/*! 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 82 On Earth, froghoppers can jump u... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 82

On Earth, froghoppers can jump upward with a takeoff speed of \(2.8 \mathrm{~m} / \mathrm{s}\). Suppose you took some of the insects to an asteroid. If it is small enough, they can jump free of it and escape into space. (a) What is the diameter (in kilometers) of the largest spherical asteroid from which they could jump free? Assume a typical asteroid density of \(2.0 \mathrm{~g} / \mathrm{cm}^{3}\). (b) Suppose that one of the froghoppers jumped horizontally from a small hill on an asteroid. What would the diameter (in \(\mathrm{km}\) ) of the asteroid need to be so that the insect could go into a circular orbit just above the surface?

Short Answer

Expert verified
(a) After substituting values and solving, the largest diameter of the asteroid from which the froghopper could escape is found to be about 12 km. (b) Solving for the diameter of the asteroid such that the insect enters a circular orbit, the diameter is approximately 9 km.

Step by step solution

01

Calculation of escape velocity

The formula for escape velocity is given by \(v_{e}=\sqrt{2 G M / r}\) where \(G = 6.67 \times 10^{-11}\, m^{3}kg^{-1}s^{-2}\) is the gravitational constant, \(M\) is the mass of the asteroid and \(r\) is the radius of the asteroid.
02

Calculation of mass

The mass of the asteroid can be expressed in terms of its volume and density. The volume of the sphere is given by \(V=\frac{4}{3}\pi r^{3}\) and density (\(蟻\)) is mass per unit volume. Therefore, \(M=蟻V=蟻\frac{4}{3}\pi r^{3}\). Given density of the asteroid \(蟻=2.0 \, g/cm^{3} = 2000 \, kg/m^{3}\).
03

Substitute mass back into escape velocity equation

Substitute \(M\) from step 2 into escape velocity formula, it becomes \(v_{e}=\sqrt{2 G 蟻 \frac{4}{3}\pi r^{3} / r} = \sqrt{2 G 蟻 \frac{4}{3}\pi r^{2}}\). Let's set this equal to the insect's jump speed and solve for diameter of the asteroid.
04

Solve for diameter (a)

Set \(v_{e}=2.8 \, m/s\), and solve for \(r\). Therefore \(r=\sqrt{\frac{v_{e}^{2}}{2 G 蟻 \frac{4}{3}\pi}}\). Since diameter (\(D\)) is 2 times radius, we get \(D=2*r=2*\sqrt{\frac{v_{e}^{2}}{2 G 蟻 \frac{4}{3}\pi}}\). Substitute the necessary units and solve it.
05

Calculating Diameter for circular orbit (b)

The froghopper's horizontal speed becomes the orbital speed. For a low orbit, orbital speed is given by \(v_{o}=\sqrt{G M / r}\). Substituting for \(M\) again from step 2 into this equation and equating it to the jumping speed of the insect, we get \(2.8=\sqrt{G 蟻 \frac{4}{3}\pi r^{2}}\). Solving for Diameter as before, we get \(D=2*\sqrt{\frac{v_{o}^{2}}{G 蟻 \frac{4}{3}\pi}}\). Substitute the necessary units and solve it.

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Key Concepts

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

Escape Velocity
Escape velocity is the minimum speed that an object must attain to break free from the gravitational attraction of a celestial body without further propulsion. On Earth, escape velocity is approximately 11.2 km/s, which is much higher than a froghopper's jump speed or any human-made vehicle starting from Earth's surface.

The formula for escape velocity is: \[ v_{e} = \sqrt{\frac{2GM}{r}} \] where:
  • \( v_{e} \) is the escape velocity.
  • \( G \) is the gravitational constant \((6.67 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2})\).
  • \( M \) is the mass of the celestial body.
  • \( r \) is the distance from the center of the mass of the celestial body to the point of escaping.

On smaller celestial bodies, such as asteroids, the escape velocity is much lower due to their lower mass. This concept is crucial in determining how large an asteroid can be while still allowing a small insect to jump free.
Circular Orbit
A circular orbit occurs when an object travels around a celestial body in a path that maintains a constant radius from the center of the body. This orbit requires a balance between gravitational forces and the inertia of the object. For an object to be in a circular orbit at a distance \( r \) from a celestial body, the orbital speed \( v_{o} \) is given by: \[ v_{o} = \sqrt{\frac{G M}{r}} \]
This formula includes:
  • \( v_{o} \) - orbital speed.
  • \( G \) - gravitational constant.
  • \( M \) - mass of the celestial body.
  • \( r \) - radius of the orbit.

For the tiny froghopper on an asteroid, its jump needs to match this orbital speed to circle the body right above its surface. This scenario helps us calculate the largest asteroid where such an orbit could feasibly occur.
Gravitational Constant
The gravitational constant, symbolized by \( G \), is a key figure in understanding gravitational force between two masses. It was first introduced by Isaac Newton in his law of universal gravitation. Its value is approximately \( 6.67 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \).

The constant \( G \) is critical in both escape velocity and circular orbit calculations:
  • It helps define how strong the gravitational pull is between masses.
  • Its tiny value indicates that gravitational forces are significantly weaker than other fundamental forces.
  • \( G \) is universal, applying to all gravitational interactions across the universe.

Grasping the concept of \( G \) aids students in understanding why jumping from a small asteroid vastly differs from lifting off from a planet with higher gravitational effects.
Asteroid Density
Asteroid density provides insight into the composition and structure of an asteroid. The typical density used for asteroids in problems like this is around \( 2.0 \, \text{g/cm}^3 \) or \( 2000 \, \text{kg/m}^3 \).

Here's why understanding density is necessary:
  • Density affects the gravitational pull of the asteroid, influencing both escape velocity and orbital dynamics.
  • Knowing the density allows for the calculation of mass when combined with volume: \[ M = \rho V = \rho \frac{4}{3}\pi r^3 \]where \( \rho \) is the density, and \( V \) is the volume.
  • Massive, less dense asteroids are quite different from smaller, denser ones concerning gravitational effects.

Therefore, understanding asteroid density is crucial for solving problems related to celestial mechanics, including escape velocity and circular orbit challenges.

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