Two satellites P and Q are moving in different circular orbits around the Earth (radius R). The heights of P and Q from the Earth surface are h_P and h_Q, respectively, where h_P = $\frac{R}{3}$. The accelerations of P and Q due to Earth’s gravity are g_P and g_Q, respectively. If $\frac{g_P}{g_Q}=\frac{36}{25}$, what is the value of h_Q?

The correct option is $\frac{3R}{5}$.

Step-by-Step Explanation:

1. Formula for Acceleration due to Gravity at a Height:
The acceleration due to gravity at a height h above the surface of the Earth of radius R is given by the formula:
$g\left(h\right)=\frac{GM}{{(R+h)}^2}$
where G is the gravitational constant and M is the mass of the Earth.

2. Expressing g_P and g_Q:
For satellite P, the height is h_P = $\frac{R}{3}$. The distance from the center of the Earth is:
$r_P=R+h_P=R+\frac{R}{3}=\frac{4R}{3}$

Therefore, the acceleration due to gravity at satellite P is:
$g_P=\frac{GM}{{\left(r_P\right)}^2}=\frac{GM}{{\left(\frac{4R}{3}\right)}^2}$

For satellite Q, the height is h_Q, so its distance from the center of the Earth is:
$r_Q=R+h_Q$

Therefore, the acceleration due to gravity at satellite Q is:
$g_Q=\frac{GM}{{(R+h_Q)}^2}$

3. Taking the Ratio of Accelerations:
The ratio of the accelerations is given as:
$\frac{g_P}{g_Q}={\left(\frac{R+h_Q}{R+h_P}\right)}^2$

Substitute the given values into the equation:
$\frac{36}{25}={\left(\frac{R+h_Q}{\frac{4R}{3}}\right)}^2$

4. Solving for h_Q:
Taking the square root on both sides:
$\frac{6}{5}=\frac{R+h_Q}{\frac{4R}{3}}$

Multiply both sides by $\frac{4R}{3}$:
$R+h_Q=\frac{6}{5}\times \frac{4R}{3}=\frac{8R}{5}$

Subtract R from both sides to find h_Q:
$h_Q=\frac{8R}{5}-R=\frac{3R}{5}$

Thus, the value of h_Q is $\frac{3R}{5}$.