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:
Show Hint
- 3 %
- 4.5 %
- 9 %
- 2.5 %
The Correct Option is B
Approach Solution - 1
To solve this problem, we need to understand the relationship between the time period of a satellite's orbit and its orbital radius. Kepler's Third Law of Planetary Motion states that the square of the period of revolution \( T \) of a satellite is directly proportional to the cube of the semi-major axis of its orbit (which is the radius \( R \) in the case of a circular orbit):
\(T^2 \propto R^3\)
Thus, for a satellite in orbit, the relationship can be written as:
\(T^2 = kR^3\)
where \( k \) is a constant of proportionality that depends on the gravitational constant and the mass of the Earth.
Let's denote the time period of the first satellite as \( T_1 \) and its orbital radius as \( R \). Thus,
\(T_1^2 = kR^3\)
For the second satellite with orbital radius \( 1.03R \), its time period \( T_2 \) is given by:
\(T_2^2 = k(1.03R)^3\)
Taking the ratio of the two time periods squared:
\(\left(\frac{T_2}{T_1}\right)^2 = \frac{k(1.03R)^3}{kR^3}\)
Simplifying this equation gives:
\(\left(\frac{T_2}{T_1}\right)^2 = (1.03)^3\)
Calculating \((1.03)^3\):
\(1.03^3 = 1.03 \times 1.03 \times 1.03 \approx 1.092727\\)
Thus,
\(\left(\frac{T_2}{T_1}\right)^2 = 1.092727\)
Taking the square root on both sides gives:
\(\frac{T_2}{T_1} = \sqrt{1.092727} \approx 1.045\)
This implies that \( T_2 \) is approximately 1.045 times the period of the first satellite, or 4.5% larger.
So, the time period of revolution of the second satellite is approximately larger than the first one by 4.5%. Hence, the correct answer is:
4.5%
Approach Solution -2
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 \( \boxed{4.5\%} \).
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