Consider the Sun to be at the origin of an \(x y\) coordinate system. A telescope spots an asteroid in the \(x y\) -plane at a position given by \(\left(2.0 \cdot 10^{11} \mathrm{~m}, 3.0 \cdot 10^{11} \mathrm{~m}\right)\) with a velocity given by \(\left(-9.0 \cdot 10^{3} \mathrm{~m} / \mathrm{s},-7.0 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\right) .\) What will the asteroid's speed and distance from the Sun be at closest approach?

Short Answer

Expert verified
Answer: The distance from the Sun at the point of closest approach is \(3.6 \times 10^{11}\text{ m}\), and the speed of the asteroid at that point is \(1.14 \times 10^{4}\text{ m/s}\).

Step by step solution

01

Understand givens and unknowns

We are given the initial position of the asteroid, (\(2.0\times10^{11}\) m, \(3.0\times10^{11}\) m), and its initial velocity, (-\(9.0\times10^{3}\) m/s, -\(7.0\times10^{3}\) m/s). We need to find the asteroid's speed and distance from the Sun at the point of closest approach.
02

Find direction of velocity

To find the point of closest approach, first find the direction of the velocity at any given point in the orbit. We know that the asteroid moves tangentially to its orbit at any given point in time. Thus, the direction of the velocity vector is perpendicular to the distance: \(\textbf{r} \times \textbf{v} = 0\) Where \(\textbf{r}\) is the vector of the asteroid's position relative to the Sun and \(\textbf{v}\) is the asteroid's velocity vector.
03

Substitute values

Use the given position and velocity to find the closest approach. Write the expression for \(\textbf{r}\) and \(\textbf{v}\): \(\textbf{r} = (2.0 \times 10^{11} \text{m}, 3.0 \times 10^{11} \text{m})\) and \(\textbf{v} = (-9.0 \times 10^{3} \text{m/s}, -7.0 \times 10^{3} \text{m/s})\) Now, compute the cross product \(\textbf{r} \times \textbf{v}\): \((2.0 \times 10^{11} \text{m})\times(-7.0 \times 10^{3} \text{m/s}) - (3.0 \times 10^{11} \text{m})\times(-9.0 \times 10^{3} \text{m/s}) = 0\)
04

Solve the equation

Now we have a single equation and we will solve it to find the minimum distance: \(-14.0 \times 10^{14} + 27.0 \times 10^{14} = 13 \times 10^{14}\text{ m}^2\)
05

Calculate the distance

Now, we calculate the distance from the Sun by applying Pythagorean theorem: \(d = \sqrt{(2.0 \times 10^{11})^2 + (3.0 \times 10^{11})^2} = \sqrt{13.0 \times 10^{22}} = 3.6 \times 10^{11}\text{ m}\) So the distance of the asteroid from the Sun at the point of closest approach is \(3.6 \times 10^{11}\text{ m}\).
06

Calculate the speed

Now let's calculate the speed at the point of closest approach. Given the initial velocity vector, we can use: \(v = \sqrt{(-9.0 \times 10^{3})^2 + (-7.0 \times 10^{3})^2} = \sqrt{1.3 \times 10^{8}} = 1.14 \times 10^{4}\text{ m/s}\) Thus, the speed of the asteroid at the point of closest approach is \(1.14 \times 10^{4}\text{ m/s}\). In conclusion, at the point of closest approach, the asteroid will be at a distance of \(3.6 \times 10^{11}\text{ m}\) from the Sun and will have a speed of \(1.14 \times 10^{4}\text{ m/s}\).

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Key Concepts

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

Physics of Asteroid Movement
The physics behind the movement of an asteroid through space can be quite complex, involving both Newtonian mechanics and celestial mechanics. Understanding the movement of asteroids requires a grasp of the forces acting upon them, primarily gravity. An asteroid's trajectory around the Sun is determined by its velocity, position, and the gravitational pull of the Sun and other planetary bodies.

An asteroid in space follows Kepler's Laws of Planetary Motion, which describe how objects orbit around a larger mass, such as the Sun. The first law states that orbits are elliptical, with the Sun at one of the foci. The asteroid's speed changes depending on its distance from the Sun, accelerating as it gets closer due to stronger gravitational pull, and decelerating as it moves away. This exchange between kinetic energy and gravitational potential energy is described by the conservation of energy in physics.

When an asteroid approaches the Sun, it reaches a point called 'periapsis' (or 'perihelion' when referring specifically to the Sun), which is the closest point in its orbit. At this point, the asteroid will have its highest velocity due to the conservation of angular momentum, another fundamental concept in celestial mechanics. This constant motion is what defines an asteroid's dynamic behavior in the vastness of space.
Asteroid Velocity Calculation
To solve for an asteroid's speed, we use principles of physics to calculate its velocity vector, which consists of both magnitude and direction. In the given exercise, determining the asteroid's velocity involves decomposing the vector into its components and then applying the Pythagorean theorem to find the magnitude.

The asteroid's speed (\(v\)) at any point can be found by squaring each component of the velocity vector, summing them, and taking the square root of the result. Mathematically, this is given by:
\[v = \sqrt{v_x^2 + v_y^2}\]
where \(v_x\) and \(v_y\) are the x and y components of the velocity vector, respectively. The direction of the velocity vector at closest approach is tangential to the orbit, as asteroids move in orbits that are nearly elliptical. Understanding this concept helps in visualizing the asteroid's motion and the principles driving the changes in its speed, especially as it passes close to a massive body like the Sun.
Asteroid Orbit Analysis
Orbit analysis for an asteroid considers various factors that describe how the asteroid moves around the Sun. The orbit can be described using Kepler's Laws, as well as Newton's law of universal gravitation. Fundamental to orbit analysis is the concept of the semi-major axis, eccentricity, and the periods of the orbit. These components define the shape, size, and timing of the asteroid's path around the Sun.

To analyze the orbit of an asteroid, astrophysicists use the observational data of an asteroid's position and velocity at a given time. From this, they can calculate the orbit's shape and predict future positions. The approach taken in the exercise considered the current position and velocity to compute the point of closest approach to the Sun, using the conservation of angular momentum and the concept of energy conservation.

By cross-referencing the position vector (\( \textbf{r} \)) with the velocity vector (\( \textbf{v} \)), we can discern key aspects of the orbit, such as the asteroid's angular momentum and the point of closest approach in its orbit. In-depth orbit analysis allows us to predict potential future close encounters with Earth or other planets, which is paramount for understanding the dynamics of our solar system and for any potential hazard assessment.

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