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The formula $v^2 = u^2+2GS$(where $v $ is final velocity, $u $ is initial velocity, $G $ is acceleration due to gravity and $S $ is the linear displacement) assumes that gravity remains constant throughout. What is the actual formula which takes into consideration the change in acceleration with height. Im in 9th grade and I don't know integration so with the formula I wish to derive the formula for escape velocity.

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  • $\begingroup$ Hi, one suggestion: identify the meaning of your variables when you post an equation. Makes it much easier for people to know what you're asking about. $\endgroup$ Jan 21, 2016 at 14:06

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You don't need acceleration. Just calculate change in potential energy with altitude and make sure your initial Kinetic energy is sufficient that your final velocity is greater than zero. You cannot use your formula without integration.

As for actual formula it would be $$g= GM/r^2$$ where M is mass of earth and r is varying distance.

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  • $\begingroup$ Not possible with your formula without integration. $\endgroup$ Jan 21, 2016 at 7:52
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The formula for acceleration at a given height above the Earth would be: \begin{equation} a = -gR^2/r^2 \end{equation} where R is the radius of the earth and r is the radius you are at. This is because gravitation varies with the square of the distance between two objects. I'm afraid you no need calculus to be able to derrive escape velocity due to the fact that the acceleration is not constant or linear. I will tell you for free though if you integrate the gravitational force that you feel from a planet to infinitely far away you get: \begin{equation} GM/R \end{equation} Where G is the gravitational constant, M the mass of the earth and R the radius you start at. Interestingly that can be re written in the form: \begin{equation} gR \end{equation} This is purely because integrating \begin{equation} (-1/r^2) dr \end{equation} gives \begin{equation} (1/r) dr \end{equation}

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