What is the escape speed of a projectile fired from Earth's surface at 8.2 km/s?

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The discussion centers on calculating the maximum height a projectile can reach when fired vertically from Earth's surface at an initial speed of 8.2 km/s. The escape speed of Earth is established as 11.2 km/s, indicating that the projectile will not escape Earth's gravitational influence. Participants recommend using conservation of energy principles, specifically the equations for gravitational potential energy (E_p) and kinetic energy (E_c), to determine the projectile's height. The formula for potential energy is given as E_p=-γ(mM)/(R+h), while kinetic energy is expressed as E_c=mv²/2.

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A projectile if fired vertically from Earth's surface with an initial speed of 8.2 km/s. Neglecting air drag, how far from the survace of Earth will it go?


Well, I know that the escape speed of Earth is 11.2 km/s. how would i go about solving this?
 
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It's not really an escape speed problem... Are you supposed to use the escape speed of the Earth to solve it?

I'd use conservation of energy methods to solve it...
 
use conservation of energy... your final velocity is 0...
the potential energy might a little bit tricky for you... just remind you... PE=mgh doesn't work in this case

PS. flying at 8.2km/s on the Earth and ignore the air drag? that is so not realistic..
 
crap, i am looking at the formula y - yo = voysin(thetao) - gt, i don't know how to put it in this equation, can someone help me?
 
As vincentcahn told you before, you can't use the laws of motion in uniform gravitational field, because the projectil will go far away from the Earth' surface.

Use
E_p=-\gamma \frac{mM}{R+h}
for the potential energy
and
E_c=\frac{mv^2}{2}
for the kinetic one and apply the conservation of total energy.
 
ahhh, ok thanks clive!
 

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