Answer
$5\times 10^9~J~~$ is the least kinetic energy that is required of a projectile launched at the surface if the projectile is to “escape” the planet.
Work Step by Step
At the planet's surface:
$U = -5\times 10^9~J$
To escape the planet:
$K+U \geq 0$
$K \geq -U$
$K \geq 5\times 10^9~J$
$5\times 10^9~J~~$ is the least kinetic energy that is required of a projectile launched at the surface if the projectile is to “escape” the planet.
