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Physics
Classical Physics
Mechanics
Exploring the Logical/Geometrical Explanation of Escape Velocity
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[QUOTE="vanhees71, post: 6345644, member: 260864"] To escape to infinity the total energy of the body must at least be ##0##, i.e., $$\frac{m}{2} v^2-\frac{G M m}{R} \geq 0,$$ where ##R## is the radius of the Earth, ##M## its mass, and ##G## Newton's gravitational constant; ##v## is the initial velocity which gets $$v \geq v_{\text{escape}}=\sqrt{\frac{2 G M}{R}}=\sqrt{2gR},$$ where ##g=M G/R^2## is the gravitational acceleration at the surface of the earth. What you call ##v_{\text{orbit}}## is the velocity of a body in a circular orbit at radius ##\tilde{R}## (I call it ##\tilde{R}## to distinguish it from the radius of the Earth). Then setting the centripetal force equal to the gravitational force indeed gives $$\frac{m v_{\text{orb}}^2}{\tilde{R}}=\frac{G M}{\tilde{R}^2} \; \Rightarrow\; v_{\text{orb}}=\sqrt{\frac{G M}{\tilde{R}}}.$$ [/QUOTE]
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Forums
Physics
Classical Physics
Mechanics
Exploring the Logical/Geometrical Explanation of Escape Velocity
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