/*! 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 5 An astronaut weighs \(700 \mathr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 5

An astronaut weighs \(700 \mathrm{~N}\) on Earth. What is the best approximation of her new weight on a planet with a radius that is two times that of Earth, and a mass three times that of Earth? A. \(200 \mathrm{~N}\) B. \(500 \mathrm{~N}\) C. \(700 \mathrm{~N}\) D. \(900 \mathrm{~N}\)

Short Answer

Expert verified
B. 500 N

Step by step solution

01

Understanding the Weight Formula

Weight is calculated using the formula: \[ W = mg \] Where \( W \) is the weight, \( m \) is the mass of the astronaut, and \( g \) is the acceleration due to gravity. Gravity on a planet is given by: \[ g = \frac{GM}{R^2} \] Where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.
02

Gravitational Acceleration on Earth

Using Earth's parameters: \[ g_{\text{earth}} = \frac{GM_{\text{earth}}}{R_{\text{earth}}^2} \] Given that the astronaut's weight on Earth is \( 700 \text{ N} \), we can express her weight as: \[ 700 = m \cdot g_{\text{earth}} \]
03

Gravitational Acceleration on the New Planet

The new planet has a mass \( M_{\text{planet}} = 3M_{\text{earth}} \) and a radius \( R_{\text{planet}} = 2R_{\text{earth}} \). The gravitational acceleration on the new planet is: \[ g_{\text{planet}} = \frac{G \times 3M_{\text{earth}}}{(2R_{\text{earth}})^2} \] Simplifying this: \[ g_{\text{planet}} = \frac{3GM_{\text{earth}}}{4R_{\text{earth}}^2} \] So, \[ g_{\text{planet}} = \frac{3}{4} g_{\text{earth}} \]
04

Calculating the New Weight

Using the ratio: \[ W_{\text{new}} = m \cdot g_{\text{planet}} \] We know that \( W_{\text{earth}} = 700 \text{ N} = m \cdot g_{\text{earth}} \), so the new weight is: \[ W_{\text{new}} = 700 \times \frac{3}{4} \] Simplifying, we get: \[ W_{\text{new}} = 525 \text{ N} \]
05

Selecting the Closest Answer

The best approximation to \( 525 \text{ N} \) given the choices is \( 500 \text{ N} \). Therefore, the correct answer is: \( B. 500 \text{ N} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

acceleration due to gravity
The acceleration due to gravity is a force that pulls objects towards the center of a planet. It's why we stay on the ground. Earth鈥檚 gravity is calculated using the formula: \[g = \frac{GM}{R^2}\] Where:
  • G is the gravitational constant
  • M is the planet鈥檚 mass
  • R is the planet鈥檚 radius
gravity on Earth is about 9.8 m/s虏. How heavy or light an object feels depends on the gravity it faces. So, when we mention gravity in space or on other planets, it might differ vastly from Earth's gravity. This gravity impacts the weight of objects on that planet.
weight calculation on different planets
Weight isn't the same everywhere in the universe. Weight is a measure of how much force gravity exerts on an object. It's calculated with: \[W = mg\] where \(W\) is weight, \(m\) is mass, and \(g\) is gravity. Even though mass stays the same, weight changes due to gravity's pull. Suppose an astronaut weighs 700 N (Newtons) on Earth. To find her weight on another planet, follow these steps:
  • Find the new gravity, using: \(g_{new} = \frac{GM_{new}}{(R_{new})^2}\)
  • Calculate her new weight: \(W_{new} = m \cdot g_{new}\)
For our given exercise, we determined the new planet's gravity and then calculated the astronaut鈥檚 new weight accordingly.
universal law of gravitation
The universal law of gravitation describes the attractive force between two masses. It's crucial in understanding how planets and objects interact in space. It's given by: \[F = \frac{GMm}{R^2}\]every mass attracts every other mass with a force. This force (\(F\)) depends on:
  • G - the gravitational constant (a universal value)
  • M - the first mass (like a planet)
  • m - the second mass (like an astronaut)
  • R - the distance between the two masses
By integrating this law into gravity and weight equations, we can solve problems like finding weight on different planets, as demonstrated in the exercise.
planetary mass and radius effects
A planet's mass and radius significantly affect gravitational pull. Let's break it down:
  • Larger mass usually means stronger gravity
  • Larger radius means weaker gravity (due to increased distance from the planet's center)
In the given exercise, the new planet had:
  • Three times Earth鈥檚 mass
  • Twice Earth鈥檚 radius
Using these values, the new gravitational acceleration was found with: \(g_{\mathrm{planet}} \ = \ \frac{3 \times G M_{\mathrm{earth}}}{(2 R_{\mathrm{earth}})^2} \ = \ \frac{3 GM_{\mathrm{earth}}}{4 R_{\mathrm{earth}}^2} \ \ = \ \frac{3}{4} g_{\mathrm{earth}} \)Therefore, compared to Earth, the new gravity is only three-fourths as strong. This adjustment directly influences the astronaut's weight on this new planet.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(1,000 \mathrm{~kg}\) rocket ship, traveling at \(100 \mathrm{~m} / \mathrm{s}\), is acted on by an average force of \(20 \mathrm{kN}\) applied in the direction of its motion for 8 \(\mathrm{s}\). What is the change in velocity of the rocket? A. \(100 \mathrm{~m} / \mathrm{s}\) B. \(160 \mathrm{~m} / \mathrm{s}\) C. \(1,600 \mathrm{~m} / \mathrm{s}\) D. \(2,000 \mathrm{~m} / \mathrm{s}\)

An elevator is accelerating down at \(4 \mathrm{~m} / \mathrm{s}^2\). How much does a \(10 \mathrm{~kg}\) fish weigh if measured inside the elevator? A. \(140 \mathrm{~N}\) B. \(100 \mathrm{~N}\) C. \(60 \mathrm{~N}\) D. \(50 \mathrm{~N}\)

A \(1 \mathrm{~kg}\) ball with a radius of \(20 \mathrm{~cm}\) rolls down a \(5 \mathrm{~m}\) high inclined plane. Its speed at the bottom is \(8 \mathrm{~m} / \mathrm{s}\). How many revolutions per second is the ball making when at the bottom of the plane? A. 6 revolutions/second B. 12 revolutions/second C. 20 revolutions/second D. 23 revolutions/second

A \(10 \mathrm{~kg}\) wagon rests on a frictionless inclined plane. The plane makes an angle of \(30^{\circ}\) with the horizontal. What is the force, \(F\), required to keep the wagon from sliding down the plane? A. \(10 \mathrm{~N}\) B. \(30 \mathrm{~N}\) C. \(49 \mathrm{~N}\) D. \(98 \mathrm{~N}\)

A distant solar system is made up one small planet of mass \(1.48 \times\) \(10^{24} \mathrm{~kg}\) revolving in a circular orbit about a large, stationary star of mass \(7.3 \times 10^{30} \mathrm{~kg}\). The distance between their centers is \(5 \times 10^{11}\) \(\mathrm{m}\). What happens to the speed of the planet if the distance between the star and the planet increases (assume the new orbit is also circular)? A. The speed of the planet increases. B. The speed of the planet decreases. C. The speed of the planet remains the same. D. It cannot be determined from the information given.
See all solutions

Recommended explanations on English Textbooks

TESOL (English)

Read Explanation

Semiotics

Read Explanation

Syntax

Read Explanation

Graphology

Read Explanation

English Grammar Summary

Read Explanation

5 Paragraph Essay

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.