/*! 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 74 The mysterious visitor that appe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The mysterious visitor that appears in the enchanting story. The Little Prince was said to come from a planet that "was scarcely any larger than a house!" Assume that the mass per unit volume of the planet is about that of Earth and that the planet does not appreciably spin. Approximate (a) the free-fall acceleration on the planet's surface and (b) the escape speed from the planet.

Short Answer

Expert verified
The free-fall acceleration is about the same as Earth's; the escape speed is much lower.

Step by step solution

01

Estimating the Planet’s Radius

First, we estimate the size of the planet relative to Earth. If the planet is the size of a house, assume its radius is approximately 10 meters, as a typical house might be around 20 meters in diameter.
02

Calculate the Planet's Mass

Using the assumption that the planet's mass density is similar to Earth's, calculate the mass of the planet using the density formula: \( \rho = \frac{m}{V} \), where \( \rho \approx 5500 \, \text{kg/m}^3 \). The volume of the planet is \( V = \frac{4}{3}\pi r^3 \). Substitute \( r = 10 \, \text{m} \) to find the mass \( m \).
03

Calculate Free-Fall Acceleration on the Surface

Use the gravitational formula for acceleration, \( g = \frac{Gm}{r^2} \). Substitute the mass \( m \) from Step 2 and radius \( r = 10 \, \text{m} \) to calculate the acceleration \( g \), with \( G \approx 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \).
04

Calculate Escape Speed from the Planet

Escape velocity \( v_{e} \) is given by the formula \( v_{e} = \sqrt{\frac{2Gm}{r}} \). Use the mass from Step 2 and radius \( r = 10 \, \text{m} \) to calculate \( v_{e} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Gravitational acceleration
One of the key concepts of planetary physics is the understanding of gravitational acceleration. In simple terms, it refers to the acceleration that an object experiences due to the gravitational pull of a planet. The strength of this acceleration depends on two main factors: the mass of the planet and the distance from the center of the planet to the object, often known as the radius.

The formula to calculate gravitational acceleration is given by \( g = \frac{Gm}{r^2} \),where:
  • \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \),
  • \( m \) is the mass of the planet,
  • \( r \) is the distance from the center of the planet to the point where the gravitational acceleration is measured (radius).
In our little story of The Little Prince, his planet is assumed to be scarcely larger than a house. If we treat the planet as a perfect sphere with a radius of around 10 meters, largely due to the average house size, the gravitational pull would be very different than on Earth.

Understanding gravitational acceleration helps us measure how "heavy" or "light" an object would feel on different celestial objects.
Escape velocity
Ever wondered how fast a spaceship would need to launch to break free from a planet's gravitational grip? That crucial speed is what scientists refer to as escape velocity. It is the minimum speed needed for an object to escape from the gravitational influence of a planet without further propulsion.

Escape velocity can be calculated with the formula:\( v_{e} = \sqrt{\frac{2Gm}{r}} \),where:
  • \( G \) is the gravitational constant,
  • \( m \) is the mass of the planet,
  • \( r \) is the radius, or distance from the planet's center.
To imagine this in the context of The Little Prince's tiny planet, where the radius is significantly smaller than Earth, the escape velocity would also be far less.

This concept is crucial in space travels and missions. It helps engineers design necessary technology to ensure spacecraft can successfully leave the planetary surface and navigate the cosmos.
Mass density
The notion of mass density is all about how much mass exists within a given volume of space. When reflecting on The Little Prince's planet, it's interesting to understand mass density, as it's often compared to Earth's in similar exercises.

The formula for mass density is straightforward:\( \rho = \frac{m}{V} \),where:
  • \( \rho \) is density,
  • \( m \) is mass,
  • \( V \) is volume.
Typically, Earth's mass density is approximately \( 5500 \, \text{kg/m}^3 \).Assuming a tiny celestial body has similar composition can simplify our calculations and give a theoretical benchmark.

Recognizing mass density allows scientists to draw conclusions about the inner properties of planetary bodies, which is essential for understanding their formation and evolution.
Radius estimation
In planetary physics, radius estimation is critical, especially when envisioning the scale of unfamiliar alien worlds. A good foundational exercise involves estimating the radius of a planet based on its visible size or contextual clues.

Given The Little Prince's planet is said to be as large as a house, we assume the diameter might be around 20 meters, translating to a 10-meter radius. This radius is incorporated into all other calculations, such as gravitational acceleration and escape velocity.

Effective radius estimation is essential when calculating crucial planetary features, as it directly affects the comprehended gravitational dynamics and escape speeds. Having the right radius ensures that all predictions about an object's behavior near such a planet are accurate and reliable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two point particles are fixed on an \(x\) axis separated by distance \(d .\) Particle \(A\) has mass \(m_{A}\) and particle \(B\) has mass \(3.00 m_{A}\). A third particle \(C,\) of mass \(75.0 m_{A},\) is to be placed on the \(x\) axis and near particles \(A\) and \(B\). In terms of distance \(d,\) at what \(x\) coordinate should \(C\) be placed so that the net gravitational force on particle \(A\) from particles \(B\) and \(C\) is zero?

Several planets (Jupiter, Saturn, Uranus) are encircled by rings, perhaps composed of material that failed to form a satellite. In addition, many galaxies contain ring-like structures. Consider a homogeneous thin ring of mass \(M\) and outer radius \(R\) (Fig. 13-52). (a) What gravitational attraction does it exert on a particle of mass \(m\) located on the ring's central axis a distance \(x\) from the ring center? (b) Suppose the particle falls from rest as a result of the attraction of the ring of matter. What is the speed with which it passes through the center of the ring?

A comet that was seen in April 574 by Chinese astronomers on a day known by them as the Woo Woo day was spotted again in May \(1994 .\) Assume the time between observations is the period of the Woo Woo day comet and its eccentricity is \(0.9932 .\) What are (a) the semimajor axis of the comet's orbit and (b) its greatest distance from the Sun in terms of the mean orbital radius \(R_{P}\) of Pluto?

Two Earth satellites, \(A\) and \(B,\) each of mass \(m,\) are to be launched into circular orbits about Earth's center. Satellite \(A\) is to orbit at an altitude of \(6370 \mathrm{~km} .\) Satellite \(B\) is to orbit at an altitude of \(19110 \mathrm{~km} .\) The radius of Earth \(R_{E}\) is \(6370 \mathrm{~km} .\) (a) \(\mathrm{What}\) is the ratio of the potential energy of satellite \(B\) to that of satellite \(A,\) in orbit? (b) What is the ratio of the kinetic energy of satellite \(B\) to that of satellite \(A,\) in orbit? (c) Which satellite has the greater total energy if each has a mass of \(14.6 \mathrm{~kg}\) ? (d) By how much?

We watch two identical astronomical bodies \(A\) and \(B\), each of mass \(m,\) fall toward each other from rest because of the gravitational force on each from the other. Their initial center-to-center separation is \(R_{i}\). Assume that we are in an inertial reference frame that is stationary with respect to the center of mass of this twobody system. Use the principle of conservation of mechanical energy \(\left(K_{f}+U_{f}=K_{i}+U_{i}\right)\) to find the following when the centerto-center separation is \(0.5 R_{i}:\) (a) the total kinetic energy of the system, (b) the kinetic energy of each body, (c) the speed of each body relative to us, and (d) the speed of body \(B\) relative to body \(A\). Next assume that we are in a reference frame attached to body \(A\) (we ride on the body). Now we see body \(B\) fall from rest toward us. From this reference frame, again use \(K_{f}+U_{f}=K_{i}+U_{i}\) to find the following when the center-to-center separation is \(0.5 R_{i}\) : (e) the kinetic energy of body \(B\) and (f) the speed of body \(B\) relative to body \(A\). (g) Why are the answers to (d) and (f) different? Which answer is correct?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.