Choose the correct alternative: (a) If the zero of potential energy is at infinity, the total energy of an orb

Exercises · 7.6

Q. Choose the correct alternative:

(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.

(b) The energy required to launch an orbiting satellite out of earth's gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth's influence.

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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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Gravitational potential energy is defined as negative when zero is at infinity, so total energy of a bound orbit is negative and equals half the potential energy (or the negative of kinetic energy). Launching a satellite already in orbit requires less additional energy than launching a stationary object at the same height, because the satellite already has kinetic energy.

Why this approach works

The key idea is that gravitational potential energy is defined relative to a chosen zero. When we set zero at infinity, any object bound to Earth (like a satellite in orbit) has negative total energy — it doesn't have enough energy to escape to infinity. The total energy $E$ of a satellite in a circular orbit is the sum of its kinetic energy $K$ and potential energy $U$ . Because the gravitational force provides the centripetal force, $K$ and $U$ are related in a fixed way. For the second part, we compare the additional energy needed to push an object from a given height to infinity — a satellite already has orbital speed, so it needs less of a boost than a stationary object at the same height.

Step-by-step solution

Part (a): Total energy of an orbiting satellite

Write the expressions for kinetic and potential energy. For a satellite 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}

From this, the orbital speed is $v = \sqrt{\frac{GM}{r}}$ , so the kinetic energy is:

K = \frac12 mv^2 = \frac{GMm}{2r}

Potential energy with zero at infinity. The gravitational potential energy of the satellite at distance $r$ from Earth's centre is:

U = -\frac{GMm}{r}

(Negative because the force is attractive and zero is at infinity.)

Total energy is the sum.

E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}

Relate $E$ to $K$ and $U$ . From above:

E = -\frac{GMm}{2r} = -\frac12 \left( \frac{GMm}{r} \right) = \frac12 U

Also, since $K = \frac{GMm}{2r}$ , we have $E = -K$ .

E = -K = \frac12 U

So the total energy is the negative of its kinetic energy (and half of its potential energy). The correct choice for part (a) is: total energy is negative of its kinetic energy.

Watch out

A common mistake is to think total energy equals potential energy. Remember: $U$ is twice as large in magnitude as $E$ , and $E$ is negative for a bound orbit.

Part (b): Energy required to escape — satellite vs. stationary object

Energy needed to escape from a given height. To escape Earth's gravity completely, an object must have total energy $\ge 0$ (since zero at infinity is the escape threshold). If an object is at distance $r$ from Earth's centre, its minimum total energy to escape is $0$ . So the additional energy needed is: $\Delta E = 0 - E_{\text{initial}}$ …

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

TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
$700~\mathrm{J}$
›Reveal solutionSolution
For a rolling sphere, total kinetic energy splits into translational and rotational parts. When it slips without rolling at the same speed, only the translational part remains. The answer is $500~\mathrm{J}$ .

The key idea here is that rolling without slipping means the sphere has both translational kinetic energy (due to its centre-of-mass motion) and rotational kinetic energy (due to spinning about its centre). When it slips without rolling — that is, it slides without rotating — the rotational part vanishes, and only the translational part survives. So we need to find how much of the $700~\mathrm{J}$ is translational.

For a solid sphere, the moment of inertia about its centre is $I = \frac{2}{5} M R^2$ . When it rolls without slipping, the angular speed $\omega$ and the centre-of-mass speed $v$ are related by $v = \omega R$ .

Write the total kinetic energy for rolling without slipping. The total KE is the sum of translational and rotational parts:

$K_{\text{total}} = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2.$
Substitute $I = \frac{2}{5} M R^2$ and $\omega = v/R$ :
$K_{\text{total}} = \frac{1}{2} M v^2 + \frac{1}{2} \left( \frac{2}{5} M R^2 \right) \left( \frac{v}{R} \right)^2 = \frac{1}{2} M v^2 + \frac{1}{5} M v^2.$

Combine the terms.

$K_{\text{total}} = \left( \frac{1}{2} + \frac{1}{5} \right) M v^2 = \frac{5+2}{10} M v^2 = \frac{7}{10} M v^2.$
So the total kinetic energy when rolling is $\frac{7}{10} M v^2$ , and we are told this equals $700~\mathrm{J}$ :
$\frac{7}{10} M v^2 = 700 \quad \Rightarrow \quad M v^2 = 1000.$

Find the translational kinetic energy alone. The translational part is $\frac{1}{2} M v^2$ . Using $M v^2 = 1000$ :

$K_{\text{trans}} = \frac{1}{2} \times 1000 = 500~\mathrm{J}.$

Interpret the slipping case. …
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TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
$2R$
›Reveal solutionSolution
The time period of a satellite in a circular orbit is determined by equating the gravitational force to the centripetal force. By substituting the given orbital radius and relating $GM$ to $g$ , the time period is found to be $\boxed{\sqrt{\dfrac{32\pi^2 R}{g}}}$ .

When a body revolves around the Earth in a circular orbit, it is continuously accelerating towards the center of the Earth. This acceleration is called centripetal acceleration, and it requires a centripetal force. In the case of an orbiting body, this necessary centripetal force is provided by the gravitational attraction between the Earth and the body. By equating these two forces, we can derive an expression for the orbital velocity or the time period of revolution.

The key idea is that for a stable circular orbit, the gravitational force acting on the satellite must exactly match the centripetal force required to keep it in that orbit.

Identify the forces involved:

Gravitational Force ( $F_g$ ): This is the attractive force between the Earth (mass $M$ ) and the orbiting body (mass $m$ ). According to Newton's Law of Universal Gravitation, it is given by $F_g = \frac{GMm}{r^2}$ , where $G$ is the universal gravitational constant and $r$ is the distance from the center of the Earth to the orbiting body.

Centripetal Force ( $F_c$ ): This is the force required to keep the body moving in a circular path. It is directed towards the center of the circle and is given by $F_c = \frac{mv^2}{r}$ , where $v$ is the orbital speed. Alternatively, in terms of the time period $T$ , the orbital speed is $v = \frac{2\pi r}{T}$ , so $F_c = m \left(\frac{2\pi r}{T}\right)^2 \frac{1}{r} = \frac{4\pi^2 m r}{T^2}$ .

Equate the gravitational force to the centripetal force:

For a stable orbit, these two forces must be equal:

$F_g = F_c$ $\frac{GMm}{r^2} = \frac{4\pi^2 m r}{T^2}$

Solve for the time period ( $T$ ): We can cancel the mass of the orbiting body ( $m$ ) from both sides and rearrange the equation to solve for $T^2$ :

$\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2}$ $T^2 = \frac{4\pi^2 r^3}{GM}$
Taking the square root of both sides gives the time period:
$T = \sqrt{\frac{4\pi^2 r^3}{GM}}$ $T = 2\pi \sqrt{\frac{r^3}{GM}}$
> [!FORMULA] > The time period of a satellite in a circular orbit of radius $r$ around a body of mass $M$ is $T = 2\pi \sqrt{\frac{r^3}{GM}}$. This is a direct consequence of Kepler's Third Law. …
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TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$h_1$
›Reveal solutionSolution
The acceleration due to gravity at a height $h$ above Earth’s surface is $g_h = g_0 \frac{R_E^2}{(R_E + h)^2}$ . Given the ratio $g_{h_1} : g_{h_2} = 16:9$ and $h_1 = 2R_E$ , we solve for $h_2$ and find $h_2 = 3R_E$ , which corresponds to option (B).

Concept & Intuition

The acceleration due to gravity decreases as we move away from Earth’s surface because the gravitational force follows an inverse-square law with distance from the Earth’s center. At a height $h$ above the surface, the distance from the center is $R_E + h$ . So the gravity at that height is $g_h = \frac{GM}{(R_E + h)^2}$ . Comparing two heights means comparing these inverse squares. The given ratio $16:9$ tells us that the gravity at $h_1$ is larger than at $h_2$ , so $h_1$ must be smaller than $h_2$ . Since $h_1 = 2R_E$ is already large, $h_2$ will be even larger — but we need the exact value.

Step-by-step solution

Write the formula for gravity at height $h$ The acceleration due to gravity at a height $h$ above Earth’s surface is:

$g_h = \frac{GM}{(R_E + h)^2}$
where $G$ is the gravitational constant and $M$ is Earth’s mass.

At the surface ( $h=0$ ), $g_0 = \frac{GM}{R_E^2}$ . So we can also write:
$g_h = g_0 \cdot \frac{R_E^2}{(R_E + h)^2}$

Set up the given ratio We are told:

$\frac{g_{h_1}}{g_{h_2}} = \frac{16}{9}$
Substitute the expression:
$\frac{g_0 \cdot \frac{R_E^2}{(R_E + h_1)^2}}{g_0 \cdot \frac{R_E^2}{(R_E + h_2)^2}} = \frac{16}{9}$
The $g_0$ and $R_E^2$ cancel, leaving:
$\frac{(R_E + h_2)^2}{(R_E + h_1)^2} = \frac{16}{9}$

Take square roots Since all distances are positive, we take the positive square root:
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TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$\frac{2M\pi^2 R^2}{T^2} + \frac{4M r^2 \pi^2}{5T^2}$
›Reveal solutionSolution

Add orbital (translational) KE and spin (rotational) KE of the Moon: $\dfrac{2M\pi^2R^2}{T^2}+\dfrac{4Mr^2\pi^2}{5T^2}$ .

Orbital (translational) kinetic energy. The Moon's orbital speed is $v=\dfrac{2\pi R}{T}$ , so
$K_{\text{orb}}=\frac12 M v^2=\frac12 M\left(\frac{2\pi R}{T}\right)^2=\frac{2M\pi^2R^2}{T^2}.$
Spin (rotational) kinetic energy. Treating the Moon as a uniform solid sphere, $I=\dfrac{2}{5}Mr^2$ , and it spins with $\omega=\dfrac{2\pi}{T}$ : …
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TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$L$
›Reveal solutionSolution
Angular momentum depends on moment of inertia and angular velocity; using the relations for kinetic energy and frequency, the new angular momentum becomes one-quarter of the original, so the answer is (D).

The key is to connect angular momentum $L$ to the given changes in frequency and kinetic energy. For a particle in uniform circular motion, angular momentum is $L = I\omega$ , where $I = mr^2$ is the moment of inertia and $\omega = 2\pi f$ is the angular velocity. Kinetic energy is $K = \frac12 I\omega^2$ . By expressing $L$ in terms of $K$ and $\omega$ (or $f$ ), we can see how it scales when those quantities change.

Write the basic relations. For a particle of mass $m$ moving in a circle of radius $r$ with angular speed $\omega$ :

$L = I\omega = mr^2\omega, \quad K = \frac12 I\omega^2 = \frac12 mr^2\omega^2.$

Express angular momentum in terms of kinetic energy and angular velocity. From $K = \frac12 I\omega^2$ , we have $I = \frac{2K}{\omega^2}$ . Substituting into $L = I\omega$ :

$L = \left(\frac{2K}{\omega^2}\right)\omega = \frac{2K}{\omega}.$
Since $\omega = 2\pi f$ , this becomes
$L = \frac{2K}{2\pi f} = \frac{K}{\pi f}.$
So $L$ is directly proportional to $K$ and inversely proportional to $f$ .

Apply the given changes. Frequency is doubled: $f' = 2f$ . Kinetic energy is halved: $K' = \frac12 K$ . The new angular momentum is
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TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
$m$
›Reveal solutionSolution
Six pairs (two at $3l_0$ , two at $4l_0$ , two at the $5l_0$ diagonal) give $|U| = \left(\frac{2}{3}+\frac{1}{2}+\frac{2}{5}\right)\frac{Gm^2}{l_0} = \frac{47}{30}\frac{Gm^2}{l_0}$ .

Concept — gravitational potential energy of a system. The total PE is the sum over all distinct pairs:
$U = -\sum_{\text{pairs}} \frac{Gm_im_j}{r_{ij}}$
Step 1 — enumerate the pairs. Four equal masses $m$ at the corners of a $3l_0 \times 4l_0$ rectangle give $\binom{4}{2} = 6$ pairs:

2 pairs separated by the shorter sides: $r = 3l_0$

2 pairs separated by the longer sides: $r = 4l_0$

2 pairs along the diagonals: $r = \sqrt{(3l_0)^2+(4l_0)^2} = 5l_0$

Step 2 — sum the pair energies. …
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TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
$400\,\mathrm{K}$
›Reveal solutionSolution
The radiated power scales as $T^4$ (Stefan–Boltzmann law). Doubling the absolute temperature multiplies the power by $2^4 = 16$ , giving $16000\,\mathrm{W}$ .

The key idea here is the Stefan–Boltzmann law: the total power radiated by a blackbody (or any body that radiates thermally) is proportional to the fourth power of its absolute temperature. That means if you double the temperature, the power doesn’t just double — it goes up by a factor of $2^4 = 16$ . This is a classic result in thermal radiation, and it’s the only physics you need for this problem.

Let’s walk through it step by step.

State the Stefan–Boltzmann law For a body at absolute temperature $T$ , the radiated power $P$ is given by

$P = \sigma \varepsilon A T^4$
where $\sigma$ is the Stefan–Boltzmann constant, $\varepsilon$ is the emissivity (a number between 0 and 1), and $A$ is the surface area. For the same body, $\sigma$ , $\varepsilon$ , and $A$ are constant — they don’t change when you heat it up.

Set up the ratio Since everything except $T$ stays the same, the power is simply proportional to $T^4$ :

$P \propto T^4$
So for two different temperatures $T_1$ and $T_2$ ,
$\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$

Plug in the numbers Here $T_1 = 400\,\mathrm{K}$ , $T_2 = 800\,\mathrm{K}$ , and $P_1 = 1000\,\mathrm{W}$ . …
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Q.Choose the correct alternative:

Why this approach works

Step-by-step solution

Part (a): Total energy of an orbiting satellite

Part (b): Energy required to escape — satellite vs. stationary object

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