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NASA has taken an increased interest in near Earth asteroids. These objects, popularized in recent blockbuster movies, can pass very close to Earth on a cosmic scale, sometimes as close as 1 million miles. Most are small-less than \(500 \mathrm{~m}\) across \(-\) and while an impact with one of the smaller ones could be dangerous, experts believe that it may not be catastrophic to the human race. One possible defense system against near Earth asteroids involves hitting an incoming asteroid with a rocket to divert its course. Assume a relatively small asteroid with a mass of \(2.10 \cdot 10^{10} \mathrm{~kg}\) is traveling toward the Earth at a modest speed of \(12.0 \mathrm{~km} / \mathrm{s}\). a) How fast would a large rocket with a mass of \(8.00 \cdot 10^{4} \mathrm{~kg}\) have to be moving when it hit the asteroid head on in order to stop the asteroid? b) An alternative approach would be to divert the asteroid from its path by a small amount to cause it to miss Earth. How fast would the rocket of part (a) have to be traveling at impact to divert the asteroid's path by \(1.00^{\circ}\) ? In this case, assume that the rocket hits the asteroid while traveling along a line perpendicular to the asteroid's path.

Short Answer

Expert verified
Question: a) Calculate the velocity of the rocket required to stop an asteroid of mass 1.5 × 10^12 kg and velocity 12.0 km/s if the rocket has a mass of 1.0 x 10^4 kg. b) Determine the speed of the rocket that can divert the asteroid's path by 1 degree from its trajectory. Answer: a) To find the velocity of the rocket, use the calculated formula: \(v_{rocket} = \frac{(m_{asteroid} + m_{rocket}) \cdot v_{final} - m_{asteroid} \cdot v_{asteroid}}{m_{rocket}}\) Plug in the given values: \(v_{rocket} = \frac{(1.5 × 10^{12} kg + 1.0 × 10^4 kg) \cdot 0 - (1.5 × 10^{12} kg)(12.0 \space km/s)}{1.0 × 10^4 kg}\) \(v_{rocket} = \frac{-1.8 × 10^{16} \space kg \cdot m/s}{1.0 × 10^4 \space kg}\) \(v_{rocket} = -1.8 × 10^{12}\) m/s b) To find the requisite vertical velocity of the rocket, use the formula: \(v_{y_{rocket}} = \frac{m_{asteroid} \cdot (v_{x_{asteroid,final}} \tan (\Delta \theta_{rad}))}{m_{rocket}}\) Plug in the given values: \(v_{y_{rocket}} = \frac{1.5 × 10^{12} kg \cdot (12.0 \space km/s \tan (\frac{\pi}{180}))}{1.0 × 10^4 kg}\) \(v_{y_{rocket}} = \frac{1.5 × 10^{12} kg \cdot (12.0 \space km/s \tan (0.01745))}{1.0 × 10^4 kg}\) \(v_{y_{rocket}} = \frac{3.14 × 10^8 \space kg \cdot m/s}{1.0 × 10^4 kg}\) \(v_{y_{rocket}} = 3.14 × 10^4\) m/s

Step by step solution

01

a) Finding the speed of the rocket to stop the asteroid

To stop the asteroid, we will use the conservation of linear momentum principle. The momentum before and after the impact will be the same: \(p_{before} = p_{after}\) The momentum of the asteroid before impact is given by: \(p_{asteroid,initial} = m_{asteroid} \cdot v_{asteroid}\) The momentum of the rocket before impact is: \(p_{rocket,initial} = m_{rocket} \cdot v_{rocket}\) After the impact, we want the asteroid to stop; hence its final velocity will be zero. Therefore, the final momentum of the asteroid is zero. The final momentum of the rocket is given by: \(p_{final} = (m_{asteroid} + m_{rocket}) \cdot v_{final}\) Using the conservation of momentum, we get: \(m_{asteroid} \cdot v_{asteroid} + m_{rocket} \cdot v_{rocket} = (m_{asteroid} + m_{rocket}) \cdot v_{final}\) We know the masses of the asteroid and the rocket, as well as the velocity of the asteroid. We can solve for the velocity of the rocket \(v_{rocket}\): \(v_{rocket} = \frac{(m_{asteroid} + m_{rocket}) \cdot v_{final} - m_{asteroid} \cdot v_{asteroid}}{m_{rocket}}\)
02

b) Finding the speed of the rocket to divert the asteroid's path by 1 degree

For this part of the question, we need to analyze the situation using conservation of momentum in two dimensions. Since the rocket impacts the asteroid perpendicular to its path, we'll have: \(p_{y_{before}} = p_{y_{after}}\) The initial vertical momentum is only due to the rocket: \(p_{y_{before}} = m_{rocket} \cdot v_{y_{rocket}}\) The final vertical momentum is shared between the asteroid and the rocket: \(p_{y_{after}} = m_{asteroid} \cdot v_{y_{asteroid,final}} + m_{rocket} \cdot v_{y_{rocket,final}}\) By conserving vertical momentum, we have: \(m_{rocket} \cdot v_{y_{rocket}} = m_{asteroid} \cdot v_{y_{asteroid,final}} + m_{rocket} \cdot v_{y_{rocket,final}}\) The change in the asteroid's path is given as 1 degree: \(\Delta \theta = 1^{\circ}\) Let's convert this angle to radians: \(\Delta \theta_{rad} = \frac{\pi}{180}\) Using the relation: \(\tan (\Delta \theta) = \frac{v_{y_{asteroid,final}}}{v_{x_{asteroid,final}}}\) We can solve for the final vertical velocity of the asteroid: \(v_{y_{asteroid,final}} = v_{x_{asteroid,final}} \tan (\Delta \theta_{rad})\) Since the asteroid keeps moving horizontally, it means that the final horizontal velocity of the asteroid remains the same: \(v_{x_{asteroid,final}} = v_{asteroid} = 12.0 \space km/s\) Plugging back into our conservation of vertical momentum equation: \(m_{rocket} \cdot v_{y_{rocket}} = m_{asteroid} \cdot (v_{x_{asteroid,final}} \tan (\Delta \theta_{rad})) + m_{rocket} \cdot v_{y_{rocket,final}}\) And finally, we can solve for the requisite vertical velocity of the rocket: \(v_{y_{rocket}} = \frac{m_{asteroid} \cdot (v_{x_{asteroid,final}} \tan (\Delta \theta_{rad}))}{m_{rocket}}\)

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Key Concepts

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

Near Earth Asteroids
Near Earth Asteroids (NEAs) are celestial bodies that have orbits bringing them close to Earth's vicinity. These fascinating objects can zoom past our planet, sometimes at distances considered close in the grand cosmos, such as 1 million miles. Although most NEAs are relatively small, often less than 500 meters across, their potential impact still sparks concern. Even a modest-sized asteroid could pose a significant threat if it were to collide with Earth. However, the movies often exaggerate; experts believe that smaller asteroids, while dangerous, are unlikely to cause humanity's extinction.

Detecting and tracking these asteroids is essential for planetary defense. Space agencies like NASA have been working on cataloging NEAs to monitor their trajectories. This is crucial for early detection of any potential impactors.

The study of NEAs not only aids in protection but also contributes to our understanding of the solar system's evolution. These objects are remnants from the solar system's formation, offering valuable insights into its primordial conditions.
Asteroid Deflection
Asteroid deflection is a strategy that aims to prevent an asteroid from colliding with Earth. The concept involves altering the asteroid's path enough to ensure a miss. One proposed method for deflection is the kinetic impactor technique. This approach involves sending a spacecraft to collide with the asteroid at high speed. The impact changes the asteroid's velocity slightly, but over millions of miles, that small change can mean a significant deviation from Earth-bound trajectories.

For instance, if a small asteroid is approaching Earth, a rocket or a space probe could be launched to meet the asteroid head-on or at a specific angle. By calculating the necessary impact speed and angle, we can plan an intervention that adjusts the asteroid's course. The change doesn't need to be large; even moving it by 1 degree could make a dramatic difference over long distances.

While the idea sounds like science fiction, it is grounded in the laws of physics, primarily Newton's laws of motion, which dictate that momentum transfer can be used to divert an object from its path. The DART mission is a real-world example of such efforts, where NASA is testing this theory on a smaller scale.
Linear Momentum
Linear momentum is a fundamental concept in physics, expressing the product of an object's mass and velocity. It is denoted by the equation:\[p = m imes v\]where \(p\) represents momentum, \(m\) is mass, and \(v\) is velocity. Momentum is a vector quantity, meaning it has both magnitude and direction.

The principle of conservation of momentum states that in a closed system, without external forces, the total momentum remains constant. This is pivotal during collisions, where the momentum before the impact equals the momentum after. This principle is what we apply when discussing asteroid deflection, ensuring the calculations correctly account for the balance of momentum between the colliding bodies.

In asteroid deflection strategies, understanding and applying these principles mean calculating the initial and final momenta of both the asteroid and the rocket (or impactor) used. For instance, to stop an asteroid, the rocket must impart sufficient momentum to counteract the asteroid's momentum, resulting in a net zero balance after collision. For a deflection, the goal is not to stop, but to tweak the direction through skilled momentum exchange.

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Most popular questions from this chapter

Consider two carts, of masses \(m\) and \(2 m\), at rest on a frictionless air track. If you push the lower-mass cart for \(3 \mathrm{~s}\) and then the other cart for the same length of time and with the same force, which cart undergoes the larger change in momentum? a) The cart with mass \(m\) has the larger change b) The cart with mass \(2 m\) has the larger change. c) The change in momentum is the same for both carts. d) It is impossible to tell from the information given.

A ball with mass \(m=0.210 \mathrm{~kg}\) and kinetic energy \(K_{1}=2.97 \mathrm{~J}\) collides elastically with a second ball of the same mass that is initially at rest. \(m_{1}\) After the collision, the first ball moves away at an angle of \(\theta_{1}=\) \(30.6^{\circ}\) with respect to the horizontal, as shown in the figure. What is the kinetic energy of the first ball after the collision?

Moessbauer spectroscopy is a technique for studying molecules by looking at a particular atom within them. For example, Moessbauer measurements of iron (Fe) inside hemoglobin, the molecule responsible for transporting oxygen in the blood, can be used to determine the hemoglobin's flexibility. The technique starts with X-rays emitted from the nuclei of \({ }^{57}\) Co atoms. These X-rays are then used to study the Fe in the hemoglobin. The energy and momentum of each X-ray are \(14 \mathrm{keV}\) and \(14 \mathrm{keV} / \mathrm{c}\) (see Example 7.5 for an explanation of the units). \(\mathrm{A}^{57}\) Co nucleus recoils as an \(\mathrm{X}\) -ray is emitted. A single \({ }^{57}\) Co nucleus has a mass of \(9.52 \cdot 10^{-26} \mathrm{~kg} .\) What are the final momentum and kinetic energy of the \({ }^{57}\) Co nucleus? How do these compare to the values for the X-ray?

A method for determining the chemical composition of a material is Rutherford backscattering (RBS), named for the scientist who first discovered that an atom contains a high-density positively charged nucleus, rather than having positive charge distributed uniformly throughout (see Chapter 39 ). In RBS, alpha particles are shot straight at a target material, and the energy of the alpha particles that bounce directly back is measured. An alpha particle has a mass of \(6.65 \cdot 10^{-27} \mathrm{~kg} .\) An alpha particle having an initial kinetic energy of \(2.00 \mathrm{MeV}\) collides elastically with atom X. If the backscattered alpha particle's kinetic energy is \(1.59 \mathrm{MeV}\), what is the mass of atom \(\mathrm{X}\) ? Assume that atom \(X\) is initially at rest. You will need to find the square root of an expression, which will result in two possible an- swers (if \(a=b^{2},\) then \(b=\pm \sqrt{a}\) ). Since you know that atom \(X\) is more massive than the alpha particle, you can choose the correct root accordingly. What element is atom X? (Check a periodic table of elements, where atomic mass is listed as the mass in grams of 1 mol of atoms, which is \(6.02 \cdot 10^{23}\) atoms.)

Cosmic rays from space that strike Earth contain some charged particles with energies billions of times higher than any that can be produced in the biggest accelerator. One model that was proposed to account for these particles is shown schematically in the figure. Two very strong sources of magnetic fields move toward each other and repeatedly reflect the charged particles trapped between them. (These magnetic field sources can be approximated as infinitely heavy walls from which charged particles get reflected elastically.) The high- energy particles that strike the Earth would have been reflected a large number of times to attain the observed energies. An analogous case with only a few reflections demonstrates this effect. Suppose a particle has an initial velocity of \(-2.21 \mathrm{~km} / \mathrm{s}\) (moving in the negative \(x\) -direction, to the left), the left wall moves with a velocity of \(1.01 \mathrm{~km} / \mathrm{s}\) to the right, and the right wall moves with a velocity of \(2.51 \mathrm{~km} / \mathrm{s}\) to the left. What is the velocity of the particle after six collisions with the left wall and five collisions with the right wall?

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