91Ó°ÊÓ

Satellite always remains in the same position over the lunar surface

Satellite will always remain on same position only if the rotational speed of moon and the satellite is same.

As we know time period of moon is 27.3 days, so the rotational period of the satellite should also be 27.3 days.

For the satellite to be in equilibrium on the orbit * doesn't get pulled by moon or thrown out of orbit by centripetal force.

$$\begin{gathered} \frac{m{v}^2}{r}=\frac{GMm}{r^2}........................\left(1\right)\text{and} \\ v=\frac{2×\mathrm{π}×r}{T}....................................\left(2\right) \end{gathered}$$

Where

G= universal gravitational constant =6.647 x 10^-11 m³kg^-1s^-2
M= mass of the moon =7.35 x 10²² kg
m= mass of the satellite
r= radius of the orbit
T = time period =27.3 days =(27.3) x 24 x 60 x 60 s=2358720 s

Solve equation (1) and (2) for r

$$\begin{gathered} Fromeq\left(1\right)v^2=\frac{GM}{r} \\ Substitutevalueofvfromequation\left(2\right) \\ {v}^2={\left(\frac{2×\mathrm{π}×r}{T}\right)}^2=\frac{GM}{r} \\ {\mathrm{r}}^3=\frac{{\mathrm{GMT}}^2}{4{\mathrm{π}}^2} \\ \mathrm{r}={\left(\frac{{\mathrm{GMT}}^2}{4{\mathrm{π}}^2}\right)}^{\frac{1}{3}} \end{gathered}$$

Now substitute the value and calculate for r

role="math" localid="1649596133244" $$\begin{gathered} ⇒r={\left(\frac{(6.647×{10}^{-11}{m}^{3}k{g}^{-1}{s}^{-2})×(7.35×{10}^{22}kg)×(2358720s{)}^2}{4×{\mathrm{π}}^2}\right)}^{\frac{1}{3}} \\ ⇒r=88.5×{10}^6\mathrm{m} \end{gathered}$$

This value of 4 includes from the center of moon so, we have to take out the radius of moon from this value to get the altitude from the surface of moon.

altitude is 88.5 x 10⁶ m - 1.74 x 0⁶ m = 86.76 x 0⁶ m