A rocket is fired vertically with a speed of $5\text{ km s}^{-1}$ from the earth's surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth $= 6.0 \times 10^{24}\text{ kg}$; mean radius of the earth $= 6.4 \times 10^{6}\text{ m}$; $G = 6.67 \times 10^{-11}\text{ N m}^{2}\text{ kg}^{-2}$.

A rocket is fired vertically with a speed of 5 km s⁻¹ from the earth's surface. How far from the earth does th

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Concept understanding — Gravitational Potential Energy

Gravitational Potential Energy

The Intuition: Energy Stored by Height

Imagine holding a heavy book above the floor. Your arm feels tired — that's because you're working against gravity. If you let go, the book falls and gains speed. Where did that motion come from? It came from the position of the book. By lifting it, you stored energy in the Earth–book system. That stored energy is gravitational potential energy.

The higher you lift, the more energy you store. The heavier the object, the more energy you store. This is the core idea: Gravitational potential energy is the energy an object has because of its position in a gravitational field.

The Precise Definition

Gravitational potential energy ( $U$ ) is the work done against gravity to bring an object from a reference point (usually the ground) to its current position.

For objects near the Earth's surface (where gravity is roughly constant), the formula is beautifully simple:

U = mgh

Where:

$U$ = gravitational potential energy (joules, J)
$m$ = mass of the object (kg)
$g$ = acceleration due to gravity (≈ 9.8 m/s² on Earth)
$h$ = height above the reference point (m)

Why "Potential"?

The word "potential" means "stored and ready to be used." The book at height $h$ has the potential to do work — it can smash a table, compress a spring, or generate sound when it hits the ground. That energy was put in when you lifted it.

The Reference Point is Arbitrary

Here's a crucial point: Only changes in gravitational potential energy matter. You can choose any height as $h = 0$ . In most problems, we take the ground as zero, but you could take the floor, the tabletop, or even the ceiling.

If you lift a 2 kg book from the floor ( $h = 0$ ) to a shelf ( $h = 2$ m), the change in potential energy is:

\Delta U = mg\Delta h = 2 \times 9.8 \times 2 = 39.2 \text{ J}

If you instead took the shelf as $h = 0$ , the book on the floor would have negative potential energy ( $-39.2$ J). The difference between the two positions is still 39.2 J — that's what matters.

Watch out

Never say "the object has $mgh$ energy" without specifying the reference level. The value is meaningless without a zero point.

The Bigger Picture: Variable Gravity

The formula $U = mgh$ works only when $g$ is constant — that is, near Earth's surface. For large distances (like a rocket leaving Earth), gravity weakens with distance. The general formula for gravitational potential energy between two masses $M$ and $m$ separated by distance $r$ is:

U = -\frac{GMm}{r}

The negative sign means that potential energy is zero at infinite separation and becomes more negative as objects come closer. This is the true definition, and $U = mgh$ is a special case of it (derived by approximating near the surface).

Key Takeaways for Exams

Gravitational potential energy is always relative — you must state or imply a reference level. …

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By energy conservation the rocket's launch kinetic energy converts fully into gravitational potential energy. It reaches a distance $r_{\max} = 8.0 \times 10^{6}\ \text{m}$ from the earth's centre, i.e. a height $h = 1.6 \times 10^{6}\ \text{m}$ (1600 km) above the earth's surface.

Concept

Gravity is conservative, so total mechanical energy is conserved between launch (speed $v$ , at the surface $r=R$ ) and the highest point (speed $0$ , at $r = r_{\max}$ ). The gravitational potential energy of a mass $m$ at distance $r$ from the earth's centre is $U = -\dfrac{GMm}{r}$ .

Energy conservation

\frac{1}{2}mv^{2} - \frac{GMm}{R} = -\frac{GMm}{r_{\max}}

The rocket's mass $m$ cancels, so the result is independent of the rocket's mass:

\frac{GM}{r_{\max}} = \frac{GM}{R} - \frac{1}{2}v^{2} \qquad\Rightarrow\qquad r_{\max} = \frac{GM}{\dfrac{GM}{R} - \dfrac{1}{2}v^{2}}

Substitute the values

$v = 5\ \text{km s}^{-1} = 5000\ \text{m s}^{-1}$ , $M = 6.0\times10^{24}\ \text{kg}$ , $R = 6.4\times10^{6}\ \text{m}$ , $G = 6.67\times10^{-11}\ \text{N m}^{2}\text{kg}^{-2}$ .

GM = (6.67\times10^{-11})(6.0\times10^{24}) = 4.002\times10^{14}\ \text{m}^{3}\,\text{s}^{-2}

$\frac{GM}{R} = \frac{4.002\times10^{14}}{6.4\times10^{6}} = 6.253\times10^{7}\ \text{J kg}^{-1}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$A$
›Reveal solutionSolution

Since $v_e=\sqrt{2gR}$ , $g\propto v_e^2/R$ , giving $g_A:g_B=16:27$ .

Escape velocity from a planet of radius $R$ with surface gravity $g$ is:
$v_e=\sqrt{2gR}\quad\Rightarrow\quad g=\frac{v_e^2}{2R}\quad\Rightarrow\quad g\propto \frac{v_e^2}{R}$
Taking the ratio for planets A and B: …
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KEAM 2026Set eng-2026-04194 marksMCQ
$m$
›Reveal solutionSolution
$|U|=GMm/r$ and $K=GMm/2r$ , giving a $2:1$ ratio.

For a satellite in orbit of radius $r$ :

Gravitational potential energy magnitude: $|U|=\dfrac{GMm}{r}$ .

Kinetic energy: $K=\dfrac12 mv^2=\dfrac{GMm}{2r}$ (since $v^2=GM/r$ ). …
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KEAM 2026Set eng-2026-04214 marksMCQ
$kg$
›Reveal solutionSolution

Gravitational PE magnitude scales as $1/(R+h)$ ; scaling from $10^7$ to $1.2\times10^7$ m gives $|U_2|=5\times10^6$ J.

$|U|=\dfrac{GMm}{R+h}$ , so $\dfrac{|U_2|}{|U_1|}=\dfrac{R+h_1}{R+h_2}$ .

$R+h_1=6.4\times10^6+3.6\times10^6=10^7$ m; $R+h_2=6.4\times10^6+5.6\times10^6=1.2\times10^7$ m. …
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KEAM 2026Set pha-2026-0418F4 marksMCQ
$1 : \sqrt{2}$
›Reveal solutionSolution

$v_e \propto R\sqrt{\rho}$ ; with $\rho_1:\rho_2 = 2:1$ and $R_1:R_2 = 1:2$ , ratio $= 1:\sqrt2$ . …

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KEAM 2026Set pha-2026-0419F4 marksMCQ
$Q. When a satellite moves from the farthest stable orbit to the nearest stable orbit towards the earth, its gravitational potential energy (A) increases (B) becomes zero (C) remains constant (D) decreases (E) first increases then decreases$
›Reveal solutionSolution

$U = -GMm/r$ ; a smaller $r$ gives a more negative $U$ , so the gravitational potential energy decreases.

The gravitational potential energy of a satellite of mass $m$ at distance $r$ from the Earth's centre is

$U = -\frac{GMm}{r}.$ …
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KEAM 2025Set eng-2025-04254 marksMCQ
$Q. The ratio between the gravitational potential energies of 1 kg of mass on the surface of two planets having masses and radii in the ratio 1 : 2 and 1 : 3 respectively, is (A) 1 : 1 (B) 2 : 3 (C) 3 : 2 (D) 9 : 4 (E) 4 : 9$
›Reveal solutionSolution

Gravitational PE of a mass on a planet's surface scales as $M/R$ . With masses $1:2$ and radii $1:3$ , the ratio is $(1/2)(3/1)=3:2$ .

Gravitational potential energy of mass $m$ on a planet's surface:
$U = -\frac{GMm}{R} \quad\Rightarrow\quad |U| \propto \frac{M}{R}.$
Form the ratio for the two planets (masses $M_1:M_2 = 1:2$ , radii $R_1:R_2 = 1:3$ ): …
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KEAM 2025Set eng-2025-04264 marksMCQ
$v_o$
›Reveal solutionSolution

The satellite orbits at radius $2R$ (height $R$ above the surface). Comparing its orbital speed with the surface escape speed gives $v_o = v_e/2$ .

Orbital velocity at orbital radius $r = R + R = 2R$ :
$v_o = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{GM}{2R}}$
Escape velocity from the earth's surface:
$v_e = \sqrt{\dfrac{2GM}{R}}$
Take the ratio: …
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KEAM 2025Set eng-2025-04274 marksMCQ
$\sqrt{2}$
›Reveal solutionSolution

Escape speed $\sqrt{2gR}$ is $\sqrt2$ times the orbital speed $\sqrt{gR}$ for a low circular orbit.

The escape velocity from a planet's surface and the orbital velocity of a satellite skimming the surface are

$v_{esc}=\sqrt{2gR},\qquad v_{orb}=\sqrt{gR}.$ …
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KEAM 2025Set eng-2025-04284 marksMCQ
$K$
›Reveal solutionSolution

Satellite energetics: $K=\tfrac{GMm}{2r}$ , $U=-\tfrac{GMm}{r}$ , so $E=K+U=-K$ .

For a satellite in a circular orbit of radius $r=R+h$ :
$K=\frac{GMm}{2r},\qquad U=-\frac{GMm}{r}.$
The total mechanical energy is …
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KEAM 2025Set eng-2025-04294 marksMCQ
$1\,m$
›Reveal solutionSolution

Gravitational PE $U=-\dfrac{Gm_1m_2}{r}$ ; with $m_1=m_2=1\,\text{kg}$ , $r=1\,\text{m}$ , $U=-G$ .

Formula. The mutual gravitational potential energy of two point masses is

$U = -\frac{Gm_1m_2}{r}.$ …
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KEAM 2024Set pha-2024-06104 marksMCQ
$G = gravitational constant, g = acceleration due to gravity$
›Reveal solutionSolution

Gravitational PE of two masses $m$ separated by $r$ . …

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KEAM 2023Set eng-2023-P1-A14 marksMCQ
$10$
›Reveal solutionSolution
From $v_{esc} = \sqrt{2gR}$ , the surface gravity is $g = 5\ \mathrm{m/s^2}$ .

Concept and Intuition

Escape speed relates to surface gravity and radius through $v_{esc} = \sqrt{2gR}$ . Rearranging isolates g in terms of the known escape speed and radius.

Step-by-Step Solution

Convert: $v_{esc} = 10\ \mathrm{km/s} = 10^4\ \mathrm{m/s}$ , $R = 10^4\ \mathrm{km} = 10^7\ \mathrm{m}$ . …
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Concept

Energy conservation

Substitute the values

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