Problem 20

Question

An astronaut's pack weighs 17.5 \(\mathrm{N}\) when she is on earth but only 3.24 \(\mathrm{N}\) when she is at the surface of an asteroid. (a) What is the acceleration due to gravity on this asteroid? (b) What is the mass of the pack on the asteroid?

Step-by-Step Solution

Verified
Answer
(a) \( g = 1.815 \, \text{m/s}^2 \), (b) Mass = 1.7857 kg.
1Step 1: Understand the Problem
We are given the weight of an astronaut's pack on Earth and on the surface of an asteroid. We need to find the acceleration due to gravity on the asteroid and the mass of the pack on the asteroid.
2Step 2: Recall the Formula for Weight
Weight is the force of gravity acting on an object and is given by the formula: \( W = m \cdot g \), where \( W \) is weight, \( m \) is mass, and \( g \) is the acceleration due to gravity.
3Step 3: Calculate Mass of the Pack
Use the weight of the pack on Earth to find its mass. On Earth, \( g = 9.8 \, \text{m/s}^2 \), and the weight \( W = 17.5 \, \text{N} \). Therefore, \( m = \frac{W}{g} = \frac{17.5}{9.8} = 1.7857 \, \text{kg}\).
4Step 4: Calculate Gravitational Acceleration on the Asteroid
Using the mass from Step 3 and the weight on the asteroid (3.24 N), use the formula \( W = m \cdot g \), where \( m = 1.7857 \, \text{kg} \) to find \( g \). Therefore, \( g = \frac{W}{m} = \frac{3.24}{1.7857} = 1.815 \text{m/s}^2 \).
5Step 5: Confirm Mass Consistency
The mass of the pack remains the same regardless of location in the universe: \( 1.7857 \, \text{kg} \). This is a constant property of matter.

Key Concepts

Understanding Acceleration Due to GravityUnderstanding Mass DeterminationUsing the Weight Formula
Understanding Acceleration Due to Gravity
Acceleration due to gravity is essentially the rate at which an object accelerates when it's falling down due to the gravitational pull of a celestial body. This acceleration is different for every planet, moon, or asteroid based on their mass and size. On Earth, the acceleration due to gravity is approximately 9.8 m/s².
  • This value is calculated based on Earth's mass and radius.
  • It represents how strong Earth's gravitational pull is.
For our asteroid, we found the gravitational acceleration to be 1.815 m/s². This is much smaller than Earth's, which is expected since asteroids generally have much less mass than planets like Earth.
When you know the weight of an object and its mass, you can use the relation between weight and gravitational force to solve any such problem involving different celestial bodies.
Understanding Mass Determination
Mass is a measure of how much matter is in an object. It remains the same regardless of where the object is located in the universe. Unlike weight, which can vary if you’re on Earth or an asteroid, mass is constant.
  • Mass is measured in kilograms (kg).
  • It can be calculated if you know the weight and gravitational acceleration.
In our scenario, the pack has a mass of 1.7857 kg, found using its weight on Earth (17.5 N) and Earth's gravitational acceleration (9.8 m/s²).
This step emphasizes the importance of distinguishing between mass and weight. While they are related, these properties are not the same. Knowing the mass allows us to find the gravitational acceleration on the asteroid by using the weight measured there.
Using the Weight Formula
Weight is defined as the force exerted on an object due to gravity. It is calculated using the weight formula: \[ W = m \cdot g \]Where:- \( W \) is the weight of the object,- \( m \) is its mass,- \( g \) is the acceleration due to gravity.
Using this formula, you can determine how much an object will weigh under different gravitational conditions. For example:
  • On Earth, the astronaut's pack weighs 17.5 N.
  • On the asteroid, it weighs only 3.24 N because the gravitational pull is weaker.
This formula helps you convert between mass and weight whenever you know one of these properties and the local gravitational acceleration. When dealing with gravity calculations, always remember to distinguish clearly between mass and weight to accurately solve problems.