Question

A satellite is launched into a circular orbit of radius \( R \) around the earth. A second satellite is launched into an orbit of radius \( 1.03R \). The time period of revolution of the second satellite is larger than the first one approximately by:

• A. 3 %
• B. 4.5 %
• C. 9 %
• D. 2.5 %

Hint:

The time period of a satellite in orbit is related to the radius of the orbit by \( T^2 \propto R^3 \). This allows us to calculate how changes in the orbital radius affect the time period.

Answer 1

The Correct Option is B

Given: A satellite is launched into a circular orbit of radius \( R \) around the Earth. A second satellite is launched into an orbit of radius \( 1.03R \). We are asked to find how much larger the time period of revolution of the second satellite is compared to the first one.

Step 1: Understanding the relationship between time period and orbit radius

The time period \( T \) of a satellite in a circular orbit around a planet is given by Kepler's third law, which can be expressed as: \[ T = 2\pi \sqrt{\frac{R^3}{GM}}, \] where: - \( T \) is the time period of revolution, - \( R \) is the radius of the orbit, - \( G \) is the gravitational constant, and - \( M \) is the mass of the Earth. This formula shows that the time period is proportional to the \( 3/2 \) power of the orbit radius: \[ T \propto R^{3/2}. \]

Step 2: Compare the time periods of the two satellites

Let \( T_1 \) be the time period of the first satellite with orbit radius \( R \), and \( T_2 \) be the time period of the second satellite with orbit radius \( 1.03R \). Using the proportionality, we can write: \[ \frac{T_2}{T_1} = \left(\frac{R_2}{R_1}\right)^{3/2} = \left(\frac{1.03R}{R}\right)^{3/2} = (1.03)^{3/2}. \]

Step 3: Calculate the percentage increase in time period

Now, calculate \( (1.03)^{3/2} \): \[ (1.03)^{3/2} \approx 1.045. \] Therefore, the time period of the second satellite is approximately 1.045 times the time period of the first satellite, or about a 4.5% increase. \[ \text{Percentage increase} = 1.045 - 1 = 0.045 \times 100 = 4.5\%. \]

Final Answer:

The time period of revolution of the second satellite is larger than the first one approximately by \( \boxed{4.5\%} \).