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ecp Show that the escape speed from the surface of a planet of uniform density is directly proportional to the radius of the planet.

Short Answer

Expert verified
The escape speed from the surface of a planet of uniform density is shown to be proportional to the radius of the planet.

Step by step solution

01

Define the Variables

Let \(v\) be the escape speed, \(r\) the radius of the planet, \(M\) the mass of the planet, and \(蟻\) the uniform density of the planet. The mass can be expressed in terms of the density and volume as \(M = 4/3 * 蟺 * r鲁 * 蟻\).
02

Equate Kinetic and Potential Energy

The gravitational potential energy of an object of mass \(m\) on the surface of the planet is:\(-G* M * m / r\). The kinetic energy of the object moving at the escape speed is: \(1/2 * m * v虏\). For the object to escape the gravitational field, the kinetic energy should be equal to the magnitude of the gravitational potential energy. This gives: \(1/2 * m * v虏 = G * M * m / r\).
03

Solve for Escape Speed

Substitute \(M = 4/3 * 蟺 * r鲁 * 蟻\) into the above equation and solve for \(v\): \(1/2 * v虏 = G * 4/3 * 蟺 * r鲁 * 蟻 / r\). Simplifying gives: \(v = 鈭(8/3 * 蟺 * G * r * 蟻)\). This shows that the escape speed is directly proportional to the radius of the planet, as required.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is an important concept when discussing escape speed. It represents the energy required to bring a mass to a given point in a gravitational field from infinity. For a planet, this energy depends on the gravitational pull it exerts on objects on its surface.
To find this energy from the surface of a planet, consider an object of mass \(m\). The gravitational potential energy \(U\) at the planet's surface can be expressed as:
  • \(U = -\frac{G M m}{r}\)
Here, \(G\) is the universal gravitational constant, \(M\) is the mass of the planet, and \(r\) is the planetary radius. The negative sign indicates that work must be done against gravity to bring the object to this position from infinity, where potential energy is defined as zero. Essentially, this energy indicates how strongly the planet's gravity holds onto the object.
In problems involving escape speed, we equate the kinetic energy of an object to its gravitational potential energy to solve for the speed necessary to overcome the gravitational pull.
Uniform Density
The concept of uniform density simplifies understanding planetary bodies. Density represents how much mass exists in a given volume. When density is constant鈥攗niform鈥攁cross a planet, it simplifies many calculations because mass can be easily expressed in terms of volume and density.
Consider a planet with uniform density \(\rho\). The planet's volume \(V\) can be determined, given its radius \(r\), using the formula for the volume of a sphere:
  • \(V = \frac{4}{3} \pi r^3\)
Since density is mass over volume, the planet's mass \(M\) becomes:
  • \(M = \rho \times V = \rho \times \frac{4}{3} \pi r^3\)
Knowing this allows us to express other quantities, like gravitational potential energy and escape speed, in terms of the planet's radius and density, making calculations more straightforward in theoretical exercises.
Planetary Radius
The radius of a planet is not only a measure of its size but also plays a crucial role in determining gravitational properties relevant to escape speed. The larger the radius, the further any point on the surface is from the planet's center of mass, which has implications for gravitational strength and potential energy.
When calculating escape speed, the planetary radius \(r\) becomes a direct factor. We use the radius in calculations of both the volume (thus mass, from density) and in gravitational equations. The escape speed \(v\), for instance, relies on formulas incorporating \(r\):
  • \(v = \sqrt{\frac{8}{3} \pi G r \rho}\)
Here, you can see that \(v\) (escape speed) is proportional to the square root of \(r\). This relationship highlights how radius influences the ability of a planet to retain atmosphere and objects with its gravitational pull. Whether calculating forces, energies, or speeds, the radius is indispensable in planetary science discussions.

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Most popular questions from this chapter

In a popular amusement park ride, a rotating cylinder of radius \(3.00 \mathrm{~m}\) is set in rotation at an angular speed of \(5.00 \mathrm{rad} / \mathrm{s}\), as in Figure \(\mathrm{P} 7.75 .\) The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping? (Hint: Recall that the magnitude of the maximum force of static friction is equal to \(\mu n\), where \(n\) is the normal force-in this case, the force causing the centripetal acceleration.)

A centrifuge in a medical laboratory rotates at an angular speed of 3600 rev/min. When switched off, it rotates through \(50.0\) revolutions before coming to rest. Find the constant angular acceleration of the centrifuge.

The tires on a new compact car have a diameter of \(2.0 \mathrm{ft}\) and are warranted for 60000 miles. (a) Determine the angle (in radians) through which one of these tires will rotate during the warranty period. (b) How many revolutions of the tire are equivalent to your answer in part (a)?

ecp A rocket is fired straight up through the atmosphere from the South Pole, burning out at an altitude of \(250 \mathrm{~km}\) when traveling at \(6.0 \mathrm{~km} / \mathrm{s}\). (a) What maximum distance from Earth's surface does it travel before falling back to Earth? (b) Would its maximum distance increase if it were fired from a launch site on the equator? Why?

ecp Because of Earth's rotation about its axis, a point on the equator has a centripetal acceleration of \(0.0340 \mathrm{~m} / \mathrm{s}^{2}\) whereas a point at the poles has no centripetal acceleration. (a) Show that, at the equator, the gravitational force on an object (the object's true weight) must exceed the object's apparent weight. (b) What are the apparent weights of a \(75.0\) -kg person at the equator and at the poles? (Assume Earth is a uniform sphere and take \(g=\) \(\left.9.800 \mathrm{~m} / \mathrm{s}^{2} .\right)\)

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