Question

The figure gives the potential energy functionU(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

Short answer

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

Step-by-step solution

The given data

Potential energy at R_sis $5\times {10}^9$ J

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

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

$$\begin{gathered} KE+U=0 \\ KE=-U \end{gathered}$$

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 $5\times {10}^{9}J$