Given:
The gravitational constant, G=6.67259×10⁻¹¹ N·m²/kg⁻²
The mass of the moon, M=7.36×10²² kg
The radius of the orbit of the satellite, R=955.2 km=955.2×10³ m
To find:
1. Satellite's speed.
2. The period of satellite.
3. The acceleration of the satellite.
Explanation:
1.
The gravitational force applied by the moon on the satellite provides the satellite with the centripetal force that is neccessary for the satellite to orbit the moon.
Thus,
[tex]\begin{gathered} F_c=F_g \\ \frac{mv^2}{R}=\frac{GMm}{R^2} \\ v=\sqrt{\frac{GM}{R}} \end{gathered}[/tex]Where F_c is the centripetal force, F_g is the gravitational force, and v is the orbital velocity of the satellite.
On substituting the known values,
[tex]\begin{gathered} v=\sqrt{\frac{6.67259\times10^{-11}\times7.36\times10^{22}}{955.2\times10^3}} \\ =2267.46\text{ m/s} \end{gathered}[/tex]2.
The orbital period of the satellite in seconds is given by,
[tex]T=\frac{2\pi R}{v}[/tex]Thus the period of the satellite in hours is given by,
[tex]T=\frac{2\pi R}{v\times3600}[/tex]On substituting the known values,
[tex]\begin{gathered} T=\frac{2\pi\times955.2\times10^3}{2267.46\times3600} \\ =0.74\text{ hr} \end{gathered}[/tex]3.
The acceleration of the satellite is given by,
[tex]a=\frac{v^2}{R}[/tex]On substituting the known values,
[tex]\begin{gathered} a=\frac{2267.46^2}{955.2\times10^3} \\ =5.38\text{ m/s}^2 \end{gathered}[/tex]Final answer:
1. The orbital velocity of the satellite is 23267.46 m/s
2. The period of the satellite is 0.74 hr
3. The acceleration of the satellite is 5.38 m/s²
