A missile is launched with a velocity less then the escape velocity. The sum of its kinetic and potential energy is

The correct option is Negative.

Let us understand this step-by-step:

1. Total Mechanical Energy: The total mechanical energy ($E$) of a missile in the gravitational field of the Earth is the sum of its kinetic energy ($K$) and gravitational potential energy ($U$).
$E=K+U$

2. Potential Energy: The gravitational potential energy of an object of mass $m$ at a distance $r$ from the center of the Earth (mass $M$) is given by:
$U=-\frac{GMm}{r}$
where $G$ is the universal gravitational constant. This potential energy is negative because the gravitational force is attractive, and the reference point of zero potential energy is at infinity.

3. Escape Velocity: Escape velocity ($v_e$) is the minimum speed needed for an object to escape from the gravitational influence of a primary body. At this velocity, the object can reach infinity with zero kinetic energy. The total energy required to just escape is zero:
$E_{\mathrm{escape}}=K_{\mathrm{escape}}+U=0$
which gives:
$\frac{1}{2}m{v}_e^2-\frac{GMm}{r}=0$

4. Launching with Velocity Less Than Escape Velocity: If the missile is launched with a velocity $v$ that is less than the escape velocity ($v<v_e$), its kinetic energy will be less than the kinetic energy required to escape:
$K=\frac{1}{2}m{v}^2<\frac{1}{2}m{v}_e^2$
Since $\frac{1}{2}m{v}_e^2=\frac{GMm}{r}$, we have:
$K<\frac{GMm}{r}$
which can be rewritten as:
$K<\left|U\right|$

5. Conclusion: Therefore, the magnitude of the negative potential energy is greater than the positive kinetic energy.
$E=K+U=K-\frac{GMm}{r}<0$
This means the total energy of the system is negative, representing a bound state where the missile cannot escape Earth's gravitational field.