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Q13-29P

Expert-verifiedFound in: Page 380

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**The figure gives the potential energy function ****U****(****r****) of a projectile, plotted outward from the surface of a planet of radius****. What least kinetic energy is required of a projectile launched at the surface if the projectile is to “escape” the planet**

Minimum kinetic energy required to escape from the surface of planet is $5\times {10}^{9}J$

Potential energy at R_{s }is $5\times {10}^{9}$ J

**The total energy of a particle is the sum of the kinetic and potential energies of the particle. To escape from the gravitational field, the supplied kinetic energy should be equal to the gravitational potential energy of the particle.**

** **

Formula:

Total energy= Kinetic Energy+ Potential Energy

TE = U + KE

Minimum kinetic energy required to escape from the surface of the planet

To escape from the surface of earth TE = 0

It means

$KE+U=0\phantom{\rule{0ex}{0ex}}KE=-U$

But, from the graph, the potential energy on surface of earth is $U=-5\times {10}^{9}J$

So, the least kinetic energy of the particle is $5\times {10}^{9}J$

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