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Problem 64

Escape Velocity The minimum velocity required for an object to escape Earth's gravitational pull is obtained from the solution of the equation

$\int v d v=-G M \int \frac{1}{y^{2}} d y$

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

$v^{2}=v_{0}^{2}+2 G M\left(\frac{1}{y}-\frac{1}{R}\right)$

where $v_{0}$ is the initial velocity of the object and $R$ is the radius of Earth.

Answer

Please see explanation

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## Discussion

## Video Transcript

So we're starting with the girl of being. You sleep well to the anti derivatives. Sorry and single. Too negative. Jean Time's intent. Derivatives of one over. Why squared do you want? Well, for the left hand side, This is going to be the squared over to minus the initial. They got squared two. And for the right hand side, you have negative chief times. Save the girl. Why? To the negatives too, Teo, to think all negative G m times negative. Why? To the negative one. My house. Negative. Why not to may of one. And that's equal GM times one over. Why minus one were are because ours the initial there we go hope for homes.

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