Question:medium

How does the orbital period of a satellite change as its distance from the Earth's center increases?

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Remember: Kepler’s 3rd Law → $T^2 \propto r^3$.
Updated On: Mar 17, 2026
  • $T \propto r$
  • $T \propto r^2$
  • $T \propto r^{3/2}$
  • $T \propto \frac{1}{r}$
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The motion of a satellite orbiting the Earth is governed by the gravitational force acting as the necessary centripetal force.
The relationship between the orbital period (\(T\)) and the orbital radius (\(r\)) is formulated in Kepler's Third Law of Planetary Motion, which applies to any satellite orbiting a central massive body.
Step 2: Key Formula or Approach:
Kepler's Third Law states that the square of the orbital period is directly proportional to the cube of the semi-major axis of its orbit. For a circular orbit of radius \(r\), this is written as:
\[ T^2 \propto r^3 \]
Step 3: Detailed Explanation:
To find the direct proportionality of \(T\) with respect to \(r\), we take the square root of both sides of Kepler's Third Law:
\[ \sqrt{T^2} \propto \sqrt{r^3} \]
\[ T \propto \left(r^3\right)^{1/2} \]
\[ T \propto r^{3/2} \]
This shows that as the distance \(r\) from the Earth's center increases, the orbital period \(T\) increases non-linearly according to the \(3/2\) power law.
Step 4: Final Answer:
The correct relationship is \(T \propto r^{3/2}\).
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