Escape Velocity - Revision history

Anguyen676: references. broken url

2025-12-03T01:15:54Z

references. broken url

← Older revision		Revision as of 21:15, 2 December 2025
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	[http://www.britannica.com/science/escape-velocity]		[http://www.britannica.com/science/escape-velocity]

	"Escape Velocity." ~~Wikipedia. Wikimedia Foundation~~, ~~n.d. Web. 05 Dec. 2015~~.		NASA. "Escape Velocity." NASA Space Math, 2018.
	[https://en.~~wikipedia~~.~~org~~/~~wiki~~/~~Escape_velocity]~~		https://www.grc.nasa.gov/www/k-12/Numbers/escapevelocity.html

	“Escape Velocity Formula - with Solved Examples.” Physicscatalyst's Blog, 4 Nov. 2022, https://physicscatalyst.com/article/escape-velocity-formula/#.Y41KpuzMKrc.		“Escape Velocity Formula - with Solved Examples.” Physicscatalyst's Blog, 4 Nov. 2022, https://physicscatalyst.com/article/escape-velocity-formula/#.Y41KpuzMKrc.

	Giancoli, Douglas C. "Physics for Scientists and Engineers with Modern Physics~~." Google Books. Google~~, ~~n.d. Web. 05 Dec. 2015~~.		Giancoli, Douglas C.
	~~[https://books.google~~.com/books?id=xz-UEdtRmzkC&pg=PA199&dq=escape+velocity+gravitational+potential+energy&hl=en&sa=X&ved=0CC0Q6AEwA2oVChMI8_PO4_PBxwIVBJmICh3T6gGl#v=onepage&q=escape%20velocity%20gravitational%20potential%20energy&f=false]		Physics for Scientists and Engineers with Modern Physics, 4th Edition.
			Pearson, 2009.

	“Gravity And Orbits.” PhET, University of Colorado Boulder, 5 Aug. 2019, phet.colorado.edu/en/simulation/gravity-and-orbits.		“Gravity And Orbits.” PhET, University of Colorado Boulder, 5 Aug. 2019, phet.colorado.edu/en/simulation/gravity-and-orbits.

Anguyen676: revising

2025-12-02T21:23:35Z

revising

← Older revision		Revision as of 17:23, 2 December 2025
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	==Escape Velocity and Black Holes==		==Escape Velocity and Black Holes==

	The concept of escape velocity also appears in discussions of black holes. In Newtonian gravity, the escape speed from the surface of a spherical body of mass <math>M</math> and radius <math>r</math> is

	~~:<math>v_e = \sqrt{\frac{2GM}{r}}.</math>~~

	~~If we formally set the escape speed equal to the speed of light <math>c</math>, we can solve for the radius at which light would just barely be able to escape:~~

	~~:<math>c = \sqrt{\frac{2GM}{r}},</math>~~

	~~:<math>r = \frac{2GM}{c^2}.</math>~~

	~~This radius is called the Schwarzschild radius <math>r_s</math>:~~

	~~:<math>r_s = \frac{2GM}{c^2}.</math>~~

	General relativity provides the full description of black holes, but it predicts the same expression for the Schwarzschild radius as this simple energy argument. For <math>r < r_s</math>, the escape speed would have to exceed <math>c</math>, so no object or signal that obeys the relativistic speed limit can escape. The surface at <math>r = r_s</math> is known as the event horizon of a non-rotating black hole.

	At sufficiently high mass density, the escape velocity can exceed the speed of light. Using Newtonian gravity, the escape speed from a spherical body of mass <math>M</math> and radius <math>r</math> is		At sufficiently high mass density, the escape velocity can exceed the speed of light. Using Newtonian gravity, the escape speed from a spherical body of mass <math>M</math> and radius <math>r</math> is

Anguyen676: more depth in the black hole section. i thought it was cool.

2025-12-02T21:23:02Z

more depth in the black hole section. i thought it was cool.

← Older revision		Revision as of 17:23, 2 December 2025
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	General relativity provides the full description of black holes, but it predicts the same expression for the Schwarzschild radius as this simple energy argument. For <math>r < r_s</math>, the escape speed would have to exceed <math>c</math>, so no object or signal that obeys the relativistic speed limit can escape. The surface at <math>r = r_s</math> is known as the event horizon of a non-rotating black hole.		General relativity provides the full description of black holes, but it predicts the same expression for the Schwarzschild radius as this simple energy argument. For <math>r < r_s</math>, the escape speed would have to exceed <math>c</math>, so no object or signal that obeys the relativistic speed limit can escape. The surface at <math>r = r_s</math> is known as the event horizon of a non-rotating black hole.

			At sufficiently high mass density, the escape velocity can exceed the speed of light. Using Newtonian gravity, the escape speed from a spherical body of mass <math>M</math> and radius <math>r</math> is

			:<math>v_e = \sqrt{\frac{2GM}{r}}.</math>

			If we set <math>v_e = c</math>, where <math>c</math> is the speed of light, we obtain the radius at which light would just be able to escape:

			:<math>c = \sqrt{\frac{2GM}{r}},</math>

			which leads to

			:<math>r = \frac{2GM}{c^2}.</math>

			This radius is called the Schwarzschild radius, denoted <math>r_s</math>:

			:<math>r_s = \frac{2GM}{c^2}.</math>

			Any mass compressed inside this radius forms a black hole. Although the above derivation is Newtonian, the same expression for <math>r_s</math> appears as the exact solution to the Einstein field equations for a non-rotating, uncharged black hole.

	==History==		==History==

Anguyen676: added something fun about black holes

2025-12-02T21:21:01Z

added something fun about black holes

← Older revision		Revision as of 17:21, 2 December 2025
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	This relationship is universal for Newtonian gravitational fields and independent of the escaping object’s mass. A satellite moving at orbital speed remains bound and repeatedly returns to the same radius. A spacecraft exceeding escape velocity has enough total energy to reach infinite separation, entering a parabolic (<math>E = 0</math>) or hyperbolic (<math>E > 0</math>) trajectory depending on its excess kinetic energy.		This relationship is universal for Newtonian gravitational fields and independent of the escaping object’s mass. A satellite moving at orbital speed remains bound and repeatedly returns to the same radius. A spacecraft exceeding escape velocity has enough total energy to reach infinite separation, entering a parabolic (<math>E = 0</math>) or hyperbolic (<math>E > 0</math>) trajectory depending on its excess kinetic energy.

			==Escape Velocity and Black Holes==

			The concept of escape velocity also appears in discussions of black holes. In Newtonian gravity, the escape speed from the surface of a spherical body of mass <math>M</math> and radius <math>r</math> is

			:<math>v_e = \sqrt{\frac{2GM}{r}}.</math>

			If we formally set the escape speed equal to the speed of light <math>c</math>, we can solve for the radius at which light would just barely be able to escape:

			:<math>c = \sqrt{\frac{2GM}{r}},</math>

			:<math>r = \frac{2GM}{c^2}.</math>

			This radius is called the Schwarzschild radius <math>r_s</math>:

			:<math>r_s = \frac{2GM}{c^2}.</math>

			General relativity provides the full description of black holes, but it predicts the same expression for the Schwarzschild radius as this simple energy argument. For <math>r < r_s</math>, the escape speed would have to exceed <math>c</math>, so no object or signal that obeys the relativistic speed limit can escape. The surface at <math>r = r_s</math> is known as the event horizon of a non-rotating black hole.

	==History==		==History==

Anguyen676: Interactive Gravity Assist Simulation

2025-12-02T19:53:20Z

Interactive Gravity Assist Simulation

← Older revision		Revision as of 15:53, 2 December 2025
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	</math>		</math>


			===Interactive Gravity Assist Simulation===

			This model uses GlowScript VPython to simulate a Voyager-style gravity assist. The spacecraft approaches a moving planet, accelerates through its gravitational field, and departs with a different heliocentric velocity.

			[https://trinket.io/glowscript/47e96432cb8f Interactive Gravity Assist Model (Click to Run)]

			'''How to use:'''
			* Press Run in the Trinket window.
			* Observe the spacecraft trajectory as it approaches the planet.
			* Vary the spacecraft’s initial velocity vector in the code.
			* If the final trajectory is open (hyperbolic), the spacecraft has gained enough energy to become unbound.
			* If the final trajectory is closed (elliptic), it remains gravitationally bound.

			[[File:Gravityassist_screenshot.png\|500px\|thumb\|center\|A spacecraft gaining energy from a planetary flyby.]]

Anguyen676: major edit. added new sections

2025-12-02T18:17:58Z

major edit. added new sections

← Older revision		Revision as of 14:17, 2 December 2025
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	v_j = \sqrt{\frac{2G(318M_r)}{11.2R_e}} = 59.7 km/s		v_j = \sqrt{\frac{2G(318M_r)}{11.2R_e}} = 59.7 km/s
	</math>		</math>



	==Connectedness==		==Connectedness==

	In real life, escape velocity is not as easy to calculate as in the above examples. The primary complicating factor is the fact that there are more than two bodies in the universe, so in systems with multiple massive attracting bodies, escape velocity can become more complicated. One example is the escape velocity of an object from Earth: an object that achieves the escape velocity of Earth could theoretically escape Earth's influence, but it would remain in orbit around the sun unless its speed were much greater. The concept of escape velocity is widely used in orbital mechanics and rocketry and is critical for the planning of space missions.		In real life, escape velocity is not as easy to calculate as in the above examples. The primary complicating factor is the fact that there are more than two bodies in the universe, so in systems with multiple massive attracting bodies, escape velocity can become more complicated. One example is the escape velocity of an object from Earth: an object that achieves the escape velocity of Earth could theoretically escape Earth's influence, but it would remain in orbit around the sun unless its speed were much greater. The concept of escape velocity is widely used in orbital mechanics and rocketry and is critical for the planning of space missions.

			==Common Misconceptions==
			Escape velocity does not mean the object stops feeling gravity.
			Gravity acts at every finite distance; the object simply never reverses direction.

			Rockets do not necessarily need to reach escape velocity.
			Continuous thrust can exceed gravitational potential without ever reaching the escape speed.

			Escape velocity is not about height, it is about total energy.
			Even at high altitude, if total mechanical energy is negative, the trajectory is still bound.

			Orbital velocity is not “half” of escape velocity.
			They differ because they represent different energy configurations:
			<math>v_\text{orbit} = \sqrt{GM/r}</math>,
			<math>v_\text{escape} = \sqrt{2GM/r}</math>.

	==Escape Velocity vs. Orbital Velocity==		==Escape Velocity vs. Orbital Velocity==

Anguyen676: Added new section. Escape Velocity vs. Orbital Velocity. And added to History section.

2025-12-02T17:46:24Z

Added new section. Escape Velocity vs. Orbital Velocity. And added to History section.

@@ Line 153: / Line 153: @@
 In real life, escape velocity is not as easy to calculate as in the above examples. The primary complicating factor is the fact that there are more than two bodies in the universe, so in systems with multiple massive attracting bodies, escape velocity can become more complicated. One example is the escape velocity of an object from Earth: an object that achieves the escape velocity of Earth could theoretically escape Earth's influence, but it would remain in orbit around the sun unless its speed were much greater. The concept of escape velocity is widely used in orbital mechanics and rocketry and is critical for the planning of space missions.
 ==History==
-Escape velocity stems from the concept of gravity, which was pioneered by Sir Issac Newton. Escape velocity became more important as people looked towards putting objects and people in space. Luna 1, launched in 1959 by the Soviets, was the first man-made object to surpass escape velocity from Earth.
+The concept of escape velocity originates from early studies of gravity. In the 17th century, Isaac Newton showed that the same gravitational force responsible for falling objects also governs planetary motion. He proposed the “Newton’s Cannonball” thought experiment, suggesting that if a projectile were launched with sufficient horizontal speed, it would never hit the Earth and instead orbit indefinitely. A higher speed would allow it to leave Earth entirely, anticipating the modern idea of escape velocity.
 ==See Also==

Anguyen676: I corrected several physics inaccuracies, including the incorrect claim of zero net force during escape and a wrong numerical value for the Moon’s escape velocity. I rewrote the introduction and mathematical model to clearly derive escape velocity using mechanical energy and to explain why gravitational force remains nonzero. I replaced vague or misleading examples with physics-appropriate problems and removed incorrect ones. The “Bound vs. Unbound” section was reorganized using total mechanical

2025-12-02T17:39:27Z

I corrected several physics inaccuracies, including the incorrect claim of zero net force during escape and a wrong numerical value for the Moon’s escape velocity. I rewrote the introduction and mathematical model to clearly derive escape velocity using mechanical energy and to explain why gravitational force remains nonzero. I replaced vague or misleading examples with physics-appropriate problems and removed incorrect ones. The “Bound vs. Unbound” section was reorganized using total mechanical

Show changes

Anguyen676: "There is no net force on an object as it escapes and zero acceleration is perceived." is a problematic statement. Current text is unclear around “without impulse.”

2025-12-02T17:28:08Z

"There is no net force on an object as it escapes and zero acceleration is perceived." is a problematic statement. Current text is unclear around “without impulse.”

← Older revision		Revision as of 13:28, 2 December 2025
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	[[File:Voyager Path czech version.jpg\|400px\|thumb\|right\| A diagram showing the paths of Voyager 1 and 2.]]		[[File:Voyager Path czech version.jpg\|400px\|thumb\|right\| A diagram showing the paths of Voyager 1 and 2.]]

	Edited by ~~Engi Alabady~~, Fall ~~2022~~		Edited by Alexander Nguyen, Fall 2025

	Escape velocity is defined as the minimum velocity required for an object to escape the gravitational force of a large object. The sum of an object's kinetic energy and its gravitational potential energy is equal to zero. The gravitational potential energy is negative due to the fact that kinetic energy is always positive. The velocity of the object will be zero at infinite distance from the center of gravity. Although a body launched at escape speed slows down as it climbs, it always experiences a gravitational pull. Its speed approaches zero asymptotically as r→∞, but never becomes negative. At no point is the force or acceleration actually zero.Escape velocity can also be thought of as the energy needed for a large body to escape the gravitational field WITHOUT impulse.

			Escape velocity is defined as the minimum velocity required for an object to escape the gravitational force of a large object. The sum of an object's kinetic energy and its gravitational potential energy is equal to zero. The gravitational potential energy is negative due to the fact that kinetic energy is always positive. The velocity of the object will be zero at infinite distance from the center of gravity. Although a body launched at escape speed slows down as it climbs, it always experiences a gravitational pull. Its speed approaches zero asymptotically as r→∞, but never becomes negative. At no point is the force or acceleration actually zero. Escape velocity is the minimum initial speed required for an object to escape the gravitational influence of a body without the need for additional propulsion. Once given this initial speed, the object will never return, even though gravity continues to act on it during its motion.
	==The Main Idea==		==The Main Idea==

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	:<math>v_e = \sqrt{\frac{2GM}{r}},</math>		:<math>v_e = \sqrt{\frac{2GM}{r}},</math>

	where <math>G</math> is the universal [[gravitational constant]] (<math>G = 6.7\times 10^{-11} \text{N}~~\cdot~~\text{m}^~~2 /~~ \text{kg}^2</math>), <math>M</math> is the mass of the large body to be escaped, and <math>r</math> the distance from the [[center of mass]] of the mass <math>M</math> to the object. This equation assumes there are no other forces acting on either body. As a side note, the escape velocity stated here could really be called escape speed due to the fact that the quantity is independent of direction. Notice that the equation does not include the mass of the orbiting body.		where <math>G</math> is the universal [[gravitational constant]] (<math>G = 6.67430\times 10^{-11},\text{m}^3\text{kg}^{-1}\text{s}^{-2}</math>), <math>M</math> is the mass of the large body to be escaped, and <math>r</math> the distance from the [[center of mass]] of the mass <math>M</math> to the object. This equation assumes there are no other forces acting on either body. As a side note, the escape velocity stated here could really be called escape speed due to the fact that the quantity is independent of direction. Notice that the equation does not include the mass of the orbiting body.

	===A Mathematical Model===		===A Mathematical Model===

Anguyen676: fixed some conceptual issues. “There is no net force as it escapes and zero acceleration is perceived.” This is not correct.

2025-12-02T17:22:38Z

fixed some conceptual issues. “There is no net force as it escapes and zero acceleration is perceived.” This is not correct.

← Older revision		Revision as of 13:22, 2 December 2025
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	Edited by Engi Alabady, Fall 2022		Edited by Engi Alabady, Fall 2022

	Escape velocity is defined as the minimum velocity required for an object to escape the gravitational force of a large object. The sum of an object's kinetic energy and its gravitational potential energy is equal to zero. The gravitational potential energy is negative due to the fact that kinetic energy is always positive. The velocity of the object will be zero at infinite distance from the center of gravity. ~~There is no net force on an object~~ as it ~~escapes and~~ zero acceleration ~~is perceived~~.Escape velocity can also be thought of as the energy needed for a large body to escape the gravitational field WITHOUT impulse.		Escape velocity is defined as the minimum velocity required for an object to escape the gravitational force of a large object. The sum of an object's kinetic energy and its gravitational potential energy is equal to zero. The gravitational potential energy is negative due to the fact that kinetic energy is always positive. The velocity of the object will be zero at infinite distance from the center of gravity. Although a body launched at escape speed slows down as it climbs, it always experiences a gravitational pull. Its speed approaches zero asymptotically as r→∞, but never becomes negative. At no point is the force or acceleration actually zero.Escape velocity can also be thought of as the energy needed for a large body to escape the gravitational field WITHOUT impulse.

	==The Main Idea==		==The Main Idea==
Line 10:		Line 10:
	:<math>v_e = \sqrt{\frac{2GM}{r}},</math>		:<math>v_e = \sqrt{\frac{2GM}{r}},</math>

	where <math>G</math> is the universal [[gravitational constant]] (<math>G = 6.7\times 10^{-11} \text{N}\cdot\text{m}^2 / \text{kg}^2</math>), <math>M</math> is the mass of the large body to be escaped, and <math>r</math> the distance from the [[center of mass]] of the mass <math>M</math> to the object. This equation assumes there are no other forces acting on either body. As a side note, the escape velocity stated here could really be called escape speed due to the fact that the quantity is independent of direction. Notice that the equation does not include the mass of the orbiting body.		where <math>G</math> is the universal [[gravitational constant]] (<math>G = 6.7\times 10^{-11} \text{N}\cdot\text{m}^2 / \text{kg}^2</math>), <math>M</math> is the mass of the large body to be escaped, and <math>r</math> the distance from the [[center of mass]] of the mass <math>M</math> to the object. This equation assumes there are no other forces acting on either body. As a side note, the escape velocity stated here could really be called escape speed due to the fact that the quantity is independent of direction. Notice that the equation does not include the mass of the orbiting body.

	===A Mathematical Model===		===A Mathematical Model===