64. Escape Velocity The minimum velocity required for an object to escape Earth's gravitational pull is obtained from the solution of the equation [v dv = - GMS 1/2 dy where v is the velocity of the object projected from Earth, y is the distance from the center of Earth, G is the gravitational constant, and M is the mass of Earth. Show that v and y are related by the equation v² = √² + 2GM (1 - 1) y R where vo is the initial velocity of the object and R is the radius of Earth.

College Physics
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Chapter1: Units, Trigonometry. And Vectors
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64. Escape Velocity The minimum velocity required for an
object to escape Earth's gravitational pull is obtained from the
solution of the equation
[v dv = - GMS ²/3 d
where v is the velocity of the object projected from Earth, y
is the distance from the center of Earth, G is the gravitational
constant, and M is the mass of Earth. Show that v and y are
related by the equation
y² = √² + 2GM (1 - 1/2)
R
where v is the initial velocity of the object and R is the radius
of Earth.
Transcribed Image Text:64. Escape Velocity The minimum velocity required for an object to escape Earth's gravitational pull is obtained from the solution of the equation [v dv = - GMS ²/3 d where v is the velocity of the object projected from Earth, y is the distance from the center of Earth, G is the gravitational constant, and M is the mass of Earth. Show that v and y are related by the equation y² = √² + 2GM (1 - 1/2) R where v is the initial velocity of the object and R is the radius of Earth.
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