An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the escape speed from the earth. If R is the radius of the earth then the height of the satellite above the surface of the earth is

The correct answer is Option R.

Let's derive the solution step-by-step.

Step 1: Understand the formulas for escape velocity and orbital velocity.
The escape speed ($v_e$) from the surface of the Earth is given by the formula:
$v_e=\sqrt{\frac{2GM}{R}}$
where:
- $G$ is the universal gravitational constant,
- $M$ is the mass of the Earth,
- $R$ is the radius of the Earth.

The orbital speed ($v_o$) of a satellite in a circular orbit at a height $h$ above the surface of the Earth is given by the formula:
$v_o=\sqrt{\frac{GM}{R+h}}$
where $R+h$ is the orbital radius (distance from the center of the Earth to the satellite).

Step 2: Relate the two velocities using the given condition.
According to the problem, the orbital speed of the satellite is equal to half the escape speed from the Earth:
$v_o=\frac{1}{2}{v}_e$

Substituting the expressions for $v_o$ and $v_e$ into the equation:
$\sqrt{\frac{GM}{R+h}}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$

Step 3: Solve for the height ($h$).
Squaring both sides of the equation to eliminate the square roots, we get:
$\frac{GM}{R+h}=\frac{1}{4}\left(\frac{2GM}{R}\right)$

We can cancel out $GM$ from both sides of the equation:
$\frac{1}{R+h}=\frac{2}{4R}$

Simplifying the fraction on the right side:
$\frac{1}{R+h}=\frac{1}{2R}$

Taking the reciprocal of both sides:
$R+h=2R$

Subtracting $R$ from both sides to find $h$:
$h=2R-R$
$h=R$

Thus, the height of the satellite above the surface of the Earth is equal to the radius of the Earth, $R$.