Energy of an orbiting satellite

Every satellite quietly carries a debt it can never pay off. A 1000 kg spacecraft circling the Earth at $7000\ \text{km}$ from the centre is moving at about $7.5\ \text{km/s}$ and has tens of billions of joules of kinetic energy sloshing through it — yet its total mechanical energy is negative. That single minus sign is the whole story of what it means to be "in orbit" rather than "in transit to nowhere": the satellite is gravitationally bound, trapped in a energy well it cannot climb out of without outside help. This is the last idea in the CBSE gravitation chapter, and it is really a bookkeeping exercise built on tools you already have — orbital velocity $v_o=\sqrt{GM/r}$ from the circular-motion condition, and gravitational potential energy $U=-GMm/r$ from earlier in the same chapter. Add the kinetic energy of orbit to the potential energy of separation and something clean falls out: $E=-GMm/2r$ , exactly half the magnitude of $U$ , with the opposite sign to both $K$ and, in a specific ratio, to $U$ itself. That neatness is not an accident of algebra — it is a fingerprint of the inverse-square law, and it is exactly the kind of relation exam-setters love to test with a one-line question that trips up anyone who has only memorised the formula. By the end of this article you will be able to derive $K$ , $U$ , and $E$ for any circular orbit from scratch, read the negative sign correctly (bound, not "no energy"), use the shortcut $E=-K=U/2$ to save calculation time, and compute the binding energy needed to send a satellite off to infinity. We will also connect this result forward to escape speed, which asks the very same question for a projectile launched from a planet's surface rather than an orbit around it.

Key takeaways

A satellite of mass $m$ in a circular orbit of radius $r$ around a planet of mass $M$ has kinetic energy $K = \dfrac{GMm}{2r}$ , potential energy $U = -\dfrac{GMm}{r}$ , and total energy $E = K+U = -\dfrac{GMm}{2r}$ .
The negative sign on $E$ means the satellite is gravitationally bound — it cannot drift to infinity without extra energy being supplied.
A neat, testable shortcut always holds for circular gravitational orbits: $E = -K$ and $U = 2E = -2K$ .
Higher orbits have a larger (less negative) total energy but a smaller speed and kinetic energy — energy and speed move in opposite directions as $r$ grows.
The binding energy, the minimum energy needed to free the satellite to infinity with zero leftover speed, equals $|E| = \dfrac{GMm}{2r}$ .
Moving a satellite to a higher orbit requires firing engines to add energy, even though the satellite ends up moving more slowly once it gets there.
Key idea: a satellite's total mechanical energy in a circular orbit is E = -GMm/2r, exactly minus its kinetic energy and half its potential energy, and this negative value measures how tightly it is gravitationally bound.

The core insight

For a satellite of mass $m$ circling a planet of mass $M$ at radius $r$ , adding the orbital kinetic energy $K=GMm/2r$ to the potential energy $U=-GMm/r$ gives a total mechanical energy $E=-GMm/2r$ that is always negative, always equal to $-K$ , and whose magnitude is the binding energy that must be supplied to send the satellite to infinity.

Where it fits in the syllabus

This is the closing topic of the CBSE Class 11 Physics chapter on Gravitation, sitting directly after orbital velocity and gravitational potential energy in the prescribed scope, with the exact result $E=-\dfrac{GMm}{2r}$ named explicitly in the syllabus. It draws on two results you should already have derived: $v_o=\sqrt{GM/r}$ from equating gravitational force to the centripetal-force requirement, and $U=-GMm/r$ from integrating the gravitational force. Nothing here goes beyond algebra and the inverse-square law — energy of elliptical orbits, orbital mechanics with angular momentum (vis-viva equation), and any treatment of relativistic or general-relativistic corrections are outside the Class 11 board syllabus and are not needed for this topic.

The physical picture

Picture the satellite as a point mass $m$ moving in a circle of radius $r$ about a planet of mass $M$ fixed at the centre. Two kinds of energy describe its state completely: the kinetic energy of its motion, and the potential energy of its position in the planet's gravitational field. The circular-motion condition — gravity supplies exactly the centripetal force needed — pins down the orbital speed, and from that speed the kinetic energy follows immediately.

v_o = \sqrt{\frac{GM}{r}} \quad\Longrightarrow\quad K = \frac12 m v_o^2 = \frac{GMm}{2r}

where: $G=6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ is the universal gravitational constant, $M$ is the mass of the planet being orbited, $m$ is the satellite's mass, $r$ is the orbital radius measured from the planet's centre (not its surface), and $v_o$ is the orbital speed. Physically, $K$ tells you how much energy is tied up purely in the satellite's motion — the faster and more massive the satellite, the larger this number, and it shrinks as the orbit gets higher because $v_o$ falls off as $1/\sqrt{r}$ . The potential energy comes from the general gravitational PE formula developed earlier in the chapter, with the zero of potential energy set at infinite separation:

U = -\frac{GMm}{r}

where: the negative sign is a convention choice, not a statement about "negative energy" in some absolute sense — it simply says that gravitational potential energy is lower (more negative) the closer the two masses are, with $U\to 0$ only as $r\to\infty$ . Two masses at a finite separation are always in a state of less potential energy than two masses infinitely far apart, because gravity does positive work as they come together.

Figure

Two concentric circles showing a planet at the centre and a satellite on a circular orbit, with an inward arrow labelled gravity providing centripetal force, a tangential arrow labelled orbital velocity, and a small energy bar chart beside it showing kinetic energy as a positive bar, potential energy as a taller negative bar, and total energy as a short negative bar exactly half the potential energy bar

Adding the two gives the total mechanical energy, and it is the sum that governs whether the orbit is bound at all:

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

where: every symbol is as before. Notice that $E$ is exactly $-K$ , and exactly half of $U$ — this pair of relations, $E=-K$ and $U=2E$ , is not a coincidence of this particular problem; it is a structural feature of any circular orbit under an inverse-square force, and it is worth memorising as a shortcut in its own right.

Derivation ATotal energy of a satellite in a circular orbit

The goal of this derivation is to combine what you already know about circular orbital motion with the general formula for gravitational potential energy to arrive at a single closed-form expression for the total mechanical energy of a satellite moving in a circular orbit of radius $r$ around a planet of mass $M$ , and along the way to expose the fixed numerical relationship between kinetic energy, potential energy, and total energy that makes circular-orbit energy problems solvable with almost no arithmetic once the relationship is known.

Step 1. Start from the condition that keeps the satellite on a circular path: gravity supplies exactly the centripetal force needed for uniform circular motion, so $\dfrac{GMm}{r^2} = \dfrac{mv_o^2}{r}$ . Multiplying both sides by $r$ and dividing by $2$ gives the kinetic energy directly, without needing to solve for $v_o$ first: $\tfrac12 mv_o^2 = \dfrac{GMm}{2r}$ , so $K = \dfrac{GMm}{2r}$ .

Step 2. Bring in the gravitational potential energy of the Earth–satellite (or planet–satellite) system, which was derived earlier in this chapter by integrating the gravitational force from infinity down to separation $r$ , with the potential energy taken to be zero at infinite separation: $U = -\dfrac{GMm}{r}$ . Note immediately that $U$ has exactly twice the magnitude of $K$ and the opposite sign — this is the key structural fact that everything else follows from.

Step 3. Add the two energies to get the total mechanical energy of the orbit: $E = K + U = \dfrac{GMm}{2r} + \left(-\dfrac{GMm}{r}\right) = \dfrac{GMm}{2r} - \dfrac{2GMm}{2r} = -\dfrac{GMm}{2r}$ .

Step 4. Read off the two shortcut relations that this algebra has produced. Since $K=GMm/2r$ and $E=-GMm/2r$ , it follows that $E=-K$ exactly. Since $U=-GMm/r$ and $E=-GMm/2r$ , it follows that $U=2E$ , equivalently $U=-2K$ . These relations hold for any circular orbit under gravity — change $M$ , $m$ , or $r$ and the individual energies change, but the ratios $E:K:U = -1:1:-2$ never do. This is a direct consequence of the inverse-square nature of the gravitational force, and the same $-1:1:-2$ pattern reappears for circular orbits in any inverse-square-force problem you meet later, such as the Bohr model of the atom.

Worked examples

Board level - energy of a satellite in low Earth orbit

A satellite of mass $1000\ \text{kg}$ orbits the Earth at $r = 7.0\times10^6\ \text{m}$ from the Earth's centre. Take $M_E = 6.0\times10^{24}\ \text{kg}$ and $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ . Find its kinetic energy, potential energy, and total energy.

Step 1. Compute the combination $GMm$ once, since it appears in every term: $GMm = (6.67\times10^{-11})(6.0\times10^{24})(1000) \approx 4.0\times10^{17}\ \text{N}\,\cdot\,\text{m}^2\cdot\text{kg}/\text{kg}$ , which has units of $\text{J}\,\cdot\,\text{m}$ once divided by a length.

Step 2. Find the kinetic energy using $K = \dfrac{GMm}{2r}$ : $K = \dfrac{4.0\times10^{17}}{2\times7.0\times10^6} = \dfrac{4.0\times10^{17}}{1.4\times10^7} \approx 2.857\times10^{10}\ \text{J}$ .

Step 3. Find the potential energy using $U = -\dfrac{GMm}{r}$ : $U = -\dfrac{4.0\times10^{17}}{7.0\times10^6} \approx -5.714\times10^{10}\ \text{J}$ . Check that this is indeed exactly twice the magnitude of $K$ with the opposite sign, as the shortcut predicts.

Step 4. Add to get the total energy: $E = K+U = 2.857\times10^{10} - 5.714\times10^{10} \approx -2.857\times10^{10}\ \text{J}$ .

The kinetic energy is about $\mathbf{2.86\times10^{10}\ J}$ , the potential energy is about $\mathbf{-5.71\times10^{10}\ J}$ , and the total energy is about $\mathbf{-2.86\times10^{10}\ J}$ , confirming $E=-K$ exactly for this circular orbit.

Competitive level - binding energy and orbit-raising energy cost

Using the same $1000\ \text{kg}$ satellite at $r=7.0\times10^6\ \text{m}$ : (a) find the minimum energy needed to send it from this orbit to infinity, starting effectively at rest relative to the orbit once free of Earth's pull; and (b) if the satellite is instead raised to a new circular orbit at $r' = 1.4\times10^7\ \text{m}$ (twice the original radius), find the energy that must be supplied by its engines, and compare the satellite's new speed with its old speed.

Step 1. The binding energy to reach infinity is simply the magnitude of the current total energy, because moving from total energy $E$ (bound) to a final state of zero kinetic and zero potential energy at infinity requires supplying exactly $|E|$ : $E_{\text{binding}} = |E| = \dfrac{GMm}{2r} = 2.857\times10^{10}\ \text{J}$ , reusing the kinetic energy value computed above, since $|E|=K$ for a circular orbit.

Step 2. For the orbit-raising case, find the new total energy at $r'=1.4\times10^7\ \text{m}$ : $E' = -\dfrac{GMm}{2r'} = -\dfrac{4.0\times10^{17}}{2\times1.4\times10^7} = -\dfrac{4.0\times10^{17}}{2.8\times10^7} \approx -1.429\times10^{10}\ \text{J}$ .

Step 3. The energy that must be supplied is the difference between the final and initial total energies: $\Delta E = E' - E = (-1.429\times10^{10}) - (-2.857\times10^{10}) \approx 1.428\times10^{10}\ \text{J}$ . This energy must come from the satellite's engines (fuel), since the orbit cannot climb to a less-negative total energy on its own.

Step 4. Compare speeds using $v_o=\sqrt{GM/r}$ : since $r'=2r$ , the new speed is $v_o' = v_o/\sqrt2 \approx 0.707\,v_o$ — the satellite at the higher orbit moves about $29\%$ slower than before, even though the engines had to add positive energy to get it there. The binding energy is about $\mathbf{2.86\times10^{10}\ J}$ ; raising the orbit to double the radius costs about $\mathbf{1.43\times10^{10}\ J}$ of supplied energy, and the satellite ends up moving slower, at about $\mathbf{71\%}$ of its original speed. Timing note: In a competitive exam, do not recompute $GMm$ or $K$ from scratch for part (b) — reuse the $K$ value from part (a) and simply rescale by the ratio of radii ( $E\propto 1/r$ ), since $E'=E\times(r/r')$ . This kind of proportional reasoning, rather than repeated full-decimal computation, is what keeps this problem inside a two-minute budget.

Common traps

Reading "negative total energy" as "the satellite has no energy" or "less than zero energy available for anything." It means the system is bound — the satellite has large kinetic and potential energies individually, but their sum is negative because escaping to infinity (where both would be zero) requires adding energy, not removing it.
Assuming a higher orbit means a faster, more energetic satellite in every sense. Speed and kinetic energy decrease with $r$ ( $v_o=\sqrt{GM/r}$ ), while total energy $E$ increases (becomes less negative) with $r$ . These two trends point opposite ways, and mixing them up is one of the most common marks lost on this topic.
Forgetting the shortcut $E=-K=U/2$ and grinding through a full $K$ -then- $U$ -then-add calculation every time, when a single value of $K$ or $U$ already fixes the other two instantly.
Dropping or duplicating the factor of $2$ : $K=GMm/2r$ has a $2$ in the denominator, $U=-GMm/r$ does not, and $E=-GMm/2r$ has the $2$ back. Writing $E=-GMm/r$ (forgetting the $2$ ) is a very common slip under exam time pressure.
Treating the binding energy calculation as if it required knowing the escape speed formula separately — for a satellite already in orbit, the binding energy is just $|E|$ of that orbit, not a fresh escape-speed computation from the surface.

You should be able to…

Derive $K = GMm/2r$ , $U=-GMm/r$ , and $E=K+U=-GMm/2r$ for a satellite in a circular orbit, starting from the centripetal-force condition and the general gravitational PE formula.
State and correctly apply the shortcut relations $E=-K$ and $U=2E=-2K$ for any circular gravitational orbit.
Explain, in words a classmate would understand, why negative total energy signals a bound orbit rather than a deficiency of energy.
Calculate the binding energy $|E|=GMm/2r$ needed to send an orbiting satellite to infinity.
Compare total energy and orbital speed across two orbits of different radii, and correctly predict that they change in opposite directions as $r$ increases.
Explain why raising a satellite's orbit requires supplying energy even though its final speed will be lower.

Ask the tutor: Want a walkthrough of how this same $-1:1:-2$ energy pattern shows up again when you meet the Bohr model of the atom, or a derivation of why $U$ is always exactly twice $E$ in magnitude for any circular orbit under an inverse-square force?

Why is potential energy negative at all — shouldn't energy just be a positive number?

Potential energy is only ever meaningful relative to a chosen reference point, and for gravity the natural, universally agreed reference is "zero potential energy when the two masses are infinitely far apart, feeling no gravitational pull from each other." Bringing two masses closer together lets gravity do positive work on them, which lowers their potential energy below that zero point — hence the negative sign at any finite separation. It is exactly like a ball in a hole: if you call ground level "zero height," the bottom of the hole is a negative height, not because height itself must be positive, but because of where you chose to measure from.

If total energy is negative, does that mean the satellite is losing energy over time and will eventually crash?

Not from this formula alone. $E=-GMm/2r$ is a constant for an ideal circular orbit with no air resistance or other losses — as the satellite goes around, kinetic and potential energy trade off between each other while their sum stays fixed at that one negative value, forever. Real low-Earth-orbit satellites do eventually decay and re-enter, but that is caused by an external effect, atmospheric drag slowly removing energy from the orbit over months or years, not by anything inherent in the negative sign of $E$ itself.

Why does raising a satellite to a higher orbit make it slower, when we had to add energy to get it there?

This is the part of the topic that trips up intuition the most, and it is worth sitting with. Adding energy $\Delta E$ moves the satellite to a state with a less negative (larger) total energy, which is achieved at a larger radius $r$ . But at that larger radius, the orbital speed formula $v_o=\sqrt{GM/r}$ gives a smaller speed. The added energy does not go into extra speed; it goes into extra potential energy, since $U=-GMm/r$ becomes less negative even faster than $K$ shrinks. Think of it like climbing a hill on a bicycle: you supply extra energy to gain height, but you arrive at the top moving slower than you were at the bottom, not faster.

How does this connect to escape speed, which we studied earlier in the chapter?

Escape speed asks a very similar question from a different starting point: what speed must an object have right at the planet's surface so that its total energy is exactly zero, letting it just barely reach infinity? That condition, $\tfrac12 mv_e^2 - \dfrac{GMm}{R}=0$ , gives $v_e=\sqrt{2GM/R}$ . The orbital energy result in this article asks the same "just reach infinity" question but for an object that starts already circling in orbit at radius $r$ , with total energy $-GMm/2r$ rather than zero — so the energy that must be supplied is $|E|=GMm/2r$ , not the full kinetic energy $\tfrac12 mv_e^2$ that a surface launch would need. Both problems are really the same idea, energy conservation between a starting state and "zero energy at infinity," applied at two different starting radii.