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Q13-29P

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Found in: Page 380

### Fundamentals Of Physics

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 $5×{10}^{9}J$

See the step by step solution

## Step 1: The given data

Potential energy at Rs is $5×{10}^{9}$ J

## Step 2: Understanding the concept of energy in a gravitational field

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

## Step 3: Calculation of the least kinetic energy

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 $U=-5×{10}^{9}J$

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