Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 4

Compared to your mass on Earth, on the Moon your mass would be a. lower because the Moon is smaller than Earth. b. lower because the Moon has less mass than Earth. c. higher because of the combination of the Moon's mass and size. d. the same, mass doesn't change.

Short Answer

Expert verified
d: Mass doesn't change.

Step by step solution

01

Understand the Concept of Mass

Mass is a measure of the amount of matter in an object and doesn't change regardless of location. Weight, however, can change depending on the gravitational pull.
02

Difference between Mass and Weight

It's key to distinguish between mass and weight. Mass remains constant wherever it is, while weight changes based on the local gravitational force.
03

Analyze the Options

a. Suggests mass changes due to size difference between Earth and Moon – incorrect.b. Suggests mass changes due to Earth's and Moon's mass difference – incorrect.c. Suggests mass increases because of Moon's mass and size – incorrect.d. Correct – mass remains constant regardless of location.
04

Choose the Correct Answer

Option (d) is correct because mass remains constant regardless of the location.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

mass
Mass is a fundamental concept in physics and measures the amount of matter in an object. It is usually measured in kilograms (kg) or grams (g). Mass itself doesn't change based on location, because the matter in the object doesn't change.
For example, if you have a 5 kg object on Earth, it will still be 5 kg on the Moon. Mass is an intrinsic property of the object, meaning it stays the same regardless of external factors such as gravitational force.
This is why, in the provided exercise, the correct answer is (d): the mass stays the same on the Moon, even though other factors might change.
weight
Unlike mass, weight is a force, and it depends on both the mass of the object and the gravitational force acting on it. Weight is calculated using the formula:
\[ \text{Weight} = \text{Mass} \times \text{Gravitational\text{ Force}} \]
On Earth, we experience more gravitational force compared to the Moon. This means a 5 kg object will weigh less on the Moon than on Earth because the gravitational force on the Moon is weaker.
Therefore, weight can change when you move from one celestial body to another, even though your mass stays the same.
The exercise clarifies this by showing weight changes with gravity, but mass remains constant.
gravitational force
Gravitational force is the attraction between two masses, such as a person and a planet. The force depends on both the masses involved and the distance separating them:
\[ F_G = G \frac{m_1 m_2}{r^2} \]
Where:
  • \( F_G \) = gravitational force
  • \( G \) = gravitational constant
  • \( m_1, m_2 \) = masses involved
  • \( r \) = distance between the masses
Gravitational force is what makes objects fall to the ground when dropped.
On Earth, this force is stronger compared to the Moon, making objects weigh more.
This explains why a 5 kg object feels heavier on Earth than it would on the Moon.
differences between mass and weight
While mass and weight are often confused, they are fundamentally different:
  • Mass is the measure of the amount of matter in an object. It doesn't change based on location.
  • Weight is a force. It is the product of mass and the gravitational force acting on that mass.
This means you could weigh less on the Moon due to its weaker gravitational force, but your mass remains unchanged.
Understanding this distinction helps clarify why the answer to the exercise is (d).
Mass remains constant irrespective of location, while weight can vary based on the local gravitational forces.

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

At the surface of Earth, the escape velocity is \(11.2 \mathrm{km} / \mathrm{s}\). What would be the escape velocity at the surface of a very small asteroid having a radius \(10^{-4}\) that of Earth and a mass \(10^{-12}\) that of Earth?

Self-gravity is a. the gravitational pull of a person. b. the force that holds objects like people and lamps together. c. the gravitational interaction of all the parts of a body. d. the force that holds objects on Earth.

Earth's average radius is \(6,370 \mathrm{km}\) and its mass is \(5.97 \times 10^{24} \mathrm{kg}\) Show that the acceleration of gravity at the surface of Earth is \(9.81 \mathrm{m} / \mathrm{s}^{2}\).

Go to NASA's "Apollo 15 Hammer-Feather Drop" Web page (http://nssdc.gsfc.nasa.gov/planetary/lunar/apollo_15_feather_ drop.html) and watch the video from Apollo 15 of astronaut David Scott dropping the hammer and falcon feather on the Moon. (You might find a better version on YouTube.) What did this experiment show? What would happen if you tried this on Earth with a feather and a hammer? Would it work? Suppose instead you dropped the hammer and a big nail. How would they fall? How does the acceleration of falling objects on the Moon compare to the acceleration of falling objects on Earth?

If the Moon were 2 times closer to Earth than it is now, the gravitational force between Earth and the Moon would be a. 2 times stronger. b. 4 times stronger. c. 8 times stronger. d. 16 times stronger.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Force

Read Explanation

Electrostatics

Read Explanation

Engineering Physics

Read Explanation

Astrophysics

Read Explanation

Fluids

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.

Sign-up for free