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David Z
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Escape How does escape velocity relate to energy and speed?

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Bernhard
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I have several confusions regarding escape velocity. I am sure I am missing something(s) obvious or maybe I am taught wrong.

  1. Lets say we throw an object of any mass at exactly escape velocity of earth calculated from $v^2=\frac{2GM}{r}$ which is almost $11Km s^{-1} $$11 \text{km s}^{-1} $ but iI am talking about exact escape speed.
    That That ball initially has $KE=\frac{1}{2}mv^2$ and $PE=mgh$. Wikipedia says and iI quote

In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.

How is that possible? 2. As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower speed will never reaches zero. Am I right? 3. A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.

  1. As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower speed will never reaches zero. Am I right?
  2. A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.

I have several confusions regarding escape velocity. I am sure I am missing something(s) obvious or maybe I am taught wrong.

  1. Lets say we throw an object of any mass at exactly escape velocity of earth calculated from $v^2=\frac{2GM}{r}$ which is almost $11Km s^{-1} $ but i am talking about exact escape speed.
    That ball initially has $KE=\frac{1}{2}mv^2$ and $PE=mgh$ Wikipedia says and i quote

In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.

How is that possible? 2. As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower speed will never reaches zero. Am I right? 3. A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.

I have several confusions regarding escape velocity. I am sure I am missing something(s) obvious or maybe I am taught wrong.

  1. Lets say we throw an object of any mass at exactly escape velocity of earth calculated from $v^2=\frac{2GM}{r}$ which is almost $11 \text{km s}^{-1} $ but I am talking about exact escape speed. That ball initially has $KE=\frac{1}{2}mv^2$ and $PE=mgh$. Wikipedia says and I quote

In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.

How is that possible?

  1. As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower speed will never reaches zero. Am I right?
  2. A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.
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Qmechanic
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I have several confusions regarding escape velocityescape velocity. I am sure I am missing something(s) obvious or maybe I am taught wrong.

  1. Lets say we throw an object of any mass at exactly escape velocity of earth calculated from $v^2=\frac{2GM}{r}$ which is almost $11Km s^{-1} $ but i am talking about exact escape speed.
    That ball initially has $KE=\frac{1}{2}mv^2$ and $PE=mgh$ Wikipedia says and i quote

In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.

How is that possible? 2. As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower speed will never reaches zero. Am I right? 3. A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.

Thanks for your help.

I have several confusions regarding escape velocity. I am sure I am missing something(s) obvious or maybe I am taught wrong.

  1. Lets say we throw an object of any mass at exactly escape velocity of earth calculated from $v^2=\frac{2GM}{r}$ which is almost $11Km s^{-1} $ but i am talking about exact escape speed.
    That ball initially has $KE=\frac{1}{2}mv^2$ and $PE=mgh$ Wikipedia says and i quote

In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.

How is that possible? 2. As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower speed will never reaches zero. Am I right? 3. A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.

Thanks for your help.

I have several confusions regarding escape velocity. I am sure I am missing something(s) obvious or maybe I am taught wrong.

  1. Lets say we throw an object of any mass at exactly escape velocity of earth calculated from $v^2=\frac{2GM}{r}$ which is almost $11Km s^{-1} $ but i am talking about exact escape speed.
    That ball initially has $KE=\frac{1}{2}mv^2$ and $PE=mgh$ Wikipedia says and i quote

In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.

How is that possible? 2. As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower speed will never reaches zero. Am I right? 3. A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.

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Suchal
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