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## Main Question or Discussion Point

I'm doing my HSC physics course and I was playing around with the algebra in space unit and came upon something that confused me (note this isn't required to know by the syllabus except knowing that escape velocity is the velocity to escape from a planet's gravitational field)

Ok, so from the Law of Conservation of Energy we can agree that in any situation: Work done = Kinetic energy = Potential energy.

Ek=W=Ep

To get our escape velocity I imagined a situation where we ignore all air resistance and assumed that the acceleration due to gravity is the same from the surface of a planet down to its centre:

Ep at the surface of the planet = Work done to lift it from the centre = Amount of kinetic energy at its peak velocity.

(Where: mE = mass of Earth, mO= mass of object, rE = radius of Earth)

Since W=Ep, by expanding the formula for work: (mgr = ( mO [GmE/rE^2] rE )

We can simplify it to give:

G mE mO / rE = Ep

since Ek=Ep

( 1/2 mO v^2 ) = ( G mE mO / rE )

Knowing this we can solve for velocity:

v^2 = 2G mE / rE

this is the escape velocity formula.

According to the maths, this should be the highest velocity any object will acquire if dropped from the surface of Earth and accelerated towards the very centre, ignoring air resistance and any change in the value of g.

Therefore, this should also be the velocity required for an object at the centre of Earth to be initially launched at to reach the surface of Earth. Now is where I am "confused": How is this also the velocity required to escape from Earths gravitational field and fly into space without falling back down? How do we know this?

Thank you in advance!

Ok, so from the Law of Conservation of Energy we can agree that in any situation: Work done = Kinetic energy = Potential energy.

Ek=W=Ep

To get our escape velocity I imagined a situation where we ignore all air resistance and assumed that the acceleration due to gravity is the same from the surface of a planet down to its centre:

Ep at the surface of the planet = Work done to lift it from the centre = Amount of kinetic energy at its peak velocity.

(Where: mE = mass of Earth, mO= mass of object, rE = radius of Earth)

Since W=Ep, by expanding the formula for work: (mgr = ( mO [GmE/rE^2] rE )

We can simplify it to give:

G mE mO / rE = Ep

since Ek=Ep

( 1/2 mO v^2 ) = ( G mE mO / rE )

Knowing this we can solve for velocity:

v^2 = 2G mE / rE

this is the escape velocity formula.

According to the maths, this should be the highest velocity any object will acquire if dropped from the surface of Earth and accelerated towards the very centre, ignoring air resistance and any change in the value of g.

Therefore, this should also be the velocity required for an object at the centre of Earth to be initially launched at to reach the surface of Earth. Now is where I am "confused": How is this also the velocity required to escape from Earths gravitational field and fly into space without falling back down? How do we know this?

Thank you in advance!