A particle moves in a circular orbit of radius ' r ' under a central attractive for | PGMN Solutions

A particle moves in a circular orbit of radius ' r ' under a central attractive force, F=-k/r, where k is a constant. The periodic time of its motion is proportional to
(A) r^(1/2)
(B) r^(2/3)
(C) r
(D) r^(3/2)


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🚀 Step-by-Step Solution to Solve This Question 🚀
📌 Chapter: Motion in a Plane
📌 Topic: Central Force Motion

🔹 Step 1: Core Concept
When an object moves in a circle due to a center-pulling force, its orbital period depends on how the force changes with distance. Here, the force weakens as distance increases.

🔹 Step 2: Application
The problem describes a force that gets weaker in exact proportion to distance (if distance doubles, force halves). We need to find how the orbital time changes when the circle's size changes.

🔹 Step 3: Reasoning

Force Balance: The inward pull must match the outward tendency of circular motion. Surprisingly, this balance makes the speed constant regardless of orbit size.

Orbital Time Calculation:

Larger orbits mean longer distances to travel (bigger circumference)

But since speed stays the same, time increases directly with orbit size

🔹 Step 4: Conclusion
The orbital period grows in direct proportion to the orbit's size. Double the radius, and the time for one revolution also doubles.

✅ Final Answer: (C) Radius itself 🚀
🔥 Pro Tip: For inverse-distance force laws, orbital speed stays constant while period scales with size - a unique feature that appears in advanced physics problems!

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