Q.40 Physics ⭐ 360° Option Elimination Active

NEET 2023 Physics Question 40 Solution

Topic: Gravitation · Official NTA Booklet Code G5

Question

A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity 3π/(Gd) represents:

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Verified Against NTA Final Answer Key

Option (3) : T²

📚 Gravitation Easy

📋 Given Parameters & Problem Setup

Orbital altitude Just above Earth’s surface (r ≈ R)

Orbital period T

Uniform density of Earth d

Universal gravitational constant G

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The 30-Second Exam Hall Shortcut: ⚡ Satellite Density-Period Fundamental: T² · d = 3π / G

⏱️ Save ~40 seconds

T = 2π√(R³/GM) = √(3π / Gd) ⟹ Squaring gives T² = 3π / (Gd). Thus, 3π / (Gd) represents T² in 10 seconds!

1 Recall the orbital period of a surface satellite: T = 2π √(R / g) = 2π √(R³ / GM).

2 Express mass M in terms of density d: M = Volume × density = (4/3) π R³ d.

3 Substitute M into period: T = 2π √[ R³ / (G · (4/3)π R³ d) ] = 2π √[ 3 / (4π G d) ].

4 Bring 2π inside the radical: √(4π² × 3 / 4πGd) = √(3π / Gd).

5 Square both sides: T² = 3π / (Gd).

6 Therefore, the quantity 3π / (Gd) identically equals T².

7 Select Option (3) T²!

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Why You Can 100% Trust This Shortcut: Gravitational Dynamics · Kepler’s Third Law for Near-Surface Orbits (The product of planetary mean density and square of near-surface orbital period is a universal constant: d · T² = 3π/G).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Orbital Velocity and Period of a Near-Surface Satellite

For a satellite orbiting just grazing the surface of a spherical planet of mass M, radius R, and uniform density d, the gravitational attraction provides the necessary centripetal force:

\frac{G M m}{R^2} = \frac{m v_0^2}{R} \implies v_0 = \sqrt{\frac{G M}{R}}

Step 2

Derivation of Orbital Period Formula

The orbital period T is the circumference of the orbit divided by the orbital velocity:

T = \frac{2\pi R}{v_0} = 2\pi R \sqrt{\frac{R}{G M}} = 2\pi \sqrt{\frac{R^3}{G M}}

Step 3

Substitution of Planet Mass in Terms of Density

For a uniform spherical planet, the mass is given by the product of its spherical volume and density d:

M = \frac{4}{3}\pi R^3 d

Step 4

Simplification and Isolation of the Target Expression

Substitute M into the period formula and simplify:

T = 2\pi \sqrt{\frac{R^3}{G \left(\frac{4}{3}\pi R^3 d\right)}} = 2\pi \sqrt{\frac{3}{4\pi G d}} = \sqrt{\frac{4\pi^2 \times 3}{4\pi G d}} = \sqrt{\frac{3\pi}{G d}}

Step 5

Squaring to Match the Target Quantity

Squaring both sides of the derived relation:

T^2 = \frac{3\pi}{G d} \implies \frac{3\pi}{G d} \text{ represents } T^2

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 (1)

✗ Incorrect

Verdict & Scientific Reason:

Incorrect power of period.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate took square root instead of squaring.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

T equals √(3π/Gd), not 3π/Gd.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate forgot that the expression inside the square root was T².

3 (3)

✓ 100% Correct

Verdict & Scientific Reason:

Accurate: Squaring T = √(3π/Gd) yields T² = 3π/(Gd).

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Why this is airtight: None. Candidate correctly squared both sides.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Keplerian power confusion.

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Student Fallacy (Why candidates pick this trap): Candidate confused Kepler’s T² ∝ R³ rule.

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NTA Trap Alert: Confusing T with T² (Forgetting the Radical)

The formula for orbital period contains a square root: T = √(3π / Gd). Consequently, (3π / Gd) without the square root equals T², not T (Option 2).

🛡️ How to avoid the -1 mark: Check dimensions: [G] = M⁻¹L³T⁻², [d] = ML⁻³ ⟹ [Gd] = T⁻² ⟹ 1/[Gd] = T²!

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Core NEET Syllabus Standard

National Eligibility cum Entrance Test (NEET UG) · Gravitation Standard Benchmark

Core Scientific Principle Tested:

"For a satellite orbiting near planetary surface, T = √(3π / Gd), demonstrating period depends solely on planetary density."

Syllabus Focus: Satellites, Orbital Speed and Period

Preserved in Rationalized Curriculum Syllabus & Academic Standards ↗

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NEET 2023 Physics Question 40 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Satellite Density-Period Fundamental: T² · d = 3π / G

Step-by-Step Rigorous Workflow & Mathematical Proof

Orbital Velocity and Period of a Near-Surface Satellite

Derivation of Orbital Period Formula

Substitution of Planet Mass in Terms of Density

Simplification and Isolation of the Target Expression

Squaring to Match the Target Quantity

360° Option Elimination Engine Key Feature

NTA Trap Alert: Confusing T with T² (Forgetting the Radical)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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