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Chapter 13: Problem 19

A planet orbiting a distant star has radius \(3.24 \times 10^{6} \mathrm{~m}\). The escape speed for an object launched from this planet's surface is \(7.65 \times 10^{3} \mathrm{~m} / \mathrm{s}\). What is the acceleration due to gravity at the surface of the planet?

Short Answer

Expert verified
The acceleration due to gravity on the surface of the planet is approximately \(7.20 \, m/s^2\).

Step by step solution

01

Apply the escape speed equation

The escape speed formula is given by \(\sqrt{2gr}\), where \(g\) represents the acceleration due to gravity and \(r\) the planet's radius. We can solve this equation for \(g\): \( g = v^2 / (2r) \). Although the acceleration due to gravity is what we ultimately want to find, it is important to note that this equation assumes that the acceleration due to gravity is the same throughout the escape path. As such, the answer provided by this formula is only approximate. In our case, the escape speed (\(v\)) is \(7.65 \times 10^{3} m/s\) and the planet's radius (\(r\)) is \(3.24 \times 10^{6} m\).
02

Substitute the values

Substitute the given values into the equation from step 1: \( g = (7.65 \times 10^{3} m/s)^2 / (2 \times 3.24 \times 10^{6} m) \)
03

Solve for g

After substituting the values, solving will yield the acceleration due to gravity on the surface of the planet.

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Key Concepts

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

Escape Speed
Escape speed, often known as escape velocity, is the minimum speed an object must reach to break free from the gravitational pull of a planet without any additional propulsion. This concept is crucial in space exploration and understanding planetary atmospheres.
The formula for escape speed is derived from the law of conservation of energy. It states that the kinetic energy of an object must be equal to the gravitational potential energy needed to escape the planet's gravity. Mathematically, this translates to:
  • Formula: \( v_e = \sqrt{2gr} \)
  • Where \( v_e \) is the escape speed, \( g \) is the acceleration due to gravity, and \( r \) is the radius of the planet.
Understanding escape speed is important for missions that require breaking away from a planet's gravitational influence, such as launching spacecraft from Earth. It helps determine the velocity needed to send a spacecraft to outer space or another planet.
Gravitational Formulas
Gravitational formulas provide ways to calculate the influence of gravity in various contexts, from understanding planetary orbits to calculating weight on different planets. In the problem at hand, we focus on the relationship between escape speed and gravitational acceleration.
In particular, the equation \( v_e = \sqrt{2gr} \) serves a dual purpose. Firstly, it calculates the escape speed for a planet. Secondly, by rearranging terms, it helps us find the acceleration due to gravity if the escape speed and radius are known:
  • Acceleration due to gravity, \( g = \frac{v^2}{2r} \), where \( v \) is the escape speed and \( r \) is the radius.
This formula is valuable not only on Earth but for any celestial body. It provides a way to calculate how strong the gravitational pull is at the surface given the known or calculated escape speed.
Planetary Radius
The planetary radius is a crucial factor in calculating both escape speed and gravitational force on a planet's surface. It represents the distance from the planet's core to its surface.
Larger planets, given their size, often exert a stronger gravitational pull. However, when talking about escape speed and acceleration, the radius is equally influential because it impacts both the gravitational potential energy and the geometry of escape paths.
In our problem, the planet's radius is provided as \(3.24 \times 10^6 \text{ m}\). This radius is plugged into the escape speed formula to determine the gravitational acceleration. Understanding the concept of planetary radius assists in grasping why different planets have different gravitational strengths and how they influence matters like atmosphere retention and satellite orbits.

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Most popular questions from this chapter

Consider the ringshaped object in Fig. \(\mathrm{E} 13.39 .\) A particle with mass \(m\) is placed a distance \(x\) from the center of the ring. along the line through the center of the ring and perpendicular to its plane. (a) Calculate the gravitational potential energy \(U\) of this system. Take the potential energy to be zero when the two objects are far apart. (b) Show that your answer to part (a) reduces to the expected result when \(x\) is much larger than the radius \(a\) of the ring. (c) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the force on the particle (see Section 7.4 ). (d) Show that your answer to part (c) reduces to the expected result when \(x\) is much larger than \(a\). (e) What are the values of \(U\) and \(F_{x}\) when \(x=0 ?\) Explain why these results make sense.

Define the gravitational field \(\vec{g}\) at some point to be equal to the gravitational force \(F_{e}\) on a small object placed at that point divided by the mass \(m\) of the object, so \(\bar{g}=\vec{F}_{e} / m .\) A spherical shell has mass \(M\) and radius \(R .\) What is the magnitude of the gravitational field at the following distances from the center of the shell: (a) \(rR ?\)

The planet Uranus has a radius of \(25,360 \mathrm{~km}\) and a surface acceleration due to gravity of \(9.0 \mathrm{~m} / \mathrm{s}^{2}\) at its poles. Its moon Miranda (discovered by Kuiper in 1948 ) is in a circular orbit about Uranus at an altitude of \(104,000 \mathrm{~km}\) above the planet's surface. Miranda has a mass of \(6.6 \times 10^{19} \mathrm{~kg}\) and a radius of \(236 \mathrm{~km}\). (a) Calculate the mass of Uranus from the given data. (b) Calculate the magnitude of Miranda's acceleration due to its orbital motion about Uranus. (c) Calculate the acceleration due to Miranda's gravity at the surface of Miranda. (d) Do the answers to parts (b) and (c) mean that an object released \(1 \mathrm{~m}\) above Miranda's surface on the side toward Uranus will fall up relative to Miranda? Explain.

An experiment is performed in deep space with two uniform spheres, one with mass \(50.0 \mathrm{~kg}\) and the other with mass \(100.0 \mathrm{~kg}\). They have equal radii, \(r=0.20 \mathrm{~m}\). The spheres are released from rest with their centers \(40.0 \mathrm{~m}\) apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres. (a) Explain why linear momentum is conserved. (b) When their centers are \(20.0 \mathrm{~m}\) apart, find (i) the speed of each sphere and (ii) the magnitude of the relative velocity with which one sphere is approaching the other. (c) How far from the initial position of the center of the \(50.0 \mathrm{~kg}\) sphere do the surfaces of the two spheres collide?

A satellite with mass \(848 \mathrm{~kg}\) is in a circular orbit with an orbital speed of \(9640 \mathrm{~m} / \mathrm{s}\) around the earth. What is the new orbital speed after friction from the earth's upper atmosphere has done \(-7.50 \times 10^{9} \mathrm{~J}\) of work on the satellite? Does the speed increase or decrease?
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