Gravitation NEET 2021: Maximum Height Problem Solution

Q: What is the maximum height in the gravitation NEET 2021 question?

The maximum height h reached by the particle is given by h = R k^2 / (1 - k^2). This is derived by applying conservation of mechanical energy between the Earth's surface and the apogee where velocity becomes zero.

Q: How do you solve the gravitation NEET 2021 maximum height problem?

Set initial total energy equal to final total energy. Initial kinetic energy plus potential at surface equals potential at R + h. Substituting v = k Ve and Ve^2 = 2GM/R simplifies to k^2 = h/(R + h), which rearranges directly to h = R k^2 / (1 - k^2).

Q: What are common mistakes in the gravitation NEET 2021 question?

Students often flip the sign when subtracting gravitational potential terms or perform incorrect cross-multiplication after reaching k^2 = h/(R + h). These errors produce distractors with 1 + k^2 or 1 + k in the denominator instead of 1 - k^2.

Q: What velocity reaches height R in related gravitation NEET questions?

To reach maximum height h = R, the launch velocity must be Ve / √2. Substituting h = R into the derived formula gives 1 = k^2 / (1 - k^2), so 2k^2 = 1 and k = 1/√2. This uses exactly the same energy conservation equation.

Why Does the Gravitation NEET 2021 Question on Maximum Height Demand Attention?

This gravitation NEET 2021 question tests application of mechanical energy conservation for a particle projected vertically at speed $v = k V_{e}$ where $k < 1$ . It appeared in NEET 2021 as one of the 45 compulsory physics questions carrying 180 marks. Students get 120 seconds to solve it in the 3-hour offline OMR exam. The same energy approach builds foundation for variable-gravity problems asked in later years.

What Does the Gravitation NEET 2021 Question Actually Ask?

A particle of mass $m$ is fired from Earth’s surface with speed equal to $k$ times the escape velocity, $k$ less than 1. Determine the maximum height $h$ it reaches above the surface, expressed using Earth’s radius $R$ and the factor $k$ . Four possible algebraic forms are given; the correct one follows from energy conservation.

Following the official sequence produces the formula in under 90 seconds because the algebra simplifies directly once the potential terms are grouped correctly.

How Do You Visualise the Physical Situation Before Starting the Algebra?

All distances for gravitational potential must be measured from the centre of the Earth, not the surface. Initial radial distance is $R$ . At maximum height the radial distance is $R + h$ and velocity drops to zero.

Spherical Earth with centre O, surface at radial distance R from O, particle of mass m launched radially outward from surface with initial velocity arrow labelled v = kVe; at apogee the radial distance from O is R + h where velocity becomes zero; labels include initial position

At maximum height kinetic energy is zero while gravitational potential energy increases from $- G M m / R$ to $- G M m / (R + h)$ . The difference equals the supplied initial kinetic energy.

What Is the Official Step-by-Step Solution for This Gravitation NEET 2021 Question?

Conservation of mechanical energy yields $h = \frac{R k^{2}}{1 - k^{2}}$ . Start with total mechanical energy at the surface and at apogee:

\frac{1}{2} m v^{2} - \frac{G M m}{R} = - \frac{G M m}{R + h}

Cancel $m$ and rearrange:

\frac{v^{2}}{2} = G M (\frac{1}{R} - \frac{1}{R + h}) = \frac{G M h}{R (R + h)}

Substitute $v = k V_{e}$ and $V_{e}^{2} = 2 G M / R$ :

\frac{1}{2} k^{2} (2 G M / R) = \frac{G M h}{R (R + h)}

This simplifies directly to

k^{2} = \frac{h}{R + h}

Rearrange: $R k^{2} = h (1 - k^{2})$

h = \frac{R k^{2}}{1 - k^{2}}

This is option A and matches the official key.

What Precise Algebraic Mistake Produces the Wrong Options in This Gravitation NEET 2021 Question?

Replacing the subtraction $(1 / R - 1 / (R + h))$ with an erroneous sum or inverting the factor before isolating $h$ yields a distractor resembling $R (k / (1 + k))^{2}$ . The error occurs at the step $k^{2} = h / (R + h)$ when cross-multiplication is careless or the binding-energy sign is flipped.

This produces the form with denominator $(1 + k^{2})$ or $(1 + k)$ after substituting $V_{e}$ . Correct isolation of $h$ on one side after $k^{2} = h / (R + h)$ prevents landing on options B, C or D.

Which Related Gravitation Questions Should You Solve Next Using the Same Method?

A particle projected vertically from Earth’s surface with velocity $(1 / \sqrt{2}) V_{e}$ reaches maximum height $R$ . The velocity needed to reach height $R$ is $V_{e} / \sqrt{2}$ . A body launched at $0.8 V_{e}$ has total energy $- 0.36 G M m / R$ , which lies below the zero value required for escape.

A particle is projected vertically from Earth surface with velocity $(1 / \sqrt{2}) V_{e}$ . Find the maximum height reached in terms of $R$ .

Here $k = 1 / \sqrt{2}$ , so $k^{2} = 1 / 2$ . Substitute directly:

h = \frac{R \times (1 / 2)}{1 - 1 / 2} = \frac{R / 2}{1 / 2} = R

Show that the velocity needed to project a body to height $R$ above surface is $V_{e} / \sqrt{2}$ and verify using the same energy equation.

Set $h = R$ in the derived formula:

R = \frac{R k^{2}}{1 - k^{2}}

Cancel $R$ :

1 = \frac{k^{2}}{1 - k^{2}} ⟹ 1 - k^{2} = k^{2} ⟹ 1 = 2 k^{2} ⟹ k = \frac{1}{\sqrt{2}}

Thus $v = V_{e} / \sqrt{2}$ .

If a satellite is given velocity $0.8 V_{e}$ from surface, compare its total energy at launch with the energy required for escape.

Total mechanical energy at launch:

E = \frac{1}{2} m (0.8 V_{e})^{2} - \frac{G M m}{R}

Substitute $V_{e}^{2} = 2 G M / R$ :

E = \frac{1}{2} m (0.64 \times 2 G M / R) - \frac{G M m}{R} = (0.64 - 1) \frac{G M m}{R} = - 0.36 \frac{G M m}{R}

Energy required for escape is zero. The launched body therefore has total energy $0.36 G M m / R$ less than the escape value and remains bound.

What Should You Remember for NEET 2025 and 2026 from This Gravitation NEET 2021 Problem?

Measure gravitational potential from the centre; surface value is $- G M / R$ . Substitute $V_{e}^{2} = 2 G M / R$ before simplifying the fraction $h / (R (R + h))$ . Algebraic rearrangement after $k^{2} = h / (R + h)$ takes 15–20 seconds but is where marks are lost.

Use the past-paper archive on this site to solve all gravitation questions from 2016–2024. That habit removes surprises on exam day.

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Frequently asked questions

What is the maximum height in the gravitation NEET 2021 question?