/*! 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 69 A \(30 \mathrm{~kg}\) satellite ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 69

A \(30 \mathrm{~kg}\) satellite has a circular orbit with a period of \(2.0 \mathrm{~h}\) and a radius of \(9.0 \times 10^{6} \mathrm{~m}\) around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is \(6.0 \mathrm{~m} / \mathrm{s}^{2}\), what is the radius of the planet?

Short Answer

Expert verified
Calculate planet mass with orbital data, then rearrange gravitational equation to find planet's radius.

Step by step solution

01

Identify Known Variables and Equation

The problem provides us with:- Satellite mass, which can be neglected for the gravitational force acting by the planet.- Orbital radius \( r = 9.0 \times 10^6 \: \mathrm{m} \).- Orbital period \( T = 2.0 \: \mathrm{h} = 7200 \: \mathrm{s} \) (converted from hours to seconds).We need the gravitational formula for satellites: \[ T^2 = \frac{4\pi^2r^3}{GM} \]where \( M \) is the planet's mass and \( G = 6.67 \times 10^{-11} \: \mathrm{Nm}^2/\mathrm{kg}^2 \), the gravitational constant.Also, on the surface, the gravitational acceleration is given by:\[ g = \frac{GM}{R^2} \] where \( R \) is the radius of the planet.
02

Solve for Planetary Mass (M)

First, rearrange the satellite orbital formula to solve for the mass of the planet:\[ M = \frac{4\pi^2r^3}{GT^2} \]Substitute the known values:\[ M = \frac{4\times (3.1415)^2 \times (9.0 \times 10^6)^3}{6.67 \times 10^{-11} \times (7200)^2} \]Calculate \( M \) using these substitutions to find the planet's mass.
03

Calculate Gravitational Constant at Surface

Using the known gravitational acceleration at the planet's surface \( g = 6.0 \, \mathrm{m/s^2} \), rearrange the equation for acceleration:\[ g = \frac{GM}{R^2} \]Solve for \( R^2 \):\[ R^2 = \frac{GM}{g} \]Insert the mass calculated from the previous step,
04

Solve for Radius (R) of the Planet

Use the result for \( R^2 \) to find \( R \):\[ R = \sqrt{\frac{GM}{g}} \]By inserting the value of \( G \), the mass \( M \) from Step 2, and \( g = 6.0 \, \mathrm{m/s^2} \), solve to find the radius \( R \) of the planet.

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 91影视!

Key Concepts

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

Satellite Orbit
A satellite's orbit refers to the path it takes as it revolves around a planet. This path can typically be circular or elliptical. In our scenario, the satellite maintains a circular orbit. Circular orbits are simpler to understand because the distance between the satellite and the planet's center, known as the orbital radius, remains constant. Key aspects of satellite orbits include:
  • Orbital Radius: Distance from the center of the planet to the satellite.
  • Orbital Period: Time it takes to complete one orbit.
  • Mass of the Satellite: Though often negligible when measuring gravitational force compared to the mass of a planet.
Here, understanding the satellite's orbit helped solve for the planet's mass, crucial for finding the planet's radius. The orbital period and radius give important clues about the forces acting on the satellite.
Gravitational Acceleration
Gravitational acceleration on a planet's surface is the acceleration due to the force of gravity acting on objects at this location. It's typically measured in meters per second squared (\( m/s^2 \)). For our particular planet, it's given as \(6.0 \: \mathrm{m/s}^2 \).Gravitational acceleration is influenced by:
  • Planetary Mass: Larger mass results in greater gravitational pull.
  • Radius of the Planet: Gravitational force decreases with distance, so a larger planet radius results in lower surface gravity.
In solving our exercise, this given gravitational acceleration enables us to calculate the planet's radius using the gravitational formula. It strongly dictates how all objects, including satellites, react to the planet's gravity.
Orbital Mechanics
Orbital mechanics involves studying the forces that govern the motion of objects in space, particularly their orbits around larger bodies like planets. The primary force in play here is gravity. Important principles include:
  • Kepler鈥檚 Laws: Describe the motion of planets but can be generalized for satellites.
  • Gravitational Force Equation: Differently adapted whether calculating for the planet from satellite data or general satellite motion.
  • Energy Conservation: Total orbital energy combines kinetic and potential energy, remaining mostly constant despite position changes.
The planet's mass in particular, calculated using the satellite's observed orbital properties, showcases how mastery of these principles is key in solving related problems.
Planetary Mass
The mass of a planet is a crucial element that influences several characteristics including gravitational acceleration and the ability to sustain orbits. In our calculation, we derived the planetary mass by considering the satellite鈥檚 circular orbit. Key factors explored include:
  • Role in Gravity: Larger masses exude a stronger gravitational influence.
  • Derivation from Orbit: Using the formula \( T^2 = \frac{4\pi^2r^3}{GM} \), we rearrange it to derive mass \( M \), based on known orbital period and radius.
  • Impact on Surface Conditions: The mass, combined with the radius, determines the gravitational acceleration felt on the surface.
Our practical derivation of planetary mass from orbital characteristics clearly highlights the intricacy of such gravitational calculations.
Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation is fundamental to understanding gravitational interactions between two masses. This law states that every mass attracts every other mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.Expressed as:\[ F = G \frac{m_1m_2}{r^2} \]where:
  • \(F\) is the gravitational force.
  • \(G\) is the gravitational constant \(6.67 \times 10^{-11} \, \mathrm{Nm^2/kg^2}\).
  • \(m_1\) and \(m_2\) are the masses involved.
  • \(r\) is the distance between their centers.
For our exercise, Newton鈥檚 law was crucial when relating the gravitational force to the planet's mass and the gravitational acceleration at its surface, showing how this consistent force shapes planetary and satellite interactions.

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

(a) What is the escape speed on a spherical asteroid whose radius is \(700 \mathrm{~km}\) and whose gravitational acceleration at the surface is \(4.5 \mathrm{~m} / \mathrm{s}^{2} ?\) (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of \(1000 \mathrm{~m} / \mathrm{s}\) ? (c) With what speed will an object hit the asteroid if it is dropped from \(1000 \mathrm{~km}\) above the surface?

Three dimensions. Three point particles are fixed in place in an \(x y z\) coordinate system. Particle \(A\), at the origin, has mass \(m_{A}\). Particle \(B\), at \(x y z\) coordinates \((2.00 d, 1.00 d, 2.00 d)\), has mass \(2.00 \mathrm{~m}_{A}\), and particle \(C\), at coordinates \((-1.00 d, 2.00 d,-3.00 d)\), has mass \(3.00 m_{A} \cdot A\) fourth particle \(D\), with mass \(4.00 \mathrm{~m}_{A}\), is to be placed near the other particles. In terms of distance \(d\), at what (a) \(x\), (b) \(y\), and (c) z coordinate should \(D\) be placed so that the net gravitational force on \(A\) from \(B, C\),and \(D\) is zero?

Assume a planet is a uniform sphere of radius \(R\) that (somehow) has a narrow radial tunnel through its center (Fig. 13-7). Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let \(F_{R}\) be the magnitude of the gravitational force on the apple when it is located at the planet's surface. How far from the surface is there a point where the magnitude is \(\frac{1}{3} F_{R}\) if we move the apple (a) away from the planet and (b) into the tunnel?

Zero, a hypothetical planet, has a mass of \(5.0 \times 10^{23} \mathrm{~kg}\), a radius of \(3.5 \times 10^{6} \mathrm{~m}\), and no atmosphere. A \(10 \mathrm{~kg}\) space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of \(5.0 \times 10^{7} \mathrm{~J}\), what will be its kinetic energy when it is \(4.0 \times 10^{6} \mathrm{~m}\) from the center of Zero? (b) If the probe is to achieve a maximum distance of \(8.0 \times 10^{6} \mathrm{~m}\) from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

At what altitude above Earth's surface would the gravitational acceleration be \(2.0 \mathrm{~m} / \mathrm{s}^{2} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Measurements

Read Explanation

Scientific Method Physics

Read Explanation

Nuclear Physics

Read Explanation

Physics of Motion

Read Explanation
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.