Show the nature of the following graph for a satellite orbiting the earth. (a) KE vs orbital radius R (b) PE v

NCERT Exemplar · Q30

Q. Show the nature of the following graph for a satellite orbiting the earth.

(a) KE vs orbital radius

R

(b) PE vs orbital radius

R

R

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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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For a satellite in a circular orbit around Earth, Kinetic Energy (KE) is positive and inversely proportional to the orbital radius $R$ , while Potential Energy (PE) and Total Energy (TE) are negative and also inversely proportional to $R$ . The graphs for KE, PE, and TE versus $R$ are all hyperbolic in nature.

When a satellite orbits the Earth, its energy is a combination of its kinetic energy due to motion and its gravitational potential energy due to its position in Earth's gravitational field. Understanding how these energies change with the orbital radius $R$ is fundamental to orbital mechanics.

The key idea here is that for a stable circular orbit, the gravitational force provides the exact centripetal force required. This balance allows us to express the satellite's speed, and consequently its kinetic energy, in terms of the orbital radius. The potential energy is defined relative to infinity, where it is considered zero.

Let's consider a satellite of mass $m$ orbiting the Earth of mass $M$ in a circular orbit of radius $R$ . $G$ is the universal gravitational constant.

Kinetic Energy (KE) of the Satellite For a satellite in a stable circular orbit, the gravitational force provides the necessary centripetal force. The gravitational force is $F_g = \frac{GMm}{R^2}$ . The centripetal force required for circular motion is $F_c = \frac{mv^2}{R}$ , where $v$ is the orbital speed. Equating these forces:

\frac{GMm}{R^2} = \frac{mv^2}{R}

From this, we can find the square of the orbital speed:

v^2 = \frac{GM}{R}

The kinetic energy (KE) of the satellite is given by $KE = \frac{1}{2}mv^2$. Substituting the expression for $v^2$:

KE = \frac{1}{2}m \left(\frac{GM}{R}\right)

KE = \frac{GMm}{2R}

*   **Nature of the graph for KE vs $R$:**
    *   KE is always positive.
    *   KE is inversely proportional to $R$ ($KE \propto \frac{1}{R}$).
    *   As $R$ increases, KE decreases.
    *   The graph will be a hyperbola in the first quadrant, approaching zero as $R$ tends to infinity.

2. Gravitational Potential Energy (PE) of the Satellite

The gravitational potential energy of a mass $m$ at a distance $R$ from a mass $M$ is defined as the work done to bring the mass $m$ from infinity to that distance $R$ . By convention, potential energy at infinity is taken as zero.

> [!FORMULA]

> The gravitational potential energy is given by:

> $PE = -\frac{GMm}{R}$

* Nature of the graph for PE vs $R$ :

* PE is always negative. This indicates that the satellite is bound to the Earth's gravitational field.

* PE is inversely proportional to $R$ ( $PE \propto -\frac{1}{R}$ ).

* As $R$ increases, the value of PE becomes less negative (i.e., it increases towards zero).

* The graph will be a hyperbola in the fourth quadrant, approaching zero from below as $R$ tends to infinity.

Total Mechanical Energy (TE) of the Satellite The total mechanical energy (TE) of the satellite is the sum of its kinetic energy and potential energy:

TE = KE + PE

Substituting the expressions for KE and PE:

TE = \frac{GMm}{2R} + \left(-\frac{GMm}{R}\right)

TE = \frac{GMm}{2R} - \frac{2GMm}{2R}

$$ TE = -\frac{GMm}{2R} $$ …

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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 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).

…

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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

…

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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. …

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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 potent …

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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 …

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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 o …

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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 con …

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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, …

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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: …

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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 ( $F = -\dfrac{dU}{dr}$ ).

…

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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).

…

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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

…

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Q.Show the nature of the following graph for a satellite orbiting the earth.

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