A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:

We begin by recalling that the speed of a satellite in a circular orbit of radius $$r$$ around a planet of mass $$M$$ is obtained by equating the gravitational attraction to the centripetal force.

Gravitational force: $$F_g=\dfrac{G M m}{r^{2}}.$$

Centripetal force needed for circular motion: $$F_c=\dfrac{m v_{\text{orbit}}^{2}}{r}.$$

Setting $$F_g=F_c$$ gives

$$\dfrac{G M m}{r^{2}}=\dfrac{m v_{\text{orbit}}^{2}}{r}.$$

The mass $$m$$ of the body cancels out, and one factor of $$r$$ in the denominator cancels as well, so we have

$$\dfrac{G M}{r}=v_{\text{orbit}}^{2}.$$

Taking the square root on both sides, we obtain the orbital speed,

$$v_{\text{orbit}}=\sqrt{\dfrac{G M}{r}}.$$

In this problem the orbit is said to be “low,” meaning its radius is practically the planet’s radius itself, $$r\approx R.$$ Substituting $$r=R$$ we get

$$v_{\text{orbit}}=\sqrt{\dfrac{G M}{R}}.$$

Next, we state the formula for escape velocity. A body will just escape the planet’s gravitational field when its kinetic energy at the surface equals the magnitude of its gravitational potential energy (taking zero potential energy at infinity):

$$\dfrac{1}{2}m v_{\text{escape}}^{2}=\dfrac{G M m}{R}.$$

Again $$m$$ cancels, and solving for $$v_{\text{escape}}$$ we obtain

$$v_{\text{escape}}=\sqrt{\dfrac{2 G M}{R}}.$$

We are asked for the ratio $$\dfrac{v_{\text{orbit}}}{v_{\text{escape}}}.$$ Substituting the two expressions derived above:

$$\dfrac{v_{\text{orbit}}}{v_{\text{escape}}}=\dfrac{\sqrt{\dfrac{G M}{R}}}{\sqrt{\dfrac{2 G M}{R}}}.$$

The common factor $$\sqrt{\dfrac{G M}{R}}$$ in numerator and denominator simplifies, leaving

$$\dfrac{v_{\text{orbit}}}{v_{\text{escape}}}=\dfrac{1}{\sqrt{2}}.$$

Hence, the correct answer is Option A.