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

Short Answer

Expert verified
The escape speed from a 300-km-diameter asteroid with a density of 2500 kg/m^3 is calculated by following the above steps.

Step by step solution

01

Calculate the Radius of the Asteroid

Given the diameter of the asteroid as 300 km, the radius \(r\) can be determined by halfing this distance. It is also necessary to convert this distance from kilometers to meters. This can be done by multiplying the value by 1000.
02

Calculate the Mass of the Asteroid

To calculate the mass \(m\) of the asteroid, the volume of the asteroid is first calculated using the formula for the volume \(V\) of a sphere: \(V = \frac{4}{3}\pi r^{3}\). The mass is then calculated by multiplying the volume with the given density \(\rho\), using the formula \(m = V \rho\).
03

Calculate the Escape Speed

Having calculated the mass and the radius, the escape speed \(v\) can now be calculated using the escape speed formula \(v = \sqrt{2gr}\), where \(g\) is the gravitational constant, which is equal to \(6.67 \times 10^{-11} m^3 kg^{-1} s^{-2}\). This formula gives the minimum speed an object must have to escape the gravitational field of the asteroid.

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Key Concepts

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

Understanding Asteroid Density
Asteroid density is a critical factor in determining the gravitational pull of an asteroid. Density is defined as mass per unit volume, typically expressed in units like kilograms per cubic meter (kg/m鲁). For this exercise, the density of the asteroid is provided as 2500 kg/m鲁. This means each cubic meter of the asteroid's material has a mass of 2500 kg.

Density is an intrinsic property, meaning it remains the same regardless of the size of the asteroid, as long as the composition is uniform. An asteroid with a high density will have a stronger gravitational pull than a less dense one of the same size. To find the entire mass of the asteroid, you would multiply this density by the asteroid's volume. By understanding density's role, we can better comprehend how different asteroids would exert gravitational forces and influence their escape speeds.
Sphere Volume Formula Demystified
The sphere volume formula is pivotal when calculating properties related to spherical objects like asteroids. The formula is given by: \[ V = \frac{4}{3} \pi r^{3} \]where \( V \) is the volume, \( \pi \approx 3.14159 \), and \( r \) is the radius of the sphere.

This formula helps determine the three-dimensional space occupied by the sphere. In the context of our exercise, it lets us find the volume of the asteroid once we know its diameter (or its radius, which is half of the diameter). By converting the asteroid's diameter from kilometers to meters and applying this formula, we can accurately calculate how much space the asteroid occupies. Understanding the volume is crucial as it directly ties to calculating mass once density is introduced. Remember, the accuracy of your volume calculation will affect subsequent steps, like determining the asteroid's mass.
Gravitational Constant Explained
The gravitational constant, denoted as \( G \), is a key figure in gravitational equations. Its value is approximately \( 6.67 \times 10^{-11} \) m鲁 kg鈦宦 s鈦宦. This constant is used universally in calculations involving gravitational forces amongst objects.

In our exercise, \( G \) is a part of the escape speed formula which helps calculate the velocity required for an object to break free from the asteroid's gravitational pull. The formula used is: \[ v = \sqrt{\frac{2Gm}{r}} \]where \( v \) is the escape velocity, \( m \) is the mass of the asteroid, and \( r \) is the radius.

Understanding \( G \) is critical as it reflects the fundamental nature of gravity in our universe. It doesn't change based on location or conditions, making it a reliable part of any calculation involving gravitational forces. Mastery of this constant allows for accurate predictions in orbital mechanics and space exploration scenarios.

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Most popular questions from this chapter

Consider the ringshaped object in Fig. \(\mathrm{E} 13.39 .\) A particle with mass \(m\) is placed a distance \(x\) from the center of the ring. along the line through the center of the ring and perpendicular to its plane. (a) Calculate the gravitational potential energy \(U\) of this system. Take the potential energy to be zero when the two objects are far apart. (b) Show that your answer to part (a) reduces to the expected result when \(x\) is much larger than the radius \(a\) of the ring. (c) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the force on the particle (see Section 7.4 ). (d) Show that your answer to part (c) reduces to the expected result when \(x\) is much larger than \(a\). (e) What are the values of \(U\) and \(F_{x}\) when \(x=0 ?\) Explain why these results make sense.

Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs \(943.0 \mathrm{~N}\) on the earth weighs \(915.0 \mathrm{~N}\) at the north pole of Planet \(X\) and only \(850.0 \mathrm{~N}\) at its equator. The distance from the north pole to the equator is \(18,850 \mathrm{~km}\), measured along the surface of Planet X. (a) How long is the day on Planet X? (b) If a \(45,000 \mathrm{~kg}\) satellite is placed in a circular orbit \(2000 \mathrm{~km}\) above the surface of Planet \(\mathrm{X}\). what will be its orbital period?

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of \(120 \mathrm{~km}\) from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of \(15.0 \mathrm{~km} .\) The astronaut is positioned inside the spaceship such that one of her \(0.030 \mathrm{~kg}\) ears is \(6.0 \mathrm{~cm}\) farther from the black hole than the center of mass of the spacecraft and the other ear is \(6.0 \mathrm{~cm}\) 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 cars 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.

A satellite with mass \(848 \mathrm{~kg}\) is in a circular orbit with an orbital speed of \(9640 \mathrm{~m} / \mathrm{s}\) around the earth. What is the new orbital speed after friction from the earth's upper atmosphere has done \(-7.50 \times 10^{9} \mathrm{~J}\) of work on the satellite? Does the speed increase or decrease?

In March \(2006,\) two small satellites were discovered orbiting Pluto, one at a distance of \(48,000 \mathrm{~km}\) and the other at \(64,000 \mathrm{~km}\). Pluto already was known to have a large satellite Charon, orbiting at \(19,600 \mathrm{~km}\) with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto.

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