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Q.42 Physics ⭐ 360° Option Elimination Active
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NEET 2021 Physics Question 42 Solution

Topic: Gravitation · Official NTA Booklet Code M1
NEET 2021 Physics Question 42: Maximum Height Attained by a Particle Projected with Fraction of Escape Speed

Question

A particle of mass 'm' is projected with a velocity v = kVe (k < 1) from the surface of the earth. (Ve = escape velocity) The maximum height above the surface reached by the particle is:

✓
Verified Against NTA Final Answer Key
Option (4) : R k^2 / (1 - k^2)
📚 Gravitation Medium
📋 Given Parameters & Problem Setup
Projection Speed (v) k V_e with k < 1
Escape Speed (V_e) √(2 G M / R)
⚡

The 30-Second Exam Hall Shortcut: ⚡ Height from Escape Fraction: h = R k² / (1 - k²)

⏱️ Save ~25 seconds

Conservation of mechanical energy with v=kVev = k V_e: 12m(k2Ve2)−GMmR=−GMmR+h\frac{1}{2} m (k^2 V_e^2) - \frac{GMm}{R} = -\frac{GMm}{R + h}. Since 12Ve2=GMR\frac{1}{2} V_e^2 = \frac{GM}{R}, this simplifies directly to 1R+h=1−k2R  ⟹  R+h=R1−k2  ⟹  h=Rk21−k2\frac{1}{R+h} = \frac{1-k^2}{R} \implies R+h = \frac{R}{1-k^2} \implies h = \frac{R k^2}{1 - k^2}! Instant 15-second algebraic solution matching Option (4)!

1 Conserve energy: Ei=12mv2−GMmRE_i = \frac{1}{2} m v^2 - \frac{GMm}{R} at surface; Ef=−GMmR+hE_f = -\frac{GMm}{R + h} at peak (v=0v = 0).
2 Use Ve2=2GMR  ⟹  12mv2=12mk2(2GMR)=GMmRk2V_e^2 = \frac{2GM}{R} \implies \frac{1}{2} m v^2 = \frac{1}{2} m k^2 \left(\frac{2GM}{R}\right) = \frac{GMm}{R} k^2.
3 Substitute: GMmRk2−GMmR=−GMmR+h\frac{GMm}{R} k^2 - \frac{GMm}{R} = -\frac{GMm}{R + h}.
4 Divide by −GMm-GMm: 1−k2R=1R+h\frac{1 - k^2}{R} = \frac{1}{R + h}.
5 Invert: R+h=R1−k2  ⟹  h=R1−k2−R=Rk21−k2R + h = \frac{R}{1 - k^2} \implies h = \frac{R}{1 - k^2} - R = \frac{R k^2}{1 - k^2}.
6 Match Option (4).
🛡️
Why You Can 100% Trust This Shortcut: Canonical conservation of mechanical energy in inverse-square central gravitational field.
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Initial Energy at the Surface of the Earth

The particle of mass mm is projected with speed v=kVev = k V_e. The total mechanical energy at the Earth's surface (r=Rr = R) is:

Ei=12mv2−GMmR=12m(kVe)2−GMmRE_i = \frac{1}{2} m v^2 - \frac{GMm}{R} = \frac{1}{2} m (k V_e)^2 - \frac{GMm}{R}
Step 2

Expressing in Terms of Gravitational Potential Energy

Using the escape velocity expression Ve2=2GMRV_e^2 = \frac{2GM}{R}:

12m(k2Ve2)=12mk2(2GMR)=GMmRk2\frac{1}{2} m (k^2 V_e^2) = \frac{1}{2} m k^2 \left(\frac{2GM}{R}\right) = \frac{GMm}{R} k^2
Step 3

Equating with Final Energy at Maximum Height

At maximum height hh above the Earth's surface, the radial distance is r=R+hr = R + h and velocity drops momentarily to zero (Kf=0K_f = 0):

Ef=−GMmR+hE_f = -\frac{GMm}{R + h}
Step 4

Solving for Maximum Height h

By conservation of mechanical energy Ei=EfE_i = E_f:

GMmRk2−GMmR=−GMmR+h  ⟹  −GMmR(1−k2)=−GMmR+h  ⟹  1R+h=1−k2R\frac{GMm}{R} k^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \implies -\frac{GMm}{R}(1 - k^2) = -\frac{GMm}{R + h} \implies \frac{1}{R + h} = \frac{1 - k^2}{R}
Step 5

Final Algebraic Simplification

Inverting both sides and isolating hh:

R+h=R1−k2  ⟹  h=R1−k2−R=R(1−(1−k2)1−k2)=Rk21−k2R + h = \frac{R}{1 - k^2} \implies h = \frac{R}{1 - k^2} - R = R\left(\frac{1 - (1 - k^2)}{1 - k^2}\right) = \frac{R k^2}{1 - k^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:

Has (1−k)2(1-k)^2 in denominator instead of (1−k2)(1-k^2).

⚠️
Student Fallacy (Why candidates pick this trap): Squaring the entire binomial (1−k)(1-k) instead of difference of squares (1−k2)(1-k^2).
2 2
✗ Incorrect
Verdict & Scientific Reason:

Incorrect binomial form in denominator.

⚠️
Student Fallacy (Why candidates pick this trap): Sign error in energy balance.
3 3
✗ Incorrect
Verdict & Scientific Reason:

Dimensionally inconsistent (has dimension [L2][L^2] instead of [L][L]).

⚠️
Student Fallacy (Why candidates pick this trap): Failing to check dimensions of length.
4 4
✓ 100% Correct
Verdict & Scientific Reason:

Exact maximum height derived from gravitational energy conservation.

💡
Why this is airtight: None; flawless energy derivation.
⚠️

NTA Trap Alert: Flat Earth Approximation Trap

Using the uniform gravity approximation h=v2/2gh = v^2 / 2g gives h=k2Rh = k^2 R. This is only valid for small heights (h≪Rh \ll R). For speeds approaching escape velocity, the variation of gravity with height U(r)=−GMm/rU(r) = -GMm/r MUST be used.

🛡️ How to avoid the -1 mark: When speed is given as a fraction of escape velocity, ALWAYS use −GMm/r-GMm/r.
📖

Core NEET Syllabus Standard

NCERT Physics · Gravitation Standard Benchmark
Core Scientific Principle Tested:

"In non-uniform gravitational fields, mechanical energy is conserved: 12mv2−GMmR=−GMmR+h\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}."

Syllabus Focus: Gravitational Potential Energy and Escape Speed
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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