Derive the escape velocity from the surface of a planet with radius, r, and mass, M.

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Derive the escape velocity from the surface of a planet with radius, r, and mass, M.

This question is about converting kinetic energy into gravitational potential energy. Escape velocity is the speed required to leave the gravitational field of a mass, in this case it's a planet. In other words the body has to be moving at such a velocity that it will reach a point infinitely far away from the planet.

The gravitational potential energy (PE) at any given point is given as:

PE = -GMm/r

Where G is Newton's gravitational constant; M is the mass of the planet; m is the mass of the body moving away from the planet and r is the distance from the centre of the planet/gravitational field.

Using this formula the potential energy at a distance infinitely far away (infinity) is 0. At the surface of the planet the potential energy is:

-GMm/r

This means that in order to get from the surface to infinity there will be a gain of

GMm/r

This will come from the kinetic energy of the body escaping.

Kinetic Energy (KE) is given as:

KE = mv²/2

Where m is the mass of the moving body and v is it's velocity.Now we have all we need to solve this problem

If we set the kinetic and potential energy equal to each other:

KE = PE

mv²/2 = GMm/r

Divide by m on both sides, this gets rid of all mentions of m. That means the final answer will not depend on the mass leaving the planet at all!

v²/2 = GM/r

Rearrange:

v = sqrt(2GM/r)

sqrt() means take the square root of what is inside the bracket.

Interesting related fact:

A black hole is an object that has an escape velocity that is greater than the speed of light. This means not even light can escape the gravitational pull of a black hole!!!

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