An artificial satellite of the earth is launched in circular orbit in the equatorial plane of the earth and the satellite is moving from west to east. With respect to a person on the equator, the satellite is completing one round trip in `24 h`. Mass of the earth is `M=6xx10^(24)kg.`For this situation the orbital radius of the satellite is

Options

A. `2.66xx10^(4)km`
B. `6400 km`
C. `36,000 km`
D. `29,600 km`

Explanation

To find the orbital radius of the satellite, we can follow these steps:

### Step 1: Understand the relationship between the orbital period and angular velocity.
The angular velocity (ω) of the satellite can be expressed in terms of its orbital period (T) using the formula:

\[ \omega = \frac{2\pi}{T} \]

Given that the satellite completes one round trip in 24 hours, we convert this time into seconds:

\[ T = 24 \text{ hours} = 24 \times 3600 \text{ seconds} = 86400 \text{ seconds} \]

Now, we can calculate the angular velocity of the satellite:

\[ \omega = \frac{2\pi}{86400} \approx 7.272 \times 10^{-5} \text{ rad/s} \]

### Step 2: Use the formula for angular velocity in terms of gravitational parameters.
The angular velocity of a satellite in circular orbit can also be expressed using the formula:

\[ \omega = \sqrt{\frac{GM}{r^3}} \]

Where:
- \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \text{ m}^3\text{kg}^{-1}\text{s}^{-2} \)
- \( M \) is the mass of the Earth, given as \( 6 \times 10^{24} \text{ kg} \)
- \( r \) is the orbital radius of the satellite.

### Step 3: Set the equations equal to each other and solve for \( r \).
Equating the two expressions for angular velocity, we have:

\[ \frac{2\pi}{86400} = \sqrt{\frac{GM}{r^3}} \]

Squaring both sides gives:

\[ \left(\frac{2\pi}{86400}\right)^2 = \frac{GM}{r^3} \]

Rearranging for \( r^3 \):

\[ r^3 = \frac{GM}{\left(\frac{2\pi}{86400}\right)^2} \]

### Step 4: Substitute the known values into the equation.
Substituting the values of \( G \) and \( M \):

\[ r^3 = \frac{(6.674 \times 10^{-11})(6 \times 10^{24})}{\left(\frac{2\pi}{86400}\right)^2} \]

Calculating the denominator:

\[ \left(\frac{2\pi}{86400}\right)^2 \approx (7.272 \times 10^{-5})^2 \approx 5.293 \times 10^{-9} \]

Now substituting this back into the equation:

\[ r^3 = \frac{(6.674 \times 10^{-11})(6 \times 10^{24})}{5.293 \times 10^{-9}} \]

Calculating \( r^3 \):

\[ r^3 \approx \frac{4.0044 \times 10^{14}}{5.293 \times 10^{-9}} \approx 7.570 \times 10^{22} \]

### Step 5: Calculate \( r \) and convert to kilometers.
Taking the cube root:

\[ r \approx (7.570 \times 10^{22})^{1/3} \approx 4.224 \times 10^{7} \text{ meters} \]

Converting to kilometers:

\[ r \approx 4.224 \times 10^{7} \text{ m} \div 1000 \approx 4.224 \times 10^{4} \text{ km} \]

### Final Answer:
The orbital radius of the satellite is approximately \( 4.224 \times 10^{4} \text{ km} \).

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