What is Escape velocity: Definition and 219 Discussions

In physics (specifically, celestial mechanics), escape velocity is the minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a massive body, that is, to eventually reach an infinite distance from it. Escape velocity rises with the body's mass (body to be escaped) and falls with the escaping object's distance from its center. The escape velocity thus depends on how far the object has already traveled, and its calculation at a given distance takes into account the fact that without new acceleration it will slow down as it travels—due to the massive body's gravity—but it will never quite slow to a stop.
A rocket, continuously accelerated by its exhaust, can escape without ever reaching escape velocity, since it continues to add kinetic energy from its engines. It can achieve escape at any speed, given sufficient propellant to provide new acceleration to the rocket to counter gravity's deceleration and thus maintain its speed.
The escape velocity from Earth's surface is about 11,186 m/s (6.951 mi/s; 40,270 km/h; 36,700 ft/s; 25,020 mph; 21,744 kn). More generally, escape velocity is the speed at which the sum of an object's kinetic energy and its gravitational potential energy is equal to zero; an object which has achieved escape velocity is neither on the surface, nor in a closed orbit (of any radius). With escape velocity in a direction pointing away from the ground of a massive body, the object will move away from the body, slowing forever and approaching, but never reaching, zero speed. Once escape velocity is achieved, no further impulse need be applied for it to continue in its escape. In other words, if given escape velocity, the object will move away from the other body, continually slowing, and will asymptotically approach zero speed as the object's distance approaches infinity, never to come back. Speeds higher than escape velocity retain a positive speed at infinite distance. Note that the minimum escape velocity assumes that there is no friction (e.g., atmospheric drag), which would increase the required instantaneous velocity to escape the gravitational influence, and that there will be no future acceleration or extraneous deceleration (for example from thrust or from gravity of other bodies), which would change the required instantaneous velocity.
For a spherically symmetric, massive body such as a star, or planet, the escape velocity for that body, at a given distance, is calculated by the formula





v

e


=




2
G
M

r





{\displaystyle v_{e}={\sqrt {\frac {2GM}{r}}}}
where G is the universal gravitational constant (G ≈ 6.67×10−11 m3·kg−1·s−2), M the mass of the body to be escaped from, and r the distance from the center of mass of the body to the object. The relationship is independent of the mass of the object escaping the massive body. Conversely, a body that falls under the force of gravitational attraction of mass M, from infinity, starting with zero velocity, will strike the massive object with a velocity equal to its escape velocity given by the same formula.
When given an initial speed



V


{\displaystyle V}
greater than the escape speed




v

e


,


{\displaystyle v_{e},}
the object will asymptotically approach the hyperbolic excess speed




v




,


{\displaystyle v_{\infty },}
satisfying the equation:







v






2


=

V

2






v

e




2


.


{\displaystyle {v_{\infty }}^{2}=V^{2}-{v_{e}}^{2}.}
In these equations atmospheric friction (air drag) is not taken into account.

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  1. Z

    Exploring GR Effects on Escape Velocity of a Star & a Black Hole

    This is an analysis involving general relativity (GR). Let a spherically symmetric, uncharged, nonrotating star (not a black hole—that comes later) have an escape velocity = infinitesimally less than c at its surface. Then its r-coordinate radius (also known as the reduced circumference and...
  2. T

    Universal Gravitation on escape velocity

    On dealing with Universal Gravitation topic, I came across this question: what is the min speed, relative to the sun, necessary for a spacecraft to escape the solar system if it starts at the Earth's orbit? My working: v=sqrt(2GM/R), where G=universal constant, R= distance between Earth...
  3. S

    How does escape velocity vary with an object's mass?

    another simple question: what is the energy needed to move an object away from the planets gravitational field at CONSTANT speed? 1) Ep 2)Ek 3)Change in Ep 4)Change in Ek
  4. B

    Escape Velocity Given Radius of 2 Objects

    Got a dosie of a problem I need checked. I've been playing around with this problem, and after I submit the right answer (the ans. I got) , it came out wrong. Someone tell me what I'm doing wrong PLEASE thanks: The prob. is to calculate escape velocity from a white dwarf and a neutron...
  5. I

    Impossible escape velocity question

    I've got a question in front of me that asks to find the escape velocity for a hypothetical celestial object that has the same density as the sun. MC answers are - 1)2.51 x 10^3 m/s 2)6.18 x 10^5 m/s 3)3.08 x 10^5 m/s 4)5.42 x 10^10 m/s As far as I can work out its not possible to find...
  6. D

    Solving Escape Velocity of the Sun: Troubleshooting an Error

    I can't seem to obtain the right answer using the escape velocity formula. I think my problem is with the gravitational constant. I would like to find the escape velocity of the sun. v_e=\sqrt{\frac{2GM}{r}} or v_e=\sqrt{\frac{2\mu}{r}} v_e_\odot=617.54(km/s)...
  7. C

    Calculating the Escape Velocity from a Neutron Star

    Question: The radius of a neutron star is 750 times smaller than the Earth's radius, and its mass is 1.8 times larger than the Earth's mass. What is the escape velocity from the surface of a neutron star? (Ignore the fact that, at high speeds, one should not really use mv^2/2 for the kinetic...
  8. T

    Escape velocity from center of earth

    How would one go about calculating the escape velocity of an object with mass m from the center of the Earth. I understand that that when launched from the surface of the Earth, mechanical energy is conserved and you end up with v escape =sqrt[2gRe] So what i did was i calculated what i...
  9. tony873004

    What is the formula for calculating delta v for escape velocity?

    Let's say I have an object orbiting the Sun in a circular orbit with an orbital velocity of 30 km/s. If I want to accelerate prograde into a hyperbolic orbit, and I'd like my velocity relative to the Sun to be 1 km/s at infinity, what is the formula I'd use to figure out how much delta v in the...
  10. Cyrus

    Escape velocity and gravitation

    Hi, me again. I have another question concerning the gravitation equation. My physics book shows through a proof that the escape velocity of an Earth satelite is just sqrt(2) greated than the orbital velocity. But I am having trouble seeing how this is so. My instinct tells me that if I increase...
  11. R

    Escape velocity outside event horizon

    Escape velocity at particular radius from the gravitational source means the initial speed that an object needs at that raduis in order to coast without limit ("to infinity") without ever falling back to the gravitational source. If the escape velocity at a radius is so large that nothing...
  12. E

    Escape Velocity from Solar System

    Here is a celestial mechanics problem I can't seem to solve:To escape the solar system, an interstellar spacecraft must overcome the gravitational attraction of both the Earth and Sun. Ignore the effects of the other bodies in the solar system. Show that the escape velocity is v =...
  13. S

    Why is the escape velocity of Earth so fast?

    Help me understand this please. Why is the escape velocity of Earth so fast? Doesn’t the gravitational pull of the Earth weaken as you get farther away from the center of the Earth? I can jump a couple feet into the air… I’m am going upwards nowhere near 27k miles an hour… if I could...
  14. P

    Why is the escape velocity for a black hole impossible to achieve?

    What is the reason why something, once it enters the event horizon of a black hole, it cannot escape? I've often heard the answer given that at the event horizon the escape velocity equals the speed of light, and from within the horizon a superluminal velocity is required, which is impossible...
  15. D

    Question about Grav. Potential Energy and Escape Velocity

    I am finding this question to be a bit challenging--- Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed Vo. (< Vesc) (Ignore air resistance and the Earth's rotation) I know I should plug in numbers only until...
  16. G

    Energy Conservation and Escape Velocity

    If a particle is launched with kinetic energy as it gets further from the mass (eg a planet) the kinetic energy decreases. However Potential Energy GMm/r also decreases the further you get from the mass. Where does all this energy go? :confused: Granted, the calculation U + K = 0 does give...
  17. S

    Escape velocity and formulas for circular orbits

    In my class, we developed a list of formulas for circular orbits. One of them is E/m=1/2v2-GM/r=constant. To derive escape velocity we find for what v does E=0. But an orbit of this nature is certainly not circular! How can we apply the formula?
  18. C

    Escape velocity and acceleration from gravity

    Ive heard that the escape velocity of a body is the same as the maximum velocity that can be achieved from gravitational acceleration from the same body. Like say it requires a spacecraft to be moving at about 11 km/sec to escape Earth's gravity, and we isolated the Earth and spacecraft , so...
  19. R

    Escape Velocity of virtual particles

    I know that the escape velocity for virtual particles being emitted from black holes is c2+ (in the black hole entropy formula, there is an escape velocity of c3). I also know that Hawking radiation and for that matter*(I don't mean matter in physics terms), any radiation being emitted from a...
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