the block if the block is to be on the verge of sliding \((a)\) up the incline and \((b)\) down the incline?

An astronaut weighs \(480 \mathrm{~N}\) on Earth. She visits the planet Krypton, which has a mass and diameter each ten times that of Earth. Determine her weight at a distance of two Kryptonian radii above that fictional planet.

A space station that weighs \(10.0 \mathrm{MN}\) on Earth is positioned at a distance of ten Earth radii from the center of the planet. What would it weigh out there in space- that is, what is the value of the gravity force pulling it toward Earth?

The radius of the Earth is about \(6370 \mathrm{~km}\), while that of Mars is about \(3440 \mathrm{~km}\). If an object weighs \(200 \mathrm{~N}\) on Earth, what would it weigh, and what would be the acceleration due to gravity, on Mars? The mass of Mars is \(0.11\) that of Earth. Neglect planetary rotations and local mass variations.

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