A 4.0 x 10 10 kg asteroid is heading directly toward the center of the earth at a steady 20 km/s. To save the planet, astronauts strap a giant rocket to the asteroid perpendicular to its direction of travel. The rocket generates 5.0 x 10 9 N of thrust. The rocket is fired when the asteroid is 4.0 x 10 6 km away from earth. You can ignore the earth's gravitational force on the asteroid and their rotation about the sun. If the mission fails, how many hours is it until the asteroid impacts the earth?

Hi, everyone in this practice problem. We are asked to determine when the rocket will crash on earth where we'll have a rocket with a mass of 5.49 times 10 to the power of five kg going back to the earth's orbit. However, when the rocket was at three times 10 to the power of five kilometers away, one of its engine actually failed due to a malfunction and it started falling back at a rate of 15 kilometers per second. We were asked to ignore the earth's gravitational force on the rocket and determine when the rocket will actually crash on the earth. The options are a eight times 10 to the power of two seconds. B, one times 10 to the power of 11 seconds. C three times 10 to the power of seven seconds and D two times 10 to the power of three seconds. So first, I'm going to, to start us off with making a list of everything that is given that are important for us to actually solve this problem. First, you'll have the displacement D which is just three times 10 to the power of 4 km because that is the position where the earth is experiencing or the rocket is experiencing the malfunction. Next, we have the velocity, the velocity is going to equals to 15 km/s. And that is essentially the rate at which the rocket is falling back to the earth. So we're being asked to uh determine the time that it takes for the rocket to essentially finally reach or fall or crash into the earth. So we want to use the equation of time equals displacement divided by velocity to essentially determine this value. So the time will then equals to D over V which will equals to three times 10 to the power of four kilometers divided by 15 kilometers per second, which will then actually correspond to two times 10 to the power of three seconds. And that will essentially be the answer to our problem, which will correspond to option D. So option D with the time of two times 10 to the power of three seconds is going to be the answer to this particular practice problem. And if you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and they'll be all for this one. Thank you.

Table of contents

Skip topic navigation

0. Math Review31m
- Math Review
  31m
1. Intro to Physics Units1h 29m
2. 1D Motion / Kinematics3h 56m
3. Vectors2h 43m
4. 2D Kinematics1h 42m
5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
7. Friction, Inclines, Systems2h 44m
8. Centripetal Forces & Gravitation7h 26m
9. Work & Energy1h 59m
10. Conservation of Energy2h 54m
11. Momentum & Impulse3h 40m
12. Rotational Kinematics2h 59m
13. Rotational Inertia & Energy7h 4m
14. Torque & Rotational Dynamics2h 5m
15. Rotational Equilibrium3h 39m
16. Angular Momentum3h 6m
17. Periodic Motion2h 9m
18. Waves & Sound3h 40m
19. Fluid Mechanics4h 27m
20. Heat and Temperature3h 7m
21. Kinetic Theory of Ideal Gases1h 50m
22. The First Law of Thermodynamics1h 26m
23. The Second Law of Thermodynamics3h 11m
24. Electric Force & Field; Gauss' Law3h 42m
25. Electric Potential1h 51m
26. Capacitors & Dielectrics2h 2m
27. Resistors & DC Circuits3h 8m
28. Magnetic Fields and Forces2h 23m
29. Sources of Magnetic Field2h 30m
30. Induction and Inductance3h 38m
31. Alternating Current2h 37m
32. Electromagnetic Waves2h 14m
33. Geometric Optics2h 57m
34. Wave Optics1h 15m
35. Special Relativity2h 10m

4. 2D Kinematics

Kinematics in 2D

Physics

4. 2D Kinematics

Kinematics in 2D

Previous problemNext problem

2:41 minutes

Problem 4a

Textbook Question

A 4.0 x 10¹⁰ kg asteroid is heading directly toward the center of the earth at a steady 20 km/s. To save the planet, astronauts strap a giant rocket to the asteroid perpendicular to its direction of travel. The rocket generates 5.0 x 10⁹ N of thrust. The rocket is fired when the asteroid is 4.0 x 10⁶ km away from earth. You can ignore the earth's gravitational force on the asteroid and their rotation about the sun. If the mission fails, how many hours is it until the asteroid impacts the earth?

Verified step by step guidance

Step 1: Calculate the time it would take for the asteroid to reach Earth if it continues traveling at its current velocity. Use the formula for time, $ t = \frac{d}{v} $, where $ d $ is the distance to Earth ($ 4.0 \times 10^6 \; \text{km} $) and $ v $ is the velocity of the asteroid ($ 20 \; \text{km/s} $). Convert the time from seconds to hours by dividing by 3600.

Step 2: Convert the given distance $ d $ from kilometers to meters to ensure consistent units. Since $ 1 \; \text{km} = 1000 \; \text{m} $, multiply $ 4.0 \times 10^6 \; \text{km} $ by 1000 to get the distance in meters.

Step 3: Substitute the converted distance (in meters) and the velocity ($ 20 \; \text{km/s} $, converted to $ 20000 \; \text{m/s} $) into the formula $ t = \frac{d}{v} $ to calculate the time in seconds.

Step 4: Convert the time from seconds to hours by dividing the result by 3600 (since there are 3600 seconds in an hour). This will give the time until the asteroid impacts the Earth in hours.

Step 5: Interpret the result to understand how much time remains before the asteroid impacts Earth if no intervention occurs. This value represents the time available for any corrective actions to be taken.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. In this scenario, understanding kinematics is essential to calculate the time it takes for the asteroid to travel a certain distance at a constant speed. The basic formula used is time = distance / speed, which allows us to determine how long it will take for the asteroid to reach Earth if no other forces act on it.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, expressed as F = ma. Although the question specifies ignoring gravitational forces, this law is crucial for understanding how the thrust from the rocket will affect the asteroid's motion if the mission were to succeed. It helps in analyzing the changes in velocity and direction due to the applied force.

Thrust and Impulse

Thrust is the force exerted by a rocket engine to propel an object, while impulse is the change in momentum resulting from a force applied over time. In this scenario, the thrust generated by the rocket can alter the asteroid's trajectory, but if the mission fails, the initial conditions of motion remain unchanged. Understanding thrust and impulse is vital for evaluating the potential effectiveness of the astronauts' intervention and the resulting motion of the asteroid.

Watch next

Master Kinematics in 2D with a bite sized video explanation from Patrick

Start learning

Related Videos

Related Practice

Multiple Choice

Bob throws a ball from his second story window to his friend Steve who is on the ground below. Assume that Bob throws the ball horizontally at

10 m / s

, and Steve will catch the ball at a point

4.5 m

below Bob. How far from the base of the building should Steve stand in order to catch the ball?

571

views

Multiple Choice

Jim pushes a penny off a

0.86 m

tall table. Assume the penny is moving horizontally as it leaves the table at 5 m/s. How long is the penny in the air? Ignore air resistance.

662

views

Multiple Choice

A baseball player hits a pop fly ball. If the ball leaves the bat at

45 m / s

at an angle of 68° above the horizontal, what is the maximum height, above the point of contact with the bat, that the ball will reach? Ignore air resistance.

679

views

Textbook Question

A 4.0 x 10¹⁰ kg asteroid is heading directly toward the center of the earth at a steady 20 km/s. To save the planet, astronauts strap a giant rocket to the asteroid perpendicular to its direction of travel. The rocket generates 5.0 x 10⁹ N of thrust. The rocket is fired when the asteroid is 4.0 x 10⁶ km away from earth. You can ignore the earth's gravitational force on the asteroid and their rotation about the sun.The radius of the earth is 6400 km. By what minimum angle must the asteroid be deflected to just miss the earth?

(II) At t = 0, a particle starts from rest at 𝓍 = 0, y = 0 and moves in the xy plane with an acceleration $\vec{a}$ = (4.0 î + 3.0 ĵ ) m/s². Determine the speed of the particle as a function of time.

471

views

Textbook Question

A skier is accelerating down a 30.0° hill at 1.80 m/s² (Fig. 3–42). How long will it take her to reach the bottom of the hill, assuming she starts from rest and accelerates uniformly, if the elevation change is 125 m?

<IMAGE>

514

views

Textbook Question

A batter hits a fly ball which leaves the bat 0.90 m above the ground at an angle of 64° with an initial speed of 28 m/s heading toward centerfield. Ignore air resistance. The ball is caught by the centerfielder who, starting at a distance of 105 m from home plate just as the ball was hit, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed.

347

views

Textbook Question

Two masses are connected by a string as shown in Fig. 8–35. Mass m_A = 3.5 kg rests on a frictionless inclined plane, while m_B = 5.0 kg is initially held at a height of h = 0.75 m above the floor. If the masses were initially at rest, use the kinematic equations (Eqs. 2–12) to find their velocity just before m_B hits the floor.

558

views

Show more practices

Download the Mobile app

Privacy

Do not sell my personal information

Accessibility

Patent notice

Sitemap