The gravitational acceleration on the Moon is a sixth of that on Earth. The weight of an apple is \(1.00 \mathrm{~N}\) on Earth. a) What is the weight of the apple on the Moon? b) What is the mass of the apple?

Short Answer

Expert verified
Answer: The weight of the apple on the Moon is approximately 0.167 N, and its mass is approximately 0.102 kg.

Step by step solution

01

1. Finding the Gravitational Acceleration on the Moon

We are given that gravitational acceleration on the Moon is a sixth of that on Earth, so we can write that as: \(g_{moon} = \frac{1}{6}g_{earth}\) The gravitational acceleration on Earth is approximately \(9.81 \mathrm{~m/s^2}\). Therefore, the gravitational acceleration on the Moon is: \(g_{moon} = \frac{1}{6}(9.81 \mathrm{~m/s^2}) = 1.635 \mathrm{~m/s^2}\)
02

2. Finding the Weight of the Apple on the Moon

To find the weight of the apple on the Moon, we will use the given weight on Earth, which is \(W_{earth} = 1.00 \mathrm{~N}\), and the relation between the gravitational accelerations on Earth and the Moon: \(W_{moon} = \frac{g_{moon}}{g_{earth}} * W_{earth}\) Using the values for \(g_{moon}\), \(g_{earth}\), and \(W_{earth}\): \(W_{moon} = \frac{1.635 \mathrm{~m/s^2}}{9.81 \mathrm{~m/s^2}} * 1.00 \mathrm{~N} = 0.167 \mathrm{~N}\) So the weight of the apple on the Moon is approximately \(0.167 \mathrm{~N}\).
03

3. Finding the Mass of the Apple

To find the mass of the apple, we can use the definition of weight: \(W = mg\). We are given the weight of the apple on Earth and the gravitational acceleration on Earth, so we can write: \(W_{earth} = m_{apple} * g_{earth}\) Solving for \(m_{apple}\): \(m_{apple} = \frac{W_{earth}}{g_{earth}}\) Now, we will substitute the values for \(W_{earth}\) and \(g_{earth}\): \(m_{apple} = \frac{1.00 \mathrm{~N}}{9.81 \mathrm{~m/s^2}} \approx 0.102 \mathrm{~kg}\) Thus, the mass of the apple is approximately \(0.102 \mathrm{~kg}\). In summary, the weight of the apple on the Moon is \(0.167 \mathrm{N}\), and its mass is \(0.102 \mathrm{kg}\).

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 Vaia!

Key Concepts

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

Weight Calculation
Weight calculation is fundamental to understanding the effect of gravity on an object. In physics, the weight of an object is the force exerted on the object due to gravity. It is calculated by the formula:
\( W = mg \),
where \(W\) is the weight, \(m\) is the mass of the object, and \(g\) is the gravitational acceleration.
For instance, if you have an object with a mass of 1 kilogram on Earth, it would weigh approximately \(9.81 \text{ Newtons} (N)\), since Earth's standard gravity is \(9.81 \text{ m/s}^2\). However, this value for \(g\) changes depending on where in the universe the object is because different celestial bodies exert different gravitational forces. An understanding of how to calculate weight is critical for tasks ranging from designing architecture on Earth to planning missions to other planets.
Moon Gravity
Let's talk about moon gravity, which is distinct from Earth's gravity. The gravitational acceleration on the Moon is only about \(\frac{1}{6}\) of that on Earth, resulting in a much weaker gravitational pull. This difference is due to the Moon's smaller mass and radius compared to Earth.
As the exercise illustrated, if the gravitational acceleration on Earth is \(9.81 \text{ m/s}^2\), then on the Moon it would be:
\(g_{moon} = \frac{1}{6}g_{earth} \),
Leading to \(g_{moon} = 1.635 \text{ m/s}^2\). This means that objects on the Moon weigh less than they do on Earth. This variance in gravitational acceleration affects everything from the escape velocity needed for a spacecraft to leave the lunar surface, to the way astronauts move when they're on the Moon. Understanding moon gravity is crucial when planning lunar missions or studying lunar geology and helps us comprehend how gravity varies across celestial bodies.
Mass of an Object
The mass of an object is a measure of how much matter it contains and is commonly measured in kilograms (kg). Unlike weight, mass is not affected by gravity or location, hence it remains constant regardless of where the object is in the universe.
Following our earlier example, the mass of the apple from the exercise was determined using its weight on Earth and the gravitational acceleration on Earth, resulting in a calculated mass of approximately \(0.102 \text{ kg}\). This mass reflects the amount of matter in the apple, which would be the same whether the apple is on Earth, the Moon, or floating in space. One way to comprehend mass is to think of it as the 'stuff' that makes up an object. Mass is a key concept in physics and plays a central role in mechanics, thermodynamics, and many other areas of science.

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

An elevator cabin has a mass of \(358.1 \mathrm{~kg},\) and the combined mass of the people inside the cabin is \(169.2 \mathrm{~kg} .\) The cabin is pulled upward by a cable, with a constant acceleration of \(4.11 \mathrm{~m} / \mathrm{s}^{2}\). What is the tension in the cable?

4.32 A hanging mass, \(M_{1}=0.50 \mathrm{~kg}\), is attached by a light string that runs over a frictionless pulley to a mass \(M_{2}=1.50 \mathrm{~kg}\) that is initially at rest on a frictionless table. Find the magnitude of the acceleration, \(a,\) of \(M_{2}\)

A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

A car without ABS (antilock brake system) was moving at \(15.0 \mathrm{~m} / \mathrm{s}\) when the driver hit the brake to make a sudden stop. The coefficients of static and kinetic friction between the tires and the road are 0.550 and 0.430 , respectively. a) What was the acceleration of the car during the interval between braking and stopping? b) How far did the car travel before it stopped?

A box of books is initially at rest a distance \(D=0.540 \mathrm{~m}\) from the end of a wooden board. The coefficient of static friction between the box and the board is \(\mu_{s}=0.320\), and the coefficient of kinetic friction is \(\mu_{k}=0.250 .\) The angle of the board is increased slowly, until the box just begins to slide; then the board is held at this angle. Find the speed of the box as it reaches the end of the board. -4.55 A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

See all solutions

Recommended explanations on Physics Textbooks

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