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.

COMEDK 2026Set 2026-M1 markMCQ
$v$
›Reveal solutionSolution
The key is to use energy conservation with variable gravity (since height is comparable to Earth’s radius). The required velocity is $v = \sqrt{\frac{20 g R}{11}}$ , which corresponds to option (C).

Concept & Intuition

When a body is projected to a height comparable to Earth’s radius, the gravitational force is no longer constant. Using $g$ as the surface value, we must account for the decrease in gravity with distance. The simplest tool is conservation of mechanical energy: the sum of kinetic and gravitational potential energy at launch equals the sum at the maximum height. The potential energy for a radial distance $r$ from Earth’s center is $-\frac{GMm}{r}$ , not $mgh$ .

Set up the energy conservation equation At the surface ( $r = R$ ), the body has kinetic energy $\frac{1}{2} m v^2$ and potential energy $-\frac{GMm}{R}$ . At the highest point ( $r = R + 10R = 11R$ ), the kinetic energy is zero, and potential energy is $-\frac{GMm}{11R}$ . Energy conservation gives:

$\frac{1}{2} m v^2 - \frac{GMm}{R} = -\frac{GMm}{11R}$

Simplify the potential energy terms Bring the potential terms together:

$\frac{1}{2} m v^2 = \frac{GMm}{R} - \frac{GMm}{11R} = \frac{GMm}{R} \left(1 - \frac{1}{11}\right) = \frac{GMm}{R} \cdot \frac{10}{11}$

Replace $GM$ with $gR^2$ At Earth’s surface, $g = \frac{GM}{R^2}$ , so $GM = gR^2$ . Substitute:
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COMEDK 2025Set 2025-A1 markMCQ
$\rho$
›Reveal solutionSolution
The escape velocity depends only on the radius for a fixed density; since surface area scales as radius squared, planet Q has twice the radius of P, and R has the same radius as Q, so V_R = V_Q > V_P.

The key concept is that for a sphere of uniform density, escape velocity is proportional to the radius. This follows from the formula $v_e = \sqrt{\frac{2GM}{R}}$ and the fact that mass $M = \frac{4}{3}\pi R^3 \rho$ . Substituting gives $v_e = \sqrt{\frac{8\pi G \rho}{3}} \, R$ , so for a fixed density, escape velocity is directly proportional to the radius. Thus, comparing escape velocities reduces to comparing the planets’ radii.

Find the radius of planet P.

Let its radius be $R_P$ . Its surface area is $A = 4\pi R_P^2$ .

Its mass is $M_P = \frac{4}{3}\pi R_P^3 \rho$ .

Find the radius of planet Q.

Its surface area is given as $4A$ . So $4A = 4\pi R_Q^2$ .

But $A = 4\pi R_P^2$ , so $4(4\pi R_P^2) = 4\pi R_Q^2$ → $16\pi R_P^2 = 4\pi R_Q^2$ → $R_Q^2 = 4R_P^2$ → $R_Q = 2R_P$ .

Its mass is $M_Q = \frac{4}{3}\pi (2R_P)^3 \rho = 8 \cdot \frac{4}{3}\pi R_P^3 \rho = 8M_P$ .

Find the radius of planet R.

Its mass is $M_R = M_P + M_Q = M_P + 8M_P = 9M_P$ .

Since density is the same, $M_R = \frac{4}{3}\pi R_R^3 \rho$ and $M_P = \frac{4}{3}\pi R_P^3 \rho$ .

Dividing: $\frac{M_R}{M_P} = \frac{R_R^3}{R_P^3}$ → $9 = \frac{R_R^3}{R_P^3}$ → $R_R = \sqrt[3]{9} \, R_P$ . …
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COMEDK 2025Set 2025-M1 markMCQ
$v$
›Reveal solutionSolution
Escape velocity depends on the ratio of mass to radius. Given the planet is 121 times heavier and 9 times larger in diameter, the moon’s escape velocity is $\frac{3v}{11}$ , so the correct option is (D).

Concept & Intuition

Escape velocity is the minimum speed needed for an object to break free from a celestial body’s gravitational pull. The formula is
$v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$
where $M$ is mass and $R$ is radius. Notice it scales as $\sqrt{M/R}$ . So if we know how mass and radius compare between two bodies, we can directly find the ratio of their escape velocities — no need for actual values.

Step-by-step reasoning

Set up the ratio

Let the moon’s mass be $M_m$ and its radius $R_m$ . The planet’s mass is $M_p = 121 M_m$ . The planet’s diameter is 9 times the moon’s, so its radius is also 9 times: $R_p = 9 R_m$ .

Write escape velocities

For the planet:

$v = \sqrt{\frac{2G M_p}{R_p}} = \sqrt{\frac{2G (121 M_m)}{9 R_m}}$
For the moon:
$v_m = \sqrt{\frac{2G M_m}{R_m}}$

Find the ratio Divide the planet’s escape velocity by the moon’s: $\frac{v}{v_m} = \sqrt{\frac{121 M_m / (9 R_m)}{M_m / R_m}} = \sqrt{\frac{121}{9}} = \frac{11}{3}$ …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{m}$
›Reveal solutionSolution
The energy required is the difference in total mechanical energy between two circular orbits. For a satellite of mass $m$ in a circular orbit of radius $r$ , total energy is $-\frac{GMm}{2r}$ . The work needed to go from radius $3R$ to $4R$ is $\frac{GMm}{24R}$ , so the correct option is (B).

The key idea is that for a satellite in a circular orbit, the total mechanical energy (kinetic + potential) is half the gravitational potential energy and negative. To move to a higher orbit, you must add energy — exactly the difference between the two total energies. This is not simply the change in potential energy, because the kinetic energy also changes (it decreases as you go higher). The net effect is a smaller positive number than you might guess.

Let’s work it out step by step.

Total energy in a circular orbit For a body of mass $m$ in a circular orbit of radius $r$ around Earth (mass $M$ ), the gravitational force provides the centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$
So kinetic energy is:
$K = \frac12 mv^2 = \frac{GMm}{2r}$
Gravitational potential energy (taking zero at infinity) is:
$U = -\frac{GMm}{r}$
Hence total mechanical energy:
$E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$
This is a standard result: for any circular orbit, $E = -K = \frac12 U$ .

Energy at the lower orbit (radius $3R$ )

$E_1 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$

Energy at the higher orbit (radius $4R$ )

$E_2 = -\frac{GMm}{2(4R)} = -\frac{GMm}{8R}$

Energy required to move from lower to higher orbit This is the change in total energy:

$\Delta E = E_2 - E_1 = \left(-\frac{GMm}{8R}\right) - \left(-\frac{GMm}{6R}\right)$ $\Delta E = -\frac{GMm}{8R} + \frac{GMm}{6R}$
Find a common denominator (24):
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COMEDK 2024Set 2024-E1 markMCQ
$1.69 \times 10^{10} \mathrm{~J}$
›Reveal solutionSolution
For a satellite in a circular orbit, the escape energy is exactly twice the kinetic energy, so the additional kinetic energy needed equals the current kinetic energy: $1.69 \times 10^{10} \, \mathrm{J}$ .

The key idea is the relationship between orbital kinetic energy and escape energy. In a circular orbit, the gravitational potential energy is $-2K$ (where $K$ is the kinetic energy), so the total mechanical energy is $-K$ . To escape, the total energy must become zero (or positive). Therefore, we need to add exactly $K$ more energy — the same amount the satellite already has.

Let’s work through it step by step.

Recall the energy of a satellite in a circular orbit. For a satellite of mass $m$ orbiting Earth (mass $M$ ) at radius $r$ , the gravitational force provides the centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$
Hence the kinetic energy is
$K = \frac12 mv^2 = \frac{GMm}{2r}.$

Find the gravitational potential energy. The potential energy (taking zero at infinity) is

$U = -\frac{GMm}{r}.$
Comparing with $K$ , we see $U = -2K$ .

Total mechanical energy in orbit.

$E_{\text{orbit}} = K + U = K - 2K = -K.$
So the satellite is bound with total energy $-K$ .

Condition for escape. To just escape to infinity (with zero speed at infinity), the total energy must be zero:

$E_{\text{escape}} = 0.$
The satellite already has $E_{\text{orbit}} = -K$ . The additional energy $\Delta E$ needed satisfies
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{X}$
›Reveal solutionSolution
The escape speed depends only on the planet’s mass and radius as $v_e = \sqrt{2GM/R}$ . Given $M_E = 9M_X$ and $R_E = 4R_X$ , the ratio $v_{e,X} / v_{e,E}$ simplifies to $\frac{2}{3}$ , so the correct option is (A).

The key concept here is escape speed — the minimum speed a rocket needs to break free from a planet’s gravitational pull without further propulsion. It comes from equating kinetic energy to gravitational potential energy:
$\frac{1}{2} m v_e^2 = \frac{G M m}{R}$
which gives
$v_e = \sqrt{\frac{2GM}{R}}.$
Notice that escape speed depends only on the planet’s mass and radius, not on the rocket’s mass. So the ratio for two planets is simply
$\frac{v_{e,X}}{v_{e,E}} = \sqrt{\frac{M_X}{M_E} \cdot \frac{R_E}{R_X}}.$
Now let’s work through it step by step.

Write the given data in ratio form.

Earth’s mass is nine times planet X’s mass: $M_E = 9 M_X$ .

Earth’s radius is four times planet X’s radius: $R_E = 4 R_X$ .

Set up the escape speed ratio.

$\frac{v_{e,X}}{v_{e,E}} = \sqrt{\frac{M_X}{M_E} \cdot \frac{R_E}{R_X}}$

Substitute the given ratios.

$\frac{v_{e,X}}{v_{e,E}} = \sqrt{\frac{M_X}{9 M_X} \cdot \frac{4 R_X}{R_X}} = \sqrt{\frac{1}{9} \cdot 4}$

Simplify the square root. $\sqrt{\frac{4}{9}} = \frac{2}{3}$ …
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COMEDK 2023Set 2023-M1 markMCQ
$11.1 \mathrm{~kmh}^{-1}$
›Reveal solutionSolution

Escape velocity scales as $\sqrt{M/R}$ . With $M_m=M_e/81$ and $R_m=R_e/4$ the ratio is $\sqrt{4/81}=2/9$ , giving $v_{moon}=11.1\times 2/9\approx2.46$ .

Escape velocity: $v_e=\sqrt{\dfrac{2GM}{R}}$ , so
$\frac{v_{moon}}{v_{earth}}=\sqrt{\frac{M_m}{M_e}\cdot\frac{R_e}{R_m}}=\sqrt{\frac{1}{81}\cdot 4}=\sqrt{\frac{4}{81}}=\frac{2}{9}.$
Therefore …
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COMEDK 2022Set 20221 markMCQ
$Q. The escape velocity of a projectile on the earth's surface is 11.2 km/s. A body is projected out with thrice this speed. The speed of the body far away from the earth will be (A) 22.4 km/s (B) 31.7 km/s (C) 33.6 km/s (D) None of these$
›Reveal solutionSolution

With u = 3 v_e: v_inf = sqrt(9 v_e^2 - v_e^2) = v_e sqrt(8) = 2 sqrt(2) v_e = 2.828 x 11.2 = 31.7 km/s

Concept: energy conservation. A body projected with speed u > v_e retains kinetic energy far from the Earth.

(1/2)m u^2 - (1/2)m v_e^2 = (1/2) m v_inf^2

v_inf = sqrt(u^2 - v_e^2)

With u = 3 v_e: …
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COMEDK 2021Set 20211 markMCQ
$\mathrm{PE}=r(\mathrm{KE})$
›Reveal solutionSolution

Therefore PE = -2 x (GMm/2r) = -2 (KE). Comparing with PE = r(KE): r = -2.

Concept: energetics of a satellite in a circular orbit of radius r.

Gravitational force provides the centripetal force:

GMm/r^2 = mv^2/r => mv^2 = GMm/r

KE = (1/2)mv^2 = GMm/(2r)

PE = -GMm/r …
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KCET 2019Set A-11 markMCQ
$K$
›Reveal solutionSolution
For a satellite in a low circular orbit, the total mechanical energy is $-K$ , so the extra kinetic energy needed to reach escape speed (zero total energy) is exactly $K$ .

The key idea here is the relationship between kinetic energy, potential energy, and total mechanical energy for a satellite in a circular orbit. When a satellite orbits close to Earth, its orbit is nearly circular, and we can use the standard orbital energy equations.

For a satellite of mass $m$ in a circular orbit of radius $r$ (where $r$ is approximately Earth's radius $R$ for a low orbit), the gravitational force provides the centripetal acceleration:
$\frac{GMm}{r^2} = \frac{mv^2}{r}$
From this, the orbital speed is $v = \sqrt{\frac{GM}{r}}$ , and the kinetic energy is:
$K = \frac{1}{2}mv^2 = \frac{GMm}{2r}$
The gravitational potential energy of the satellite is:
$U = -\frac{GMm}{r}$
So the total mechanical energy is:
$E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} = -K$
This is a crucial result: for a circular orbit, the total energy is negative and exactly equal to $-K$ .

Now, to "just overcome" Earth's gravity means the satellite should reach escape speed — the speed at which it can go to infinity with zero kinetic energy remaining. At escape speed, the total mechanical energy becomes zero (kinetic energy exactly balances the negative potential energy at that point). So the satellite needs to go from total energy $-K$ to total energy $0$ .

Let's work through the steps:

Current state: The satellite has kinetic energy $K$ and total energy $E = -K$ . It is bound to Earth.

Target state for escape: The satellite needs total energy $E' = 0$ . This means its new kinetic energy $K'$ must satisfy $K' + U = 0$ , so $K' = -U = \frac{GMm}{r} = 2K$ . …
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KCET 2018Set A-11 markMCQ
$q$
›Reveal solutionSolution

Combine $qV = \frac12 mv^2$ with $r = mv/(qB)$ to get $r \propto \sqrt{V}$ ; doubling $r$ therefore needs $4V$ .

Step 1 — Speed gained in the accelerating field.

All the electrical work done on the charge becomes kinetic energy:
$qV = \frac{1}{2}mv^{2} \;\Rightarrow\; v = \sqrt{\frac{2qV}{m}}$
Step 2 — Radius in the magnetic field.

Entering $\vec B$ normally, the magnetic force supplies the centripetal force:
$qvB = \frac{mv^{2}}{r} \;\Rightarrow\; r = \frac{mv}{qB}$
Step 3 — Eliminate $v$ .
$r = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$
With $m$ , $q$ and $B$ all fixed, the only variable is $V$ :
$\boxed{r \propto \sqrt{V}}$
Step 4 — Apply the required change. …
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KCET 2018Set A-11 markMCQ
$v_E$
›Reveal solutionSolution

Escape speed depends on the distance from the centre of the Earth as $v \propto 1/\sqrt{r}$ ; at a height $h = R$ the distance is $2R$ , so the escape speed falls by a factor $\sqrt2$ .

Step 1 — Derive the escape speed (why the formula).

To just escape, the body's total mechanical energy must be zero (it reaches infinity with zero speed):
$\frac12 m v^{2} - \frac{GMm}{r} = 0 \;\Longrightarrow\; v_{esc}(r) = \sqrt{\frac{2GM}{r}} .$
Note it depends on $r$ , the distance from the centre of the Earth, not on the height alone.

Step 2 — Escape speed at the Earth's surface.

Here $r = R$ :
$v_E = \sqrt{\frac{2GM}{R}} \quad (\approx 11.2\ \mathrm{km\,s^{-1}}).$
Step 3 — Escape speed at the space station.

The station is at height $h = R$ , so its distance from the centre is
$r = R + h = R + R = 2R.$ $v_{station} = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{GM}{R}} .$
Step 4 — Take the ratio. …
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Q.Show the nature of the following graph for a satellite orbiting the earth.

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