Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 13: Problem 35
A "treetop satellite" (Fig. Pl3.35) moves in a circular orbit just above the surface of a planet, assumed to offer no air resistance. Show that its orbital speed \(v\) and the escape speed from the planet are related by the expression $$ v_{\operatorname{exc}}=\sqrt{2} v $$
Short Answer
Expert verified
\( v_{\operatorname{exc}} = \sqrt{2} v \)
Step by step solution
01
Understand Orbital Speed
The orbital speed (v) is the speed at which the satellite must travel to stay in a circular orbit. The gravitational force provides the necessary centripetal force to keep the satellite in orbit. This force is given by the formula: \( F = \frac{m v^2}{r} \), where \( m \) is the mass of the satellite, \( v \) is its orbital speed, and \( r \) is the radius of the orbit.
02
Equation for Gravitational Force
The gravitational force can also be described by Newton's law of universal gravitation: \( F = \frac{G m M}{r^2} \), where \( G \) is the gravitational constant, \( m \) is the mass of the satellite, \( M \) is the mass of the planet, and \( r \) is the distance between their centers which, for a treetop orbit, is approximately equal to the radius of the planet.
03
Relate Orbital Speed to Gravitational Force
Since the satellite stays in orbit due to the gravitational force, we can equate the expressions from Step 1 and Step 2: \( \frac{m v^2}{r} = \frac{G m M}{r^2} \). Simplifying, we get the orbital speed formula \( v = \sqrt{\frac{G M}{r}} \).
04
Understand Escape Speed
Escape speed (\(v_{\operatorname{exc}}\)) is the speed an object needs to be moving to break free from a planet's gravitational pull without further propulsion. The formula for escape speed is derived by equating kinetic energy to gravitational potential energy: \( \frac{1}{2}m v_{\operatorname{exc}}^2 = \frac{G m M}{r} \), solving for escape speed gives \( v_{\operatorname{exc}} = \sqrt{\frac{2 G M}{r}} \).
05
Relate Orbital Speed to Escape Speed
By comparing the orbital speed and escape speed formulas from Step 3 and Step 4, we can see the relationship between them: \( v_{\operatorname{exc}} = \sqrt{2} \sqrt{\frac{G M}{r}} = \sqrt{2} v \), which shows that the escape speed is \(\sqrt{2}\) times the orbital speed.
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.
Gravitational Force
Gravitational force is a fundamental interaction that attracts two bodies with mass. According to Newton's law of universal gravitation, every mass attracts every other mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This can be summarized with the formula:
\[ F = \frac{G \times m \times M}{r^2} \]
where \( F \) is the gravitational force, \( G \) is the gravitational constant (6.67430(15) 脳 10^(-11) m^3 kg^(-1) s^(-2)), \( m \) and \( M \) are the masses of the two objects, and \( r \) is the distance between their centers.
For objects close to a planetary surface, such as satellites in a 'treetop' orbit, the gravitational force acts as the centripetal force needed to keep them in a circular path. That is why the satellite's mass and velocity, as well as the planet's mass and radius, play crucial roles in defining the satellite's orbital speed. The balance between this gravitational pull and the inertia from the satellite's motion determines whether it stays in orbit or escapes into space.
\[ F = \frac{G \times m \times M}{r^2} \]
where \( F \) is the gravitational force, \( G \) is the gravitational constant (6.67430(15) 脳 10^(-11) m^3 kg^(-1) s^(-2)), \( m \) and \( M \) are the masses of the two objects, and \( r \) is the distance between their centers.
For objects close to a planetary surface, such as satellites in a 'treetop' orbit, the gravitational force acts as the centripetal force needed to keep them in a circular path. That is why the satellite's mass and velocity, as well as the planet's mass and radius, play crucial roles in defining the satellite's orbital speed. The balance between this gravitational pull and the inertia from the satellite's motion determines whether it stays in orbit or escapes into space.
Escape Speed
Escape speed is the minimum speed an object must have to overcome a celestial body's gravitational pull without further propulsion. It is a crucial concept in astrophysics and space exploration as it determines the energy required to send probes and other spacecraft from Earth to other planets or out of the solar system.
The escape speed from a planet is derived from the energy conservation principle. The idea is that for the satellite to escape, its kinetic energy must at least equal the gravitational potential energy imposed by the planet. This relationship is captured by the equation:
\[ \frac{1}{2}mv_{\text{exc}}^2 = \frac{G m M}{r} \]
Solving this equation for escape speed \(v_{\text{exc}}\) gives us:
\[ v_{\text{exc}} = \sqrt{\frac{2 G M}{r}} \]
This way, the escape speed is geometry-independent and relies solely on the planet's mass and radius, as well as the gravitational constant. Understanding the escape speed is vital for mission planning and ensuring spacecraft can enter or exit orbits as intended.
The escape speed from a planet is derived from the energy conservation principle. The idea is that for the satellite to escape, its kinetic energy must at least equal the gravitational potential energy imposed by the planet. This relationship is captured by the equation:
\[ \frac{1}{2}mv_{\text{exc}}^2 = \frac{G m M}{r} \]
Solving this equation for escape speed \(v_{\text{exc}}\) gives us:
\[ v_{\text{exc}} = \sqrt{\frac{2 G M}{r}} \]
This way, the escape speed is geometry-independent and relies solely on the planet's mass and radius, as well as the gravitational constant. Understanding the escape speed is vital for mission planning and ensuring spacecraft can enter or exit orbits as intended.
Circular Orbit
A circular orbit is a type of orbit where an object, such as a satellite, moves around a planet or another celestial body along a circular path. This kind of orbit is stabilized by the balance between gravity acting as a centripetal force and the satellite's inertia from its tangential velocity.
When an object maintains a circular orbit, its velocity is constant, and it's referred to as the orbital speed (\(v\)). The formula for calculating the orbital speed is derived by equating gravitational force to centripetal force:
\[ \frac{m v^2}{r} = \frac{G m M}{r^2} \]
After rearranging and solving for \(v\), we get:
\[ v = \sqrt{\frac{G M}{r}} \]
Understanding orbits is essential for satellite deployment and ensuring they follow the right path around a planet. It also helps us calculate other crucial variables such as period and altitude of the satellites. Operating within a circular orbit allows satellites to provide consistent data and communication services to and from Earth.
When an object maintains a circular orbit, its velocity is constant, and it's referred to as the orbital speed (\(v\)). The formula for calculating the orbital speed is derived by equating gravitational force to centripetal force:
\[ \frac{m v^2}{r} = \frac{G m M}{r^2} \]
After rearranging and solving for \(v\), we get:
\[ v = \sqrt{\frac{G M}{r}} \]
Understanding orbits is essential for satellite deployment and ensuring they follow the right path around a planet. It also helps us calculate other crucial variables such as period and altitude of the satellites. Operating within a circular orbit allows satellites to provide consistent data and communication services to and from Earth.
