• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Fundamentals Of Physics
Found in: Page 380

Answers without the blur.

Just sign up for free and you're in.


Short Answer

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 5×109 J

See the step by step solution

Step by Step Solution

Step 1: The given data

Potential energy at Rs is 5×109 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.


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 = 0KE =-U

But, from the graph, the potential energy on surface of earth is U=-5×109 J

So, the least kinetic energy of the particle is 5×109 J


Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.