A satellite is arbithng a planet of a fathes of 5500 km from the planet's centen If Its ofbital veloelty $$ 18250 \mathrm{~m} / \mathrm{s} $$ deternine the inme pethat of the sotellite's motlon in hours

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To determine the time period of the satellite's motion, we can use the formula for the orbital period $$ T $$ of a satellite, which is given by:

$$
T = \frac{2\pi r}{v}
$$

where:
- $$ T $$ is the orbital period,
- $$ r $$ is the radius of the orbit (distance from the center of the planet to the satellite),
- $$ v $$ is the orbital velocity.

### Step 1: Extract Known Conditions
- The radius $$ r $$ is given as $$ 5500 $$ km, which we need to convert to meters:
  $$
  r = 5500 \, \text{km} = 5500 \times 1000 \, \text{m} = 5,500,000 \, \text{m}
  $$
- The orbital velocity $$ v $$ is given as $$ 18250 \, \text{m/s} $$.

### Step 2: Calculate the Orbital Period
Now we can substitute the values into the formula:

$$
T = \frac{2\pi (5,500,000)}{18250}
$$

Let's calculate this value.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(2\pi \times 5500000\right)}{18250}$$
- step1: Remove the parentheses:
  $$\frac{2\pi \times 5500000}{18250}$$
- step2: Multiply the terms:
  $$\frac{11000000\pi }{18250}$$
- step3: Reduce the fraction:
  $$\frac{44000\pi }{73}$$
The calculated orbital period $$ T $$ is approximately $$ 1893.56 $$ seconds.

### Step 3: Convert Seconds to Hours
To convert the period from seconds to hours, we use the conversion factor $$ 1 \, \text{hour} = 3600 \, \text{seconds} $$:

$$
T_{\text{hours}} = \frac{T_{\text{seconds}}}{3600} = \frac{1893.56}{3600}
$$

Now, let's calculate this value.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1893.56}{3600}$$
- step1: Convert the expressions:
  $$\frac{\frac{47339}{25}}{3600}$$
- step2: Multiply by the reciprocal:
  $$\frac{47339}{25}\times \frac{1}{3600}$$
- step3: Multiply the fractions:
  $$\frac{47339}{25\times 3600}$$
- step4: Multiply:
  $$\frac{47339}{90000}$$
The orbital period of the satellite in hours is approximately $$ 0.526 $$ hours.

### Final Answer
The time period of the satellite's motion is approximately $$ 0.526 $$ hours.
The satellite's orbital period is approximately 0.526 hours.

A satellite is arbithng a planet of a fathes of 5500 km from the planet's centen If Its ofbital veloelty \( 18250 \mathrm{~m} / \mathrm{s} \) deternine the inme pethat of the sotellite's motlon in hours

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