When calculating escape speed why is v_f =0

[Saladsamurai]

That's all. I am just trying to derive the reason why when using comservation of mechanical energy $$v_{final}$$ goes to zero?

 

[Doc Al]

The escape speed is the minimum speed to just barely make it to infinity with no energy left. So the final speed is zero.

 

[jedisoccer]

well escape speed is the minimum speed to break the gravitational pull, so if you look at it like a derivative, the minima will b eequal to 0 where you get Vf is 0
i think

can someone help me with my questions as well?
https://www.physicsforums.com/showthread.php?t=167738
and
https://www.physicsforums.com/showthread.php?p=1449317#post1449317

 

[AbedeuS]

Because escape speed is the minimum velocity at distance $$r=x$$ that the object has to be moving in order to break away from the gravitational pull of a planet at a velocity of 0.

As the object moving at speed $$V_{initial}$$ moves from $$r = x$$ to $$r=\frac{1}{0}$$ the force of Earth's gravity decellerated the object (use equation: $$F=ma$$) and this causes the object to, of course, lose speed, the escape velocity therefore is a velocity that puts you JUST out of reach of a planets gravitational field, but fully decellerated, moving faster than the escape velocity will mean that you will have additional, yet reduced speed after escaping the gravitational field.

 

[Saladsamurai]

Because escape speed is the minimum velocity at distance $$r=x$$ that the object has to be moving in order to break away from the gravitational pull of a planet at a velocity of 0.

As the object moving at speed $$V_{initial}$$ moves from $$r = x$$ to $$r=\frac{1}{0}$$ the force of Earth's gravity decellerated the object (use equation: $$F=ma$$) and this causes the object to, of course, lose speed, the escape velocity therefore is a velocity that puts you JUST out of reach of a planets gravitational field, but fully decellerated, moving faster than the escape velocity will mean that you will have additional, yet reduced speed after escaping the gravitational field.

Ahhh...yes. I completely overlooked that little word minimum...This makes perfect sense..except for the whole r=1/0. I could see it being 1/infinity though.

well escape speed is the minimum speed to break the gravitational pull, so if you look at it like a derivative, the minima will b eequal to 0 where you get Vf is 0
i think

can someone help me with my questions as well?
https://www.physicsforums.com/showthread.php?t=167738
and
https://www.physicsforums.com/showthread.php?p=1449317#post1449317

Patience grasshopper. Someone will surely reply to your posts. Please do not go around hijacking threads.

Casey