A satellite is orbiting a planet in a circular orbit with a radius of 7.0×106
 m. The satellite has a constant angular velocity of 1.7×10−3
rad/s.
 
T = 2πω

What is the period of its orbit?
A.5236
 s
B.3696
 s
C.2859
 s
D.8765
 s

To find the period \( T \) of the satellite's orbit, we can use the formula:

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

where \( \omega \) is the angular velocity.

Given:
\[
\omega = 1.7 \times 10^{-3} \, \text{rad/s}
\]

Substituting the value into the formula:

\[
T = \frac{2\pi}{1.7 \times 10^{-3}}
\]

Calculating this gives:

\[
T \approx \frac{6.2832}{1.7 \times 10^{-3}} \approx 3696 \, \text{s}
\]

Thus, the period of the satellite's orbit is approximately **3696 s**.

The correct answer is **B. 3696 s**.

Do you want further details or have any questions? Here are some related questions for you:

1. How does the radius of the orbit affect the period?
2. What would happen to the period if the angular velocity increased?
3. How can we relate this concept to satellites in geostationary orbit?
4. What are the gravitational forces acting on the satellite in this orbit?
5. How do we calculate the centripetal acceleration for the satellite? 

**Tip:** Remember that in circular motion, the period is inversely related to angular velocity; increasing the velocity decreases the period.

A satellite is orbiting a planet in a circular orbit with a radius of 7.0×10^6 m. The satellite has a constant angular velocity of 1.7×10−3 rad/s. What is the period of its orbit?

Discover how to calculate the period of a satellite orbiting a planet with a radius of 7.0×10^6 m and an angular velocity of 1.7×10−3 rad/s. The calculated period is approximately 3696 seconds. Ideal for high school students studying circular motion.

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