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Chapter 6: Problem 27

Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a 1.30 \(\mathrm{kg}\) wrench from 5.00 \(\mathrm{m}\) above the ground and measure that it hits the ground 0.811 s later. You also do enough surveying to determine that the circumference of the planet is \(62,400 \mathrm{km}\) . (a) What is the mass of the planet, in kilograms? (b) Express the planet's mass in terms of the earth's mass.

Short Answer

Expert verified
The planet's mass is approximately \(1.013 \times 10^{25} \text{ kg}\), or \(1.697\) times Earth's mass.

Step by step solution

01

Calculate Gravitational Acceleration

Use the formula for the distance fallen under gravity: \[ d = \frac{1}{2}gt^2 \]where \(d = 5.00 \text{ m}\) and \(t = 0.811 \text{ s}\). Solve for \(g\):\[ 5 = \frac{1}{2}g(0.811)^2 \]Now solve for \(g\):\[ g = \frac{2 \times 5}{(0.811)^2} \approx 15.19 \text{ m/s}^2 \]
02

Calculate the Planet's Radius

The circumference of the planet is given as \(62,400 \text{ km} = 62,400,000 \text{ m}\). The formula for circumference is: \[ C = 2\pi R \]Rearrange to solve for \(R\):\[ R = \frac{62,400,000}{2\pi} \approx 9,936,249 \text{ m} \approx 9.936 \times 10^6 \text{ m} \]
03

Use Gravitational Formula to Find Planet's Mass

The gravitational force equation is: \[ g = \frac{GM}{R^2} \]where \(G = 6.674 \times 10^{-11} \text{ m}^3\text{ kg}^{-1}\text{ s}^{-2}\), \(g = 15.19 \text{ m/s}^2\), and \(R = 9,936,249 \text{ m}\).Solve for \(M\):\[ M = \frac{gR^2}{G} \]Plugging in values:\[ M = \frac{15.19 \times (9,936,249)^2}{6.674 \times 10^{-11}} \approx 1.013 \times 10^{25} \text{ kg} \]
04

Express the Planet's Mass in Terms of Earth's Mass

Earth's mass is approximately \(5.972 \times 10^{24} \text{ kg}\). To express the planet's mass in terms of Earth's mass:\[ M_{\text{planet}} = \frac{1.013 \times 10^{25}}{5.972 \times 10^{24}} \approx 1.697 \]So, the planet's mass is about 1.697 times Earth's mass.

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Key Concepts

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

gravity formula
When we drop an object and it falls under the influence of gravity, its motion can be described by a well-known equation. The distance an object falls is given by the formula \(d = \frac{1}{2}gt^2\). Here, \(d\) is the distance fallen, \(g\) is the acceleration due to gravity, and \(t\) is the time it takes to fall. By manipulating this formula, you can calculate the gravitational acceleration on any planet. This formula is a cornerstone of many physics problems.
  • It helps determine how fast an object will accelerate towards the ground.
  • It's crucial in physics problem solving, especially in scenarios involving falling objects.
Understanding the gravity formula is essential for calculations involving distances, times, and the forces at play when an object falls.
planetary mass calculation
To determine the mass of a planet, we use the gravitational force equation: \(g = \frac{GM}{R^2}\). This equation relates the acceleration due to gravity \(g\), the gravitational constant \(G\), the planet's mass \(M\), and the radius \(R\) of the planet.
  • Given the gravitational acceleration, you can solve for the planet's mass if the radius is known.
  • The formula requires simple algebraic manipulation to isolate \(M\).
By substituting the known values, you can accurately compute the mass of the planet, demonstrating its importance in planetary mass calculation tasks in physics.
physics problem solving
Physics problems often involve multiple steps and require critical thinking and a methodical approach. In this scenario, you identify and apply different physical laws and formulas successively:

- Start by determining the gravitational force acting on an object. - Use known data to find unknown variables like the planet's gravitational acceleration.

  • Problem-solving involves breaking down complex problems into manageable steps.
  • Each step usually results in a piece of the puzzle, leading to a complete solution when integrated.
With practice, this approach helps build confidence in tackling complex physics exercises, making concepts like force and motion more approachable.
space exploration education
Understanding the physics of different planets is crucial for space exploration. It requires applying principles of gravity and motion consistently. Learning about gravitational acceleration and mass calculations aids in several space-related inquiries:
  • Determining the habitability of other planets.
  • Planning landing and takeoff procedures for spacecraft.
Educating about these concepts emphasizes their relevance beyond classroom settings, supporting the broader goal of space exploration education. These principles not only help in the understanding of known celestial bodies but also prepare us for exploring new ones. Science enthusiasts can thus appreciate the practical applications of physics in real-world space missions.

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Most popular questions from this chapter

\(\bullet\) The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 \(\mathrm{m}\) . Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 \(\mathrm{s} )\) . (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 \(\mathrm{N}\) at the weight- guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

Artificial gravity. One way to create artificial gravity in a space station is to spin it. If a cylindrical space station 275 \(\mathrm{m}\) in diameter is to spin about its central axis, at how many revolutions per minute (rpm) must it turn so that the outermost points have an acceleration equal to \(g\) ?

What is the period of revolution of a satellite with mass \(m\) that orbits the earth in a circular path of radius 7880 \(\mathrm{km}\) (about 1500 \(\mathrm{km}\) above the surface of the earth \() ?\)

A flat (unbanked) curve on a highway has a radius of 220 \(\mathrm{m}\) .A car rounds the curve at a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) . (a) Make a free body diagram of the car as it rounds this curve. (b) What is the minimum coefficient of friction that will prevent sliding?

A 2150 kg satellite used in a cellular telephone network is in a circular orbit at a height of 780 \(\mathrm{km}\) above the surface of the earth. What is the gravitational force on the satellite? What fraction is this force of the satellite's weight at the surface of the earth?
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