/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 What is the escape speed from a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
The escape speed is approximately 192 m/s.

Step by step solution

01

Calculate the Radius of the Asteroid

The problem states that the asteroid has a diameter of 300 km. To find the radius, which we need for further calculations, we divide the diameter by 2. Thus, the radius \( r \) is: \[ r = \frac{300}{2} = 150 \text{ km} = 150,000 \text{ m} \]
02

Calculate the Mass of the Asteroid

We need to calculate the mass of the asteroid using its density and volume. The formula for mass \( m \) from density \( \rho \) and volume \( V \) is: \[ m = \rho \times V \]The volume \( V \) of a sphere is given by:\[ V = \frac{4}{3} \pi r^3 \]Substituting the radius in meters:\[ V = \frac{4}{3} \pi (150,000)^3 \]Now, calculate the volume:\[ V \approx 1.414 imes 10^{16} \text{ m}^3 \]Using the density \( \rho = 2500 \text{ kg/m}^3 \), calculate the mass:\[ m = 2500 \times 1.414 \times 10^{16} \approx 3.535 \times 10^{19} \text{ kg} \]
03

Use the Formula for Escape Velocity

The escape velocity \( v_e \) can be calculated using:\[ v_e = \sqrt{\frac{2GM}{r}} \]Where:- \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2} \).- \( M \) is the mass we calculated in the previous step.- \( r \) is the radius in meters.Substitute the known values:\[ v_e = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 3.535 \times 10^{19}}{150,000}} \]Calculate the result:\[ v_e \approx 192 \text{ m/s} \]
04

Final Step: Conclusion

The escape speed from the 300-km-diameter asteroid with a density of 2500 kg/m\(^3\) is approximately 192 m/s.

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 91Ó°ÊÓ!

Key Concepts

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

Asteroid Density
When analyzing the escape speed of an asteroid, one important factor is its density. Asteroid density refers to the mass of the asteroid per unit volume and is typically measured in kilograms per cubic meter (kg/m\(^3\)). It gives us insight into the materials the asteroid is composed of.A higher density means the asteroid is more tightly packed with material. In our example, a density of 2500 kg/m\(^3\) suggests it is fairly dense, perhaps composed of rock-like materials. By knowing the density, you can determine the mass more easily if the volume is known. This makes the calculation of other properties, such as escape velocity, more straightforward.
Gravitational Constant
The gravitational constant, often denoted by the symbol \( G \), is a key factor in calculating escape velocity. It represents the gravitational attraction between two masses and is a universal constant.- Its approximate value is \( 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2} \).This constant allows us to relate the force of gravity to the masses involved and the distance between them. In our escape velocity problem, it is used in the equation \( v_e = \sqrt{\frac{2GM}{r}} \), where it helps to link the asteroid's mass and radius to the speed needed to break free from its gravitational pull.
Sphere Volume Formula
When calculating the mass of a spherical object like an asteroid, knowing its volume is essential. For a sphere, the volume \( V \) is determined using the formula:\[ V = \frac{4}{3} \pi r^3 \]Here, \( r \) stands for the radius of the sphere. This formula arises because the shape of a sphere distributes its volume across three dimensions. Plugging in the asteroid's calculated radius, you can find its total volume.Understanding this formula is crucial for computing other characteristics like mass when density is provided because it gives the spatial context needed for such calculations.
Mass Calculation
To determine the mass of an asteroid, we need both its volume and its density. The mass \( m \) is calculated using the straightforward formula:\[ m = \rho \times V \]Where:- \( \rho \) is density, which in this context is given as 2500 kg/m\(^3\).- \( V \) is the volume, calculated using the sphere volume formula.By multiplying the asteroid's volume with its density, you can determine its mass, a critical value for calculating the escape velocity. In our scenario, the mass helps us find out how massive the asteroid is, which influences the gravitational force it exerts. This concept is key for many astronomical calculations and understanding how gravity operates with mass and space.

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

The dwarf planet Pluto has an elliptical orbit with a semimajor axis of 5.91 \(\times\) 10\(^{12}\) m and eccentricity 0.249. (a) Calculate Pluto's orbital period. Express your answer in seconds and in earth years. (b) During Pluto's orbit around the sun, what are its closest and farthest distances from the sun?

A planet orbiting a distant star has radius 3.24 \(\times\) 10\(^6\) m. The escape speed for an object launched from this planet's surface is 7.65 \(\times\) 10\(^3\) m/s. What is the acceleration due to gravity at the surface of the planet?

The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth. (a) Compute the acceleration due to gravity on the surface of Venus from these data. (b) If a rock weighs 75.0 N on earth, what would it weigh at the surface of Venus?

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030-kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer. (a) What is the tension between her ears? Would the astronaut find it difficult to keep from being torn apart by the gravitational forces? (Since her whole body orbits with the same angular velocity, one ear is moving too slowly for the radius of its orbit and the other is moving too fast. Hence her head must exert forces on her ears to keep them in their orbits.) (b) Is the center of gravity of her head at the same point as the center of mass? Explain.

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.980 m/s\(^2\) if the acceleration due to gravity at the surface has magnitude 9.80 m/s\(^2\)?

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.