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.
While reading a similar and deservedly closed post a contradiction came to my mind. The supposed contradiction is related to Statistical Physics where my understanding is only conceptual so correct me where I might be wrong.
I remember reading that lightweight gasses can escape Earth's...
I have a difficulty when making the energy-conservation-equation for the second step.
When making the equation, we need to know the exact position (measured from the sun) of the rocket after it is freed from the Earth gravitation.
But, where exactly does the rocket free from Earth...
I'm pretty confused by this but I have a few thoughts. Since the sun takes up most of the mass of the solar system, I was thinking maybe I'm really looking for the escape velocity of the sun? So I would use the mass of the sun for M and the radius of the sun for r. My other thought was to add up...
He explain escape velocity in example where rocket goes straight up,isnt escacpe velocity ,velocity where centrifugal forces and gravity are equal,so refers only when rocket going in circle/orbit?
Can rocket really leave Earth in straight line like he show in video once reach this velocity and...
So Ekf-Eki+Epf-Epi=0. I understand that the final potential energy is 0 (distance away approaches infinity), but don't get why the final kinetic energy becomes 0. If the final kinetic energy was 0, wouldn't that mean the object no longer has any velocity and would start being effected by the...
I was surprised to read that the formula for escape velocity — at least for a spherical mass like the Earth — is the same in relativity as it is in classical physics:
v_e = (2GM/r)^{1/2}
I'm wondering if someone can give me a good source for deriving this. (I assume one takes a radial...
1. The satellite would be in a jovian-synchronous orbit,
Rearranging the formula for the orbital period in terms of r, since T^2 is proportional to r^3:
T^2=4π^2r^3/GM which becomes r^3=(GM/4π^2) T^2
M(mass of Jupiter)=1.89 x 10^27
G=6.67*10^-11 m^3kg^-1s^-2
T=9 hours and 55 minutes =...
In the last step of the derivation of escape velocity, the two sides of the equation seem to have opposite signs.
$$-1/2mv_0^2=-mgR_e^2\,\lim_{r\to\infty}(1/r-1/R_e)$$
$$-1/2mv_0^2=mgR_e^2 \frac{1}{R_e}$$
Since the mass and the square of the velocity are positive, the left side of the equation...
If you derive the equation for orbital velocity you get
\begin{equation}
v_{orbit} = \sqrt{\frac{GM}{R}}
\end{equation}
and for escape velocity you get
\begin{equation}
v_{escape} = \sqrt{\frac{2GM}{R}}=\sqrt{2}\,v_{orbit}
\end{equation}
I'm wondering if there is a logical/geometrical...
We have 2 different formulas for escape velocity. and . If we look at the first formula we see that escape velocity is inversely proportional to the square root of Radius of Earth. While in the second formula, escape velocity is directly proportional to the square root of Radius of Earth.
We...
Below is the work I've attempted. I used 2 PE b'c there were 2 point charges, and only one KE b'c only the proton is moving. The final equation in case it's hard to see is V(esc) = sqrt (4kQq / mr).
I'm not sure if I did it right. Did I set up this equation right? and I am also not sure what...
I really cannot understand where this is going wrong...
Plugging in the constants, I get
vescape=Sqrt(2(6.67x10^-11)(5.976x10^24kg)/6378).
(6.67x10^-11)(5.976x10^24kg) gives me 3.99x10^14, and multiplied by 2 gives me 7.97x10^14.
7.97x10^14/6378=1.25x10^11.
The square root of 1.25x10^11 would...
The is a question about gravitational time dilation and escape velocity. As others have pointed out, you may use escape velocity to calculate gravitational time dilation in a gravitational field. (Interestingly, you can't use gravity to calculate gravitational time dilation, which makes...
Homework Statement: Jack can jump upwards a distance 1.4 meters when he is on the surface of the Earth. What is his initial velocity when he jumps?
Most of the time, however, Jack is in space. He is an asteroid miner: he looks for asteroids made out of useful metals. His job involves landing on...
Is it possible to derive escape velocity say using momentum and force balance considerations? or using angular momentum consideration?
Namely, any other approach then energy consideration that utilizes gravitation potential energy and kinetic energy?
we all know escape velocity is a velocity in which a body can escape orbit around the Earth and fly off into space. this needs to happen in space our just out of our atmosphere, due to drag that could bring an object back into orbit and cause it's velocity to degrade to a point where it would...
Hi!
I have a question about escape velocity. If a planet is bigger and have a greater escape velocity than another planet. Do this effect the density of the bigger planet in any way? Or do we have to know the mass of the bigger planet to know if the density is larger or lower for this planet?
Hey there,
If body 1, mass M1 has escape velocity V_e1 = (2GM1/r)**.5 but M2 is more massive than M1 is this relation still valid? In this case, the subordinate body really isn't the subordinate body so does this still hold? And r (distance b/t the two) changes not only due to the motion of M2...
Hello
When one calculates the speed of liberation of an object with respect to a star, one takes into account the force exerted by the star on the object. But not the force exerted by the object on the star. For small objects, this is valid. But what is the speed of release of a 1000 kg rocket...
Homework Statement
Gravitational force exerted on mass m is GMm/r^2
2. Relevant equations
Orbital velocity at distance R from Earth = ##\sqrt gR##
Escape velocity = ##\sqrt 2gR##
gR = GM/R
Fc = mV^2/R
F =m.aThe Attempt at a Solution
1) express acceleration of gravity in terms of G, M , and R...
This is just a reality that I have stumbled upon that I'm sure was well-known, but I still found it interesting. I apologize if this is second-nature to physics experts.
I was responding to a post on a different thread that claimed you could make a tunnel around the Earth and if you sent a...
I may have a fundamental misunderstanding of the concept, but I was wondering, how does the accelerating expansion of the universe calculate for the time dilation in light travel?
From my understanding, we know that the universe expansion is accelerating because the farthest galaxies that we...
If you had two masses, m_{1} and m_{2}, and you released them in space infinitely far apart, their kinetic energies would satisfy \frac{1}{2}m_{1}v_{1}^2+\frac{1}{2}m_{2}v_{2}^2=\frac{Gm_{1}m_{2}}{r} if they met with a distance r between their centres of mass. This equation therefore tells you...
Homework Statement
solve for escape velocity from Earth's surface
Homework Equations
just either use the line integral, and the tangential term disappears, or just use the energy equations
The Attempt at a Solution
I've solved it, but I'm having some trouble just coming to grips with this...
If I drive a plane and the force of engine is bigger than force of gravity of it , if the engine is turn on always ,and assuming no air , will the plane continue moving up and escape from the gravity ?
Is it necessary to be the kinetic energy greater than gpe to move ( I don't talk about in orbits)
Example :
Is it will be harder to move an object has a bigger gpe ( same mass but bigger hight from ground).
And thanks.
Hey people, I just want to ask that what will happen to the total mechanical force of the rocket if its speed is less than escape velocity?
a. KE+PE=0
b. KE+PE>0
c. KE+PE<0
d. Depends upon initial speed of the rocket
Pick one. And Why??
Homework Statement
A missile is launched from the surface of a planet with the speed v0 at t=0.
According to the theory of universal gravitation, the speed v of the missile after launch
is given by
v2 = 2gr0(r0 / r -1) + v02
where g is the gravitational...
Can you please direct me to ref that shows the derivation of the escape velocity from a spherical object that moves in velocity v~c with respect to rest frame?
I suspect the escape velocity is increasing (intuitively since the mass increases).
Please comment and suggest alternatives.
So, in preparation to the Portuguese Astronomy Olympiads, I've stumbled upon this problem (exercise):
The sun, which is 8 kpc away from the centre of the Milky Way, has a rotation speed of approximately 220 kms-1 . Whereas a a star that is 15 kpc from the centre of the Galaxy orbits at a speed...
Homework Statement
The gravitational potential energy of a certain rocket at the surface of the Earth is -1.9x10^12 J. The gravitational potential energy of the same rocket 300km above the Earth's surface is -1.8x10^12 J. Assume the mass of the rocket is constant for this problem.
A) How much...
In another forum, the question was raised, "could a ship with 1G acceleration escape the gravity well of a planet with 1G gravity?"
A popular response is, if the craft is aerodynamic, it could accelerate laterally until it reached escape velocity and then manage to get to space.
I don't...
Hey, orbital mechanics!
I can't find what I need to figure this out on the internet, and I don't do calculus so I don't understand all that I find. Help me make my next sci-fi novel plausible?
I just did an Oberth maneuver around Sol, 21 radii (.0977 AU, 14,616,000 km) from center...
Text books ordinarily give the escape velocity of a mass-M body (in the center of mass frame of the system of the body and the escaping projectile, whose mass I'll label m) as
(*) v2 = 2GM/r
where r is the distance between the body and the escaping projectile.
it doesn’t seem to me that (*)...
Going through several definitions, it appears that escape velocity is equal to the potential energy. That is:$$\frac{1}{2}m v^2=-\frac{G M m}{r}$$but if I solve for velocity, $v$, I get:$$v=\sqrt{-2\frac{G M}{r}}$$So how do I get an escape velocity that isn't imaginary?
Homework Statement
Calculate the escape velocity on the surface of the neutron star in the previous problem (##m = \frac{2}{3} \cdot 2,1 \cdot M_{\odot}##; ##R = 15km##).
Hint: Basic physics. Note, however, that the escape velocity is not going to be small when compared to the speed of light...
Homework Statement
Rockets are propelled by the momentum of the exhaust gases expelled from the tail. Since these gases arise from the reaction of the fuels carried in the rocket, the mass of the rocket is not constant, but decreases as the fuel is expended. Show that for a Rocket starting...
According to this video, , if a black hole is large enough you could actually travel for some time within the event horizon without dying because the event horizon is so far from the actual singularity. So, assuming that's true, what would you see while you were inside the black hole?
Here's...
Why does a planet's kinetic energy become 0 when it reaches infinity? And why does a planet's kinetic energy get converted to gravitational potential energy when subjected under another planet's gravitational field?
Escape velocity seems very abstract to me!
It is my understanding of black holes that nothing can escape them because their escape velocity is higher than the speed of light. The place at which the escape velocity becomes higher than the speed of light is known as the event horizon. My question is what happens if there is a normal force...
I have been reading D.E. Littlewood's book "The Skeleton Key of Mathematics", and near the beginning he says that if a projectile weighing one (long) ton were given a velocity of 44 miles a second, this would be "sufficient to raise it to a height of 1000 miles above the Earth's surface."...
Homework Statement
[A rocket has landed on Planet X, which has half the radius of Earth. An astronaut onboard the rocket weighs twice as much on Planet X as on Earth. If the escape velocity for the rocket taking off from Earth is v , then its escape velocity on Planet X is
a) 2 v
b) (√2)v
c) v...
is it right to say, "when all the potential energy is converted in kinetic energy the object is moving at the escapevelocity.
and "when the change in potential energy and kinetic energy is constant at the same time it is laying still on the ground or in a perfect circulair orbit.
and the last...