What Determines the Escape Speed from Earth's Gravitational Pull?

[hadronthunder]

Hi friends.

Somewhere in a reference book I read about escape speed of a particle for earth.
Let a particle is projected from the Earth surface. Let its mass be m and speed of projection be u. Let mass of Earth be M and its radius be R.

According to law of conservation of energy,
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The problem is that,

If the Ist term becomes negative also but its magnitude is less than the IInd term, then also final Kinetic energy will be positive. And the particle will never doesn't give the proper answer. Friend isn't it so?
Please help me in understanding this.
Thank you all in advance.

 

[Simon Bridge]

The first term in what? You mean term I in the first relation?
That's kinetic energy - does it make physical sense to have a negative kinetic energy?

 

[hadronthunder]

The first term in what? You mean term I in the first relation?
That's kinetic energy - does it make physical sense to have a negative kinetic energy?

The first term is in the second relation. the complete 1/2(mu²) - (GMm)R

The complete term can be negative due to less value of 1/2 (mu²).

 

[Bandersnatch]

For escape velocity you're comparing the total energy at the surface of a massive body and at infinity from it.

This means that the second term \frac{GMm}{R+h} will always be zero. So the first term can not be less in magnitude than the second one.If the first term becomes negative due to too low a value of initial velocity U, it simply means that the velocity U is too low to escape the gravity of the massive body.