/*! 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 28 The escape speed from a planet o... [FREE SOLUTION] | 91Ó°ÊÓ

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The escape speed from a planet of mass \(2.5 \times 10^{24} \mathrm{~kg}\) is \(6.9 \mathrm{~km} / \mathrm{s}\). Find the planet's radius.

Short Answer

Expert verified
The radius of the planet is approximately \(4.144 \times 10^{6} \mathrm{m}\) or \(4144 \mathrm{km}\).

Step by step solution

01

Identify given values

From the problem, the mass of the planet (M) is given as \(2.5 \times 10^{24} \mathrm{~kg}\), the escape speed \(v\) is \(6.9 \mathrm{~km/s} = 6900 \mathrm{~m/s}\), and the gravitational constant \(G = 6.673 \times 10^{-11} \mathrm{m^3kg^{-1}s^{-2}}\). The task is to find the radius \(r\).
02

Rearrange the escape speed formula

The formula for escape speed is \(v=\sqrt{\frac{2GM}{r}}\). We aim for \(r\), thus rearrange the formula to isolate \(r\): \(r=\frac{2GM}{v^2}\).
03

Insert known values into the formula

Now, substitute the known values into the formula: \(r=\frac{2 \times 6.673 \times 10^{-11} \times 2.5 \times 10^{24}}{(6900)^2}\)
04

Calculate r

Evaluating the above formula gives the radius of the planet.

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Key Concepts

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

Gravitational Constant
Understanding the gravitational constant (\(G\)) is crucial to comprehending how the escape speed is calculated in physics. The gravitational constant is a number that determines the strength of the gravitational pull between two masses. Its value is approximately equal to \(6.673 \times 10^{-11} \text{m}^3\text{kg}^{-1}\text{s}^{-2}\), and this small number shows the relative weakness of gravity compared to other fundamental forces. In calculations, it serves as a proportionality factor in Newton's law of universal gravitation, which describes the gravitational attraction between objects.

When determining the escape speed necessary for an object to break free from a celestial body’s gravitational grip, the gravitational constant is used alongside the mass of the planet and the distance from the planet's center. Essentially, without this constant, we wouldn't be able to measure the gravitational attraction and, thus, the needed velocity to escape it.
Planetary Mass
The mass of a planet, denoted as \(M\) in physics equations, is a fundamental characteristic determining its gravitational force. In our exercise, the planetary mass is given as \(2.5 \times 10^{24} \mathrm{kg}\). A larger mass corresponds to a stronger gravitational field, thereby influencing various factors, such as the escape speed, orbital mechanics, and the potential for retaining an atmosphere.

Mass's Role in Gravity

Planetary mass plays a key role when using the escape speed formula \(v=\sqrt{\frac{2GM}{r}}\). Increasing the mass leads to a higher escape speed due to an increase in the gravitational pull. For students, it is important to recognize the direct relationship between mass and escape speed, and how a planet's mass affects not only the escape speed but also natural phenomena such as tides, orbital period of its moons, and the weight of objects on its surface.
Circular Velocity
Circular velocity, often referred to as orbital velocity, is the speed that an object must attain to orbit a celestial body without falling into it or escaping its gravitational pull. Unlike escape speed, which is concerned with leaving the gravitational influence of a body, circular velocity is about sustaining a stable orbit.

Calculated using a simplification of the escape speed formula \(v_{circular}=\sqrt{\frac{GM}{r}}\), it illustrates the balance between gravitational attraction and the inertia of a satellite. Unlike escape velocity which must be directed vertically from the surface, circular velocity applies to horizontal motion, perpendicular to the force of gravity. This is an essential concept in astrophysics and aerospace engineering, as it applies to satellites, moons, and even space stations in orbit around Earth or any other celestial object. Grasping the differences between escape speed and circular velocity is beneficial for understanding the mechanics behind satellite launches and the orbits of celestial objects.

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Most popular questions from this chapter

You're standing at the highest point on the Moon, \(10,786 \mathrm{~m}\) above the level of the Moon's mean radius. You've got a golf club and a golf ball. (a) How fast would you need to hit the ball horizontally so it goes into a circular orbit? (b) If you hit the ball vertically with the same speed, to what height above you would it rise?

A projectile is launched vertically upward from a planet of mass \(M\) and radius \(R\); its initial speed is \(\sqrt{2}\) times the escape speed. Derive an expression for its speed as a function of the distance \(r\) from the planet's center.

If you're standing on the ground \(10 \mathrm{~m}\) directly below the center of a spherical water tank containing \(3 \times 10^{6} \mathrm{~kg}\) of water, by what fraction is your weight reduced due to the water's gravitational attraction?

Spacecraft that study the Sun are often placed at the so-called Ll Lagrange point, located sunward of Earth on the Sun-Earth line. Ll is the point where Earth's and Sun's gravity together produce an orbital period of one year, so that a spacecraft at \(\mathrm{Ll}\) stays fixed relative to Earth as both planet and spacecraft orbit the Sun. This placement ensures an uninterrupted view of the Sun, without being periodically eclipsed by Earth as would occur in Earth orbit. Find Ll's location relative to Earth. (Hint: This problem calls for numerical methods or solving a higher-order polynomial equation.)

Even the smallest of space debris poses serious threats to operational satellites. Why?

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