Equation for Final velocity: Projectile Fired from Earth

[sweetpete28]

Suppose a projectile is fired upward from the surface of the Earth. If the initial speed of the projectile is v = 12.0 km/s (more than the escape velocity): how fast will it be moving it is so far from Earth that the force of gravity is approximately zero?

The equation I was given to solve is:

1/2 v(initial)^2 - G(Mass of Earth) / radius of Earth = 1/2v(final)^2

I do NOT think this equation is correct. Can someone please help/advise? Many thanks.

 

[PeterO]

Suppose a projectile is fired upward from the surface of the Earth. If the initial speed of the projectile is v = 12.0 km/s (more than the escape velocity): how fast will it be moving it is so far from Earth that the force of gravity is approximately zero?

The equation I was given to solve is:

1/2 v(initial)^2 - G(Mass of Earth) / radius of Earth = 1/2v(final)^2

I do NOT think this equation is correct. Can someone please help/advise? Many thanks.

"so far from Earth that the force of gravity is approximately zero" means basically at infinite separation.

At all times, the sum of Kinetic and Potential energy is constant.

Kinetic energy is given by 1/2 m v²

Potential energy is given by -GM_Em/R

when r = ∞ , Potential Energy = zero

thus comparing "at the surface" - the initial position - to infinite separation - the final position - we have

1/2 m v_i² + -GM_Em/R_E = 1/2 m v_f²

Once you cancel the common term - the mass of the satellite - you get the equation you were given.

As an aside:
If you wanted to find out how fast it was traveling 4 Earth radii above the surface [which means a centre - centre separation of 5R_E] you would use:

1/2 m v_i² + -GM_Em/R_E = 1/2 m v_f²+ -GM_Em/5R_E