Problem 19 - University Physics with Modern Physics

Question

At the surface of Jupiter's moon Io, the acceleration due to gravity is \(g=1.81 \mathrm{m} / \mathrm{s}^{2} .\) A watermelon weighs 44.0 \(\mathrm{N}\) at the surface of the earth. (a) What is the watermelon's mass on the earth's surface? (b) What are its mass and weight on the surface of Io?

Step-by-Step Solution

Answer

(a) 4.49 kg; (b) Mass: 4.49 kg, Weight: 8.13 N on Io.

Step 1: Understand the Relation between Weight and Mass on Earth

The weight of an object on Earth's surface is given by the formula \[ W = m imes g \]where \( W \) is the weight in Newtons, \( m \) is the mass in kilograms, and \( g \) is the acceleration due to gravity, which is approximately \( 9.8 \, \text{m/s}^2 \) on Earth. In this problem, we know the weight of the watermelon: \( W = 44.0 \, \text{N} \). We need to rearrange the formula to find the mass \( m \):\[ m = \frac{W}{g} \]

Step 2: Calculate the Mass of the Watermelon on Earth

Using the equation from Step 1, substitute \( W = 44.0 \, \text{N} \) and \( g = 9.8 \, \text{m/s}^2 \):\[ m = \frac{44.0 \, \text{N}}{9.8 \, \text{m/s}^2} = 4.49 \, \text{kg} \] Thus, the mass of the watermelon on Earth's surface is approximately \( 4.49 \, \text{kg} \).

Step 3: Understand the Mass on Io's Surface

Mass does not change with location; it is an intrinsic property of the watermelon. Therefore, the mass of the watermelon on the surface of Io remains \( 4.49 \, \text{kg} \), the same as on Earth.

Step 4: Calculate the Watermelon's Weight on Io

The weight on Io can be calculated using the same weight formula, but substituting Io's gravitational acceleration:\[ W_{\text{Io}} = m \times g_{\text{Io}} \]where \( g_{\text{Io}} = 1.81 \, \text{m/s}^2 \). Using the previously calculated mass:\[ W_{\text{Io}} = 4.49 \, \text{kg} \times 1.81 \, \text{m/s}^2 = 8.13 \, \text{N} \] Hence, the weight of the watermelon on Io is approximately \( 8.13 \, \text{N} \).

Key Concepts

Weight and Mass Calculation

Understanding weight and mass is essential in gravitational physics. Weight is the force exerted on an object due to gravity. It depends on both the object's mass and the gravitational acceleration acting on it. In equations, we describe weight using the formula:

• \( W = m \times g \)

where \( W \) denotes weight in Newtons (N), \( m \) signifies mass in kilograms (kg), and \( g \) is the gravitational acceleration in meters per second squared (m/s²).

Mass, on the other hand, remains constant regardless of location. It is a fundamental property and isn't influenced by external factors like gravity. For instance, in the solution, the watermelon's mass is calculated based on its weight and Earth's gravity, which gives:

• \( m = \frac{W}{g} = \frac{44.0 \, \text{N}}{9.8 \, \text{m/s}^2} = 4.49 \, \text{kg} \)

This mass remains the same whether the watermelon is on Earth, Io, or anywhere else in the universe.

Gravitational Acceleration

Gravitational acceleration varies based on the celestial body, influencing the weight but not the mass of an object. On Earth, this value is approximately \( 9.8 \, \text{m/s}^2 \). However, other celestial bodies have different gravitational accelerations. For example, Io, one of Jupiter’s moons, has a gravitational acceleration of \( 1.81 \, \text{m/s}^2 \).

This difference explains why an object can weigh differently on Io compared to Earth. The formula used to explore this variance in weight is:

• \( W = m \times g \)
• On Io: \( W = 4.49 \, \text{kg} \times 1.81 \, \text{m/s}^2 = 8.13 \, \text{N} \)

The lower gravitational pull on Io results in the watermelon weighing less despite having the same mass as on Earth. Understanding gravitational acceleration helps in solving physics problems on different cosmic terrains.

Physics Problems on Different Celestial Bodies

Solving physics problems on different celestial bodies involves understanding how physical concepts change with environment. When considering gravity, it's crucial to adjust calculations to account for different gravitational accelerations.

Let's use the watermelon example again to clarify. While its mass is constant at \( 4.49 \, \text{kg} \) everywhere, its weight changes based on the gravitational pull of the celestial body. If we transported the watermelon from Earth to Io:

• On Earth, the weight calculation: \( W = 44.0 \, \text{N} \)
• On Io, with less gravity: \( W = 8.13 \, \text{N} \)

By using Earth's gravity (\( 9.8 \, \text{m/s}^2 \)) and Io's (\( 1.81 \, \text{m/s}^2 \)), you see how gravitational differences affect the calculations. This conceptual shift is essential for physics students when they're tasked with similar hypotheticals about items on the Moon, Mars, or other cosmic entities.