Question

The image I've attached holds the question.What is the satellite’s speed?Answer in units of m/s.What is the period of the satellite’s orbit?Answer in units of h.

Answer 1

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.

Step-by-step 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,

`\begin{gathered} F_c=F_g \\ (mv^2)/(R)=(GMm)/(R^2) \\ v=\sqrt{(GM)/(R)} \end{gathered}`

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,

`\begin{gathered} v=\sqrt{(6.67259*10^(-11)*7.36*10^(22))/(955.2*10^3)} \\ =2267.46\text{ m/s} \end{gathered}`

2.

The orbital period of the satellite in seconds is given by,

`T=(2\pi R)/(v)`

Thus the period of the satellite in hours is given by,

`T=(2\pi R)/(v*3600)`

On substituting the known values,

`\begin{gathered} T=(2\pi*955.2*10^3)/(2267.46*3600) \\ =0.74\text{ hr} \end{gathered}`

3.

The acceleration of the satellite is given by,

`a=(v^2)/(R)`

On substituting the known values,

`\begin{gathered} a=(2267.46^2)/(955.2*10^3) \\ =5.38\text{ m/s}^2 \end{gathered}`

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