Problem 53 - University Physics with Modern Physics

Question

What is the escape speed from a 300-km-diameter asteroid with a density of 2500 \(\mathrm{kg} / \mathrm{m}^{3} ?\)

Step-by-Step Solution

Answer

Escape speed is approximately 144 m/s.

Step 1: Determine the Volume of the Asteroid

Use the formula for the volume of a sphere, which is \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius. Given the diameter of 300 km, the radius \( r \) is 150 km, or 150,000 meters. Calculate the volume by substituting \( r = 150,000 \) m into the volume formula. The volume \( V = \frac{4}{3} \pi (150,000)^3 \approx 1.414 \times 10^{16} \) cubic meters.

Step 2: Calculate the Mass of the Asteroid

The mass \( M \) can be found using the formula \( M = \text{Density} \times \text{Volume} \). Given the density of the asteroid is 2500 \( \mathrm{kg/m^3} \), multiply it by the volume calculated in Step 1. Thus, the mass \( M = 2500 \times 1.414 \times 10^{16} = 3.535 \times 10^{19} \) kg.

Step 3: Apply the Escape Velocity Formula

Use the formula for escape velocity: \( v_e = \sqrt{\frac{2GM}{r}} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \mathrm{m^3/kg/s^2} \), \( M \) is the mass of the asteroid, and \( r \) is the radius of the asteroid in meters. Substitute \( M = 3.535 \times 10^{19} \) kg and \( r = 150,000 \) m into the formula. Calculate \( v_e = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 3.535 \times 10^{19}}{150,000}} \approx 144 \) m/s.

Key Concepts

Understanding Asteroids

Asteroids are fascinating celestial objects that orbit the Sun, primarily found in the asteroid belt between Mars and Jupiter. These space rocks vary significantly in size, shape, and composition. Some asteroids can be as small as a few meters across, while others stretch to hundreds of kilometers in diameter.

When dealing with astrophysical problems, such as escape velocity, it's important to understand the physical properties of an asteroid. Its diameter, mass, and density all play crucial roles in calculating other related values. Knowing these values helps in determining how strong the gravitational pull of the asteroid is and what forces are needed to overcome it for an object to escape its gravitational field.

Role of Density in Astrophysics

Density is a fundamental property that describes how much mass is contained within a specific volume. It is calculated using the formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). In the context of an asteroid, knowing the density allows us to determine its total mass when its volume is known.

For this particular problem involving escape velocity, density helps in calculating the mass of the asteroid, which is a crucial component of the escape velocity formula. For example, an asteroid with a higher density would result in a higher mass, subsequently affecting the gravitational pull it has on surrounding objects. This is why density is often a key focus in astrophysics when studying celestial objects.

Gravitational Constant and Its Uses

The gravitational constant, represented by \( G \), is a vital constant in physics that quantifies the strength of gravitational attraction between two masses. It is used in Newton's law of universal gravitation, which calculates the force of attraction between masses. The value of \( G \) is approximately \( 6.674 \times 10^{-11} \ \mathrm{m^3/kg/s^2} \).

In the escape velocity formula \( v_e = \sqrt{\frac{2GM}{r}} \), the gravitational constant \( G \) helps to determine the speed needed for an object to break free from the gravitational pull of an asteroid. Any changes in \( G \) would significantly affect gravitational interactions, making it an essential parameter in all calculations involving gravity.

Calculating the Volume of a Sphere

The volume of a sphere is an important concept used in many scientific calculations, including astrophysics. To find the volume, you can use the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere.

For an asteroid with a given diameter, the radius would be half of the diameter. For instance, if an asteroid has a diameter of 300 km, its radius would be 150 km, or 150,000 meters. Using the formula, you plug in the radius to find the volume. This calculation is key to determining other properties such as the mass of the asteroid when density is known, as it directly influences the strength of its gravitational field which is crucial for assessing escape velocity.