The work done by the Sun on the Earth at any finite interval of time is (a) positive, negative or zero (b) str

V. Conceptual Questions · Q3

Q. The work done by the Sun on the Earth at any finite interval of time is

(a) positive, negative or zero

(b) strictly positive

(d) it is always zero

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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.
It depends on height, not path — lifting straight up or along a ramp stores the same energy (if friction is ignored).
It converts to kinetic energy when the object falls: $mgh = \frac{1}{2}mv^2$ (ignoring air resistance).
The formula $U = mgh$ is for near-Earth problems only. For orbital mechanics, use $U = -GMm/r$ .

Tip

In numerical problems, always write $\Delta U = mg\Delta h$ rather than $U = mgh$ — this reminds you that only changes matter and forces you to define your zero level.

A Simple Example

A 5 kg stone is on a cliff 20 m high. Take the cliff base as $h = 0$ .

Potential energy of stone: $U = 5 \times 9.8 \times 20 = 980$ J
If the stone falls, just before hitting the ground, all this becomes kinetic energy: $\frac{1}{2} \times 5 \times v^2 = 980 \implies v = \sqrt{392} \approx 19.8$ m/s

That's the energy of position, converted to energy of motion.

For quick revision, remember that Gravitational Potential Energy is drawn directly from the Gravitation coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers, which is exactly why "Gravitational Potential Energy important questions" shows up so often in Physics question banks. The clearest way to build exam confidence here is to combine this explanation with the NCERT Physics textbook's own solved examples and chapter-end questions.

Step 1. The work done equals $W=U(r_i)-U(r_f)$ , which depends on how $r$ (the Earth-Sun distance) changes between the start and end of the chosen interval.

Step 2. If the interval covers the Earth moving from a larger to a smaller $r$ (approaching the Sun, e.g. aphelion toward perihelion), $W>0$ ; if $r$ increases over the interval (receding from the Sun), $W<0$ .

Step 3. If the interval happens to be exactly one full orbit (or any other interval where $r_i=r_f$ ), $W=0$ , as in the previous question.

Step 4. Since the interval in this question is left arbitrary (any finite interval, not necessarily a full year), all three outcomes are genuinely possible depending on which part of the orbit is chosen.

✓Final answer

(a) positive, negative or zero -- the sign depends entirely on whether the Earth's distance from the Sun decreases, increases, or returns to its starting value over the chosen interval.

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 16 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Relation between orbital velocity (Vo) of a satellite revolving near the earth and escape velocity (Ve) of a body from the earth's surface is$
(a) Ve = √2 Vo
(b) Vo = √2 Ve
(c) Ve = √(Vo/2)
(d) Vo = √(Ve/2)
›Reveal solutionSolution

Ve = sqrt(2) Vo. Answer (A).

Orbital velocity of a satellite close to the earth: Vo = sqrt(gR).

Escape velocity from the earth's surface: Ve = sqrt(2gR).

Taking the ratio: Ve/Vo = sqrt(2gR)/sqrt(gR) = sqrt(2).

So Ve = sqrt(2) Vo.

✓Final answer
$(A) Ve = sqrt(2) Vo.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Gravitational potential at a distance r from a body of mass M is$
(a) GM / r
(b) GM / r^2
(c) -GM / r
(d) -GM / r^2
›Reveal solutionSolution

Gravitational potential V(r) = -GM/r; the negative sign reflects that gravity does positive work as a mass falls inward from infinity (where potential is taken as zero).

Gravitational potential at a point is defined as the work done per unit mass in bringing a test mass from infinity to that point, against/with the gravitational field:

V(r) = -Integral (from infinity to r) of (GM/r^2) dr = -GM/r

The negative sign is essential -- gravity is always attractive, so potential energy (and potential) decreases as you move closer to the mass M, approaching -infinity as r -> 0, and rising to zero as r -> infinity.

✓Final answer
$(c) -GM/r.$
CBSE 2025Set ANNUAL1 mark
$Q. Answer in one word or one sentence: Calculate the gravitational potential energy of a stone of mass 5 kg placed at a height 10 m above the earth surface. (g = 9.8 m/s^2)$
›Reveal solutionSolution

Using PE = mgh, a 5 kg stone at 10 m height has gravitational potential energy of 490 J.

Near the Earth's surface, gravitational potential energy (relative to the surface) is given by PE = mgh, where m is mass, g is acceleration due to gravity, and h is height above the reference level.

Given: m = 5 kg, h = 10 m, g = 9.8 m/s^2.

PE = 5 x 9.8 x 10 = 490 J.

✓Final answer
$The gravitational potential energy of the stone is 490 J.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The gravitational potential inside a hollow sphere (A) is not uniform (B) is uniform (C) is zero (D) none of these$
›Reveal solutionSolution

Gravitational potential inside a hollow sphere is uniform (constant), equal to its surface value.

For a thin uniform spherical shell, the gravitational field at any point inside the cavity is exactly zero (contributions from all parts of the shell cancel). Since $E=-dV/dr$ , a zero field means $V$ does not vary with position inside the shell — the potential is the same everywhere inside, equal to the potential value right at the surface, $V=-GM/R$ . It is not zero, only uniform.

✓Final answer
$(B) is uniform.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The gravitational potential V at a point P, outside a solid sphere is V = -GM/x, where G is gravitational constant and M is mass of the sphere. 'x' represents (A) radius of the sphere (B) distance of point P from the surface of the sphere (C) distance of point P from the centre of the sphere (D) square of distance of point P from the centre of the sphere$
›Reveal solutionSolution

In $V=-GM/x$ , $x$ is measured from the centre of the sphere.

For a uniform solid sphere, treated as a point mass for external points (by the shell theorem), the gravitational potential at a distance $x$ from its centre is $V=-GM/x$ , valid for $x\geq R$ (the sphere's radius). The variable $x$ is always the centre-to-point distance, not the surface-to-point distance, and it is not squared here (that's the field formula, not potential).

✓Final answer
$(C) distance of point P from the centre of the sphere.$
CBSE 2023Set ANNUAL1 mark
$Q. If the earth revolves around the Sun in a circular orbit, then what would be the work done by the gravitational force?$
›Reveal solutionSolution

Work done by a force is $W = \vec{F}\cdot\vec{d}$ ; since gravity on Earth always points radially toward the Sun while Earth's velocity (and hence its displacement) is always tangential to the circular path, the force is always perpendicular to the motion, so $W = 0$ .

For a body moving in a circle, work done by any force is $dW = \vec{F}\cdot d\vec{s} = F\,ds\cos\theta$ , where $\theta$ is the angle between the force and the displacement. The Sun's gravitational pull on the Earth acts along the radius, i.e. it is centripetal — always directed from the Earth toward the Sun. The Earth's instantaneous displacement at any point of a circular orbit is always tangent to the circle, i.e. perpendicular to the radius. So the angle between the gravitational force and the displacement is always $\theta = 90^\circ$ , and $\cos 90^\circ = 0$ . Hence the elementary work done in each small displacement is zero, and integrating over the whole orbit, the total work done by the gravitational force in one revolution (or any part of a truly circular orbit) is zero. This is consistent with the fact that the Earth's speed (and hence its kinetic energy) stays constant in a circular orbit.

✓Final answer
$The work done by the gravitational force is zero, because the force is always perpendicular to the Earth's velocity/displacement in a circular orbit.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Gravitational potential inside a hollow spher is:$
(a) not uniform
(b) uniform
(c) zero
(d) none of these
›Reveal solutionSolution

Inside a hollow sphere the gravitational potential is uniform.

By the shell theorem, the gravitational field inside a uniform hollow sphere is zero everywhere. Since the field E = −dV/dr = 0, the potential V does not change with position: it is constant (uniform) throughout the interior and equals the value at the surface, −GM/R.

✓Final answer
$(B) uniform.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Escape velocity for the object from the earth is:$
(a) 11.2 km/s
(b) 11.2 m/s
(c) 112 km/s
(d) 112 km/h
›Reveal solutionSolution

Earth's escape velocity ≈ 11.2 km/s.

Escape velocity v_e = √(2gR) = √(2GM/R). Using g = 9.8 m/s^2 and R = 6.4 × 10^6 m,

v_e = √(2 × 9.8 × 6.4 × 10^6) ≈ 1.12 × 10^4 m/s = 11.2 km/s.

✓Final answer
$(A) 11.2 km/s.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. If R is the radius of the earth and g is acceleration due to granity, then expression for escape velocity is:$
(a) R/g
(b) 2R/g
(c) √(2Rg)
(d) 2Rg
›Reveal solutionSolution

Escape velocity = √(2gR).

To just escape Earth's gravity, the kinetic energy given must equal the gravitational potential energy magnitude:

(1/2)m v_e^2 = GMm/R = mgR (since g = GM/R^2).

Solving, v_e^2 = 2gR, so v_e = √(2gR) = √(2Rg).

✓Final answer
$(C) \sqrt(2Rg).$
CBSE 2022Set ANNUAL1 markMCQ
$Q. Gravitational force is:$
(a) a conservative force
(b) Pseudo force
(c) a non-conservative force
(d) None of the above
›Reveal solutionSolution

Gravitational force is conservative because the work it does depends only on the start and end points, not the path.

Concept. A force is called conservative if the work done by it in moving an object between two points is independent of the path taken, and depends only on the initial and final positions. Equivalently, the work done over any closed path is zero, and the force can be derived from a potential energy function ( $F = -\dfrac{dU}{dr}$ ).

Gravitational force satisfies this: the work done by gravity in lifting or lowering a mass depends only on the height difference (or the change in $r$ from the source mass), never on the actual trajectory taken. This is exactly why we can define a gravitational potential energy $U = -\dfrac{GMm}{r}$ .

✓Final answer
$Gravitational force is a conservative force.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. A body enters in the gravitational field of the earth from external atmosphere. The potential energy of the earth-body system:$
(a) will increase
(b) will decrease
(c) will remain unchanged
(d) None of the above
›Reveal solutionSolution

As a body falls toward Earth from far away, gravitational PE decreases (becomes more negative) while KE increases.

Reasoning. Gravitational potential energy between Earth (mass M) and a body (mass m) at separation r is:

$U(r) = -\frac{GMm}{r}$

This is taken as zero at $r=\infty$ and becomes increasingly negative as $r$ decreases (i.e. as the body gets closer to Earth).

When a body "enters the gravitational field from the external atmosphere" — moving from far away toward Earth — $r$ decreases, so $U(r)$ becomes more negative, i.e. it decreases. Correspondingly, by conservation of energy, kinetic energy increases as the body speeds up while falling.

✓Final answer
$The potential energy of the earth-body system will decrease as the body falls toward Earth.$
CBSE 2022Set sz1 mark
$Q. Evaluate: the definite integral of (GMm / x^2) dx, taken from x = R to x = infinity (i.e. Integral[R to infinity] (GMm/x^2) dx).$
›Reveal solutionSolution

Integrating GMm/x^2 with respect to x from R to infinity gives GMm/R -- this is the magnitude of work required to take a mass m from the surface (radius R) of a body of mass M to infinity against gravity.

We evaluate:

Integral[R to infinity] (GMm/x^2) dx

Since GMm is a constant, take it outside:

= GMm * Integral[R to infinity] x^-2 dx

= GMm * [ -1/x ] from R to infinity

= GMm * [ (-1/infinity) - (-1/R) ]

= GMm * [ 0 + 1/R ]

= GMm/R

Physically, this quantity is the negative of the gravitational potential energy at distance R (or equivalently, the work done against gravity to move the mass from R to infinity), and it is exactly the expression used while deriving escape velocity, since (1/2)mv_esc^2 = GMm/R.

✓Final answer
$Integral[R to infinity] (GMm/x^2) dx = GMm/R.$

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Q.The work done by the Sun on the Earth at any finite interval of time is

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