# what is the value of escape speed on the surface of earth

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## Escape velocity

## Escape velocity

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For other uses, see Escape Velocity.

Not to be confused with Orbital speed.

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In celestial mechanics, **escape velocity** or **escape speed** is the minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a primary body, thus reaching an infinite distance from it. It is typically stated as an ideal speed, ignoring atmospheric friction. Although the term "escape velocity" is common, it is more accurately described as a speed than a velocity because it is independent of direction; the escape speed increases with the mass of the primary body and decreases with the distance from the primary body. The escape speed thus depends on how far the object has already traveled, and its calculation at a given distance takes into account 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 speed, 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.

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;[nb 1] 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.[1] 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.

Escape speed at a distance from the center of a spherically symmetric primary body (such as a star or a planet) with mass is given by the formula[2]

{\displaystyle v_{e}={\sqrt {\frac {2GM}{d}}}={\sqrt {2gd}}}

where is the universal gravitational constant ( ≈ 6.67×10−11 m3·kg−1·s−2)[nb 2] and is the local gravitational acceleration (or the surface gravity, when = ). The escape speed is independent of the mass of the escaping object. For example, the escape speed from Earth's surface is about 11.186 km/s (40,270 km/h; 25,020 mph; 36,700 ft/s)[3] and the surface gravity is about 9.8 m/s2 (9.8 N/kg, 32 ft/s2).

When given an initial speed

{\displaystyle V}

greater than the escape speed

{\displaystyle v_{e},}

the object will asymptotically approach the

{\displaystyle v_{\infty },}

satisfying the equation:[4]

{\displaystyle {v_{\infty }}^{2}=V^{2}-{v_{e}}^{2}.}

## The value of escape velocity on earth is:

Click here👆to get an answer to your question ✍️ The value of escape velocity on earth is:

Question

## The value of escape velocity on earth is:

**A**11.2 kms

−1

**B**11.2 kms

−2

**C**10.2 ms

−1

**D**11.2 ms

−2 Medium Open in App Solution Verified by Toppr

Correct option is A)

Escape velocity is the minimum speed needed for an object to escape from the gravitational influence earth.For escape velocity KE+U=0 2 1 mv 2 = r GMm v escape = r 2GM V escape =11.2KmS −1

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## Terminal velocity

escape velocity, in astronomy and space exploration, the velocity needed for a body to escape from a gravitational centre of attraction without undergoing any further acceleration. The escape velocity vesc is expressed as vesc = 2GM r ,where G is the gravitational constant, M is the mass of the attracting mass, and r is the distance from the centre of that mass. Escape velocity decreases with altitude and is equal to the square root of 2 (or about 1.414) times the velocity necessary to maintain a circular orbit at the same altitude. At Earth’s surface, if atmospheric resistance could be

## escape velocity

physics

By The Editors of Encyclopaedia Britannica Last Updated: Dec 30, 2022 Article History

Related Topics: velocity orbital velocity

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**escape velocity**, in astronomy and space exploration, the velocity needed for a body to escape from a gravitational centre of attraction without undergoing any further acceleration. The escape velocity vesc is expressed as vesc = Square root of√2GM/r,where G is the gravitational constant, M is the mass of the attracting mass, and r is the distance from the centre of that mass. Escape velocity decreases with altitude and is equal to the square root of 2 (or about 1.414) times the velocity necessary to maintain a circular orbit at the same altitude. At Earth’s surface, if atmospheric resistance could be disregarded, escape velocity would be about 11.2 km (6.96 miles) per second. The velocity of escape from the less massive Moon is about 2.4 km (1.5 miles) per second at its surface. A planet or moon cannot long retain an atmosphere if its escape velocity is low enough to be near the average velocity of the gas molecules making up the atmosphere. Inside the event horizon of a black hole, the escape velocity exceeds the speed of light, so not even rays of light can escape into space.

The Editors of Encyclopaedia BritannicaThis article was most recently revised and updated by Erik Gregersen.

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