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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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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