Escape Velocity

created by Varun Rajagopal

space x

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 be zero at infinite distance from the centre of gravity. There is no net force on an object as it escapes and zero acceleration is perceived.

The Main Idea

The formula for escape velocity at a certain distance from a body is calculated by the formula Template:Cite book

[math]\displaystyle{ v_e = \sqrt{\frac{2GM}{r}}, }[/math]

where G is the universal gravitational constant (G = 6.67×10⁻¹¹ m³ kg⁻¹ s⁻²), M is the mass of the large body to be escaped, and r the distance from the center of mass of the mass M to the object.<ref group="nb"> This equation assumes there is no atmospheric friction and is an ideal scenario with sending an object on a trajectory. In fact, the escape velocity stated here should actually be called escape speed due to the fact that the quantity to be calculated is completely independent of direction. Notice that the equation does not include the mass of the object escaping a large body as escape velocity is only dependent on gravitational force. We also assume that an object is escaping from a uniform body.

A Mathematical Model

When we model escape velocity or speed, we have to assume that an object must have a velocity an infinite distance away from a large body allowing us to find the very minimum speed to escape that large body a certain distance away. "Escape velocity" is the speed needed to go from some distance away from a large body to an infinite distance, ending at infinity with a final speed of zero. This dismisses any initial acceleration. This means that a modern spacecraft for example with propellers does not follow these assumptions.

[math]\displaystyle{ (K + U_g)_i = (K + U_g)_f \, }[/math]

K_ƒ = 0 because final velocity is zero, and U_gƒ = 0 because its final distance is expressed as infinity, therefore

[math]\displaystyle{ \frac{1}{2}mv_e^2 + \frac{-GMm}{r} = 0 + 0 }[/math]

[math]\displaystyle{ v_e = \sqrt{\frac{2GM}{r}} }[/math]

A Computational Model

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