Problem 82 - Physics

Question

Neutron stars consist only of neutrons and have unbelievably high densities. A typical mass and radius for a neutron star might be \(2.7 \times 10^{28} \mathrm{~kg}\) and \(1.2 \times 10^{3} \mathrm{~m}\). (a) Find the density of such a star. (b) If a dime \(\left(V=2.0 \times 10^{-7} \mathrm{~m}^{3}\right)\) were made from this material, how much would it weigh (in pounds)?

Step-by-Step Solution

Answer

The density is approximately \(3.73 \times 10^{18} \text{ kg/m}^3\) and the weight of the dime is about \(1.64 \times 10^{12}\) pounds.

Step 1: Understanding Density Formula

Density is calculated by the formula \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume. We will use this formula to find the density of the neutron star.

Step 2: Calculate Volume of Neutron Star

The volume of a sphere is given by \( V = \frac{4}{3}\pi r^3 \). We substitute \( r = 1.2 \times 10^3 \) meters into the formula to calculate the volume: \[ V = \frac{4}{3} \pi (1.2 \times 10^3)^3 \].

Step 3: Calculate Density of Neutron Star

Using the volume calculated in the previous step and the given mass \( m = 2.7 \times 10^{28} \) kg, we find the density: \[ \rho = \frac{2.7 \times 10^{28} \text{ kg}}{7.2382 \times 10^9 \text{ m}^3} \]. The result is \( \rho \approx 3.73 \times 10^{18} \text{ kg/m}^3 \).

Step 4: Convert Density to Weight of Dime

Given that the volume of the dime is \( 2.0 \times 10^{-7} \text{ m}^3 \), we calculate the mass using \( m = \rho \cdot V = 3.73 \times 10^{18} \cdot 2.0 \times 10^{-7} \). Mass of the dime is \( 7.46 \times 10^{11} \) kg.

Step 5: Convert Mass to Weight in Pounds

To convert the mass of the dime to weight in pounds, we use the conversion factor: 1 kg = 2.20462 lbs. Multiply the mass by this factor to get \( 1.64 \times 10^{12} \) lbs.

Key Concepts

Density Calculation

Density is a measure of how much mass is contained within a given volume. It tells us how tightly matter is packed together. For a neutron star, which is made up of densely packed neutrons, understanding density is essential. This concept is captured by the formula \( \rho = \frac{m}{V} \), where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume. In our problem, to find the density of a neutron star with a mass \( 2.7 \times 10^{28} \text{ kg}\), we first need to determine its volume. Once we know the volume, we divide the mass by the volume to get the density. This step is crucial in understanding just how incredibly dense a neutron star really is.

Stellar Mass and Radius

When examining celestial objects like neutron stars, one of the first things to note is their mass and radius. The mass of the neutron star in this exercise is given as \( 2.7 \times 10^{28} \text{ kg} \), a value that places it among the densest objects in the universe. The radius is stated as \( 1.2 \times 10^{3} \text{ m} \), which, while seemingly large, is minuscule compared to stars like the Sun. These measurements, mass and radius, allow us to calculate other important properties, such as volume, which is necessary for density calculations. They also help astronomers understand the powerful gravitational forces at play within neutron stars.

Volume of a Sphere

Understanding the volume of a geometric shape is key to many areas of physics, including astrophysics. For a sphere, the volume is calculated using the formula:

• \( V = \frac{4}{3}\pi r^3 \)

where \( r \) is the radius of the sphere. In the context of a neutron star with a radius of \( 1.2 \times 10^{3} \text{ m} \), we plug this value into the formula to find its volume. The calculation involves raising the radius to the power of three, making the operation sensitive to errors, so accuracy is important. Once the volume is known, it sets the stage for determining the density of the neutron star, illustrating how even massive objects can be profoundly compact.

Weight Conversion

Converting weight from one unit to another is a practical skill, especially when working in different scientific domains. In our exercise, we need to convert the mass of a hypothetical dime made of neutron star material from kilograms to pounds. Given that the mass calculation yields \( 7.46 \times 10^{11} \text{ kg} \), the conversion to pounds is done using the factor:

• 1 kg = 2.20462 lbs.

Thus, multiplying the mass in kilograms by this conversion factor gives us the weight in pounds. This conversion demonstrates the immense weight nutrient star material would have, and underscores the utility of being able to convert between units in practical measurements.