Energy of an Orbiting Satellite and Binding Energy

The Three Energies of an Orbit

Section 7 told us how fast a satellite goes and how long it takes. It deliberately said nothing about energy. This section is that missing half, and it is where most of the exam questions in the satellite topic actually live.

Everything follows from two things we already have: the orbital speed from Section 7, and the gravitational potential energy from Section 5.

Kinetic energy

A satellite of mass $m$ in a circular orbit of radius $r$ moves at $v_o = \sqrt{\frac{GM_E}{r}}$. So

$$K = \frac{1}{2}mv_o^{2} = \frac{1}{2}m\left(\frac{GM_E}{r}\right)$$

Key Point — kinetic energy in a circular orbit: $$K = +\frac{GM_Em}{2r}$$ Always positive, and it falls as the orbit gets larger, because the higher satellite moves more slowly.

Potential energy

Nothing new here. Section 5 established that, with the zero of potential energy taken at infinity,

Key Point — potential energy: $$U = -\frac{GM_Em}{r}$$ Always negative, and it rises towards zero as the orbit gets larger.

Note carefully that this is $U$, the potential energy of the satellite, in joules. It is not $V$, the gravitational potential of the field, in joules per kilogram. The two differ by a factor of $m$ and are the single most confused pair in this chapter. In this section every energy carries the satellite's mass in it.

Total energy

Add them:

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

Key Point — total energy of a circular orbit: $$E = -\frac{GM_Em}{2r}$$ Negative, and it also rises towards zero as the orbit gets larger.

The three relations — pure exam currency

Put the three side by side and the pattern is unmissable:

$$K = +\frac{GM_Em}{2r} \qquad\qquad U = -\frac{GM_Em}{r} \qquad\qquad E = -\frac{GM_Em}{2r}$$

Every one of them is $\frac{GM_Em}{2r}$ with a different sign or a factor of two. So:

Key Point — memorise these three relations: $$E = -K \qquad\qquad U = 2E \qquad\qquad U = -2K$$ Equivalently, $K : U : E = 1 : -2 : -1$. The kinetic energy is half the magnitude of the potential energy, and the total energy has the same magnitude as the kinetic energy but the opposite sign.

These are worth committing to memory exactly as written, signs included. A very large fraction of satellite questions consist of giving you one of $K$, $U$ or $E$ and asking for another, and if you know the ratios the answer takes one line and no formula sheet.

[NEET Important] The one-liners that get asked verbatim: the kinetic energy of a satellite is positive; its potential energy is negative and twice as large in magnitude; its total energy is negative and equal in magnitude to the kinetic energy.

Why the factor of two is not an accident

The relation $U = -2K$ for a circular gravitational orbit is a special case of a general result called the virial theorem, which holds for any inverse-square force. You do not need it by name, but it is worth knowing that the same $1 : -2$ ratio appears for an electron in a Bohr orbit in the atom chapter. When you meet it there, you will already know where it comes from.

[Board Important] A standard three-mark question asks you to derive the total energy of a satellite. Write $K = \frac{1}{2}mv_o^2$, substitute $v_o^2 = \frac{GM_E}{r}$, write $U = -\frac{GM_Em}{r}$ stating the zero-of-potential convention, add, and conclude $E = -\frac{GM_Em}{2r}$. The mark for the convention is real; do not skip it.

Why the Total Energy Must Be Negative

Look again at

$$E = -\frac{GM_Em}{2r}$$

Every symbol on the right is positive, and there is a minus sign in front. So the total energy of any circular orbit is negative, no matter the satellite, no matter the height. That is not an accident of the algebra; it is the definition of being trapped.

Bound and unbound, in one line

Section 6 built the whole escape argument on a single inequality. Let us re-read it with the satellite in mind.

At infinity, $U = 0$ and $K \geq 0$, so a body that reaches infinity arrives with $E \geq 0$. Energy is conserved, so a body that is going to reach infinity must have $E \geq 0$ now. Turn that round:

Key Point — the meaning of the sign of $E$: With the zero of potential energy at infinity,

• $E < 0$ — the body is bound. It can never reach infinity. It is a satellite, on a circle or an ellipse.
• $E = 0$ — the borderline. The body just escapes, arriving at infinity with exactly zero speed. Its path is a parabola.
• $E > 0$ — the body is free. It escapes and still has speed left over, $v_\infty = \sqrt{\frac{2E}{m}}$. Its path is a hyperbola.

So the negativity of $E$ is not a curiosity to be memorised. It is the statement that the satellite is a satellite. If someone hands you an orbit with a positive total energy, they have not handed you an orbit at all.

What about elliptical orbits?

For an ellipse both $K$ and $U$ change from point to point — fastest and most negative at perigee, slowest and least negative at apogee — but the sum stays constant, and stays negative. The formula generalises in exactly the way Section 7's period formula did: replace $r$ by the semi-major axis $a$.

Key Point: For an elliptical orbit of semi-major axis $a$, $$E = -\frac{GM_Em}{2a}$$ so two orbits with the same semi-major axis have the same total energy and the same period, however different their shapes.

That is a strikingly useful fact. A nearly circular orbit and a long thin cigar of an orbit, if they have the same $a$, cost exactly the same energy to reach.

A sanity check on the signs

It is easy to write $-\frac{GM_Em}{2r}$ and never look at it again, so let us check it means something.

Take a 1000 kg satellite at 300 km, where $r = 6.67 \times 10^{6}$ m and $GM_E = 3.982 \times 10^{14}$ m$^3$/s$^2$:

$$K = \frac{3.982 \times 10^{14} \times 1000}{2 \times 6.67 \times 10^{6}} = +2.985 \times 10^{10} \text{ J}$$ $$U = -\frac{3.982 \times 10^{14} \times 1000}{6.67 \times 10^{6}} = -5.970 \times 10^{10} \text{ J}$$ $$E = K + U = -2.985 \times 10^{10} \text{ J}$$

Check the relations: $E = -K$ — yes. $U = 2E$ — yes. And $E$ is negative, so the satellite stays. A cross-check on $K$: the orbital speed there is 7727 m/s, and $\frac{1}{2} \times 1000 \times 7727^{2} = 2.985 \times 10^{10}$ J. The two routes agree exactly, as they must.

[JEE Tip] If a question ever produces a positive total energy for something described as "in orbit", go back and look for a sign error in $U$. Writing $U = +\frac{GMm}{r}$ is the single commonest slip in this section, and it makes every subsequent line wrong in a way that is hard to spot.

Binding Energy: What It Would Cost to Set It Free

This topic sits outside the rationalised syllabus body text, but Boards, JEE Main, JEE Advanced and NEET ask about it every year, so it is developed here from first principles.

We know the satellite is trapped, because $E < 0$. The obvious next question is: trapped by how much?

The definition

To free the satellite completely means to raise its total energy to zero — to get it infinitely far away with nothing left over. Its total energy now is $-\frac{GM_Em}{2r}$. So the energy that must be supplied is

$$\text{energy needed} = 0 - E = 0 - \left(-\frac{GM_Em}{2r}\right)$$

Key Point — binding energy of a satellite in orbit: $$\text{B.E.} = -E = +\frac{GM_Em}{2r}$$ The binding energy is the minimum extra energy that must be given to a satellite in a circular orbit of radius $r$ to free it from the planet's gravity altogether. It is positive, it is numerically equal to the kinetic energy the satellite already has, and it is half the magnitude of the potential energy.

Three ways of saying the same thing, all examinable:

$$\text{B.E.} = -E = K = \frac{|U|}{2} = \frac{GM_Em}{2r}$$

Careful: two different "binding energies"

Examiners exploit the following distinction constantly, so get it straight now.

A body at rest on the ground. Its kinetic energy is zero and its potential energy is $-\frac{GM_Em}{R_E}$, so its total energy is $-\frac{GM_Em}{R_E}$. To free it you must supply

$$\frac{GM_Em}{R_E} = \frac{1}{2}mv_e^{2}$$

which is exactly the escape energy of Section 6 — $6.25 \times 10^{7}$ J for every kilogram.

A satellite already in a surface-skimming orbit. Its total energy is $-\frac{GM_Em}{2R_E}$, so to free it you need only

$$\frac{GM_Em}{2R_E} = 3.13 \times 10^{7} \text{ J per kilogram}$$

Key Point: A satellite skimming the surface needs only half as much extra energy to escape as a body sitting at rest on the ground at the same place. Being in orbit has already done half the job — which is another way of saying that the orbital speed is $\frac{1}{\sqrt{2}}$ of the escape speed, since the energies go as the squares of the speeds.

Read that table downwards. The higher the orbit, the less tightly the satellite is bound, and the less it would cost to set it free — but the more it cost to put it there in the first place.

Binding energy of the Earth to the Sun, and other big numbers

The same formula applies to any bound pair. The Moon, mass $7.34 \times 10^{22}$ kg, orbits the Earth at $3.84 \times 10^{8}$ m, so its total energy is

$$E = -\frac{3.982 \times 10^{14} \times 7.34 \times 10^{22}}{2 \times 3.84 \times 10^{8}} = -3.81 \times 10^{28} \text{ J}$$

and its binding energy to the Earth is $+3.81 \times 10^{28}$ J. For scale, the whole world uses about $6 \times 10^{20}$ J of energy in a year, so freeing the Moon would take some 60 million years of total human energy consumption. The Moon is going nowhere.

[Board Important] Define binding energy in words as "the minimum energy required to remove a satellite from its orbit to infinity, so that its total energy becomes zero", then quote $\text{B.E.} = \frac{GM_Em}{2r}$ and note that it equals the satellite's kinetic energy. Those three steps are the full three marks.

Raising an Orbit, and the Paradox That Comes With It

Now the question that separates the students who have understood this section from those who have memorised it.

The cost of moving up

A satellite is in a circular orbit of radius $r_1$ and we want it in a circular orbit of radius $r_2 > r_1$. How much energy must be supplied?

Energy in equals the change in total energy:

$$\Delta E = E_2 - E_1 = -\frac{GM_Em}{2r_2} - \left(-\frac{GM_Em}{2r_1}\right)$$

Key Point — energy to raise a circular orbit: $$\Delta E = \frac{GM_Em}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$$ Positive whenever $r_2 > r_1$, as it must be: you have to pay to go up.

And now the paradox

Along with $\Delta E$, work out what happens to the two pieces separately. Since $K = -E$ and $U = 2E$ at every radius, the same must be true of the changes:

Key Point — the orbit-raising ledger: $$\Delta E = +\text{(energy supplied)} \qquad \Delta U = +2\,\Delta E \qquad \Delta K = -\,\Delta E$$ The potential energy goes up by twice the energy you supplied. The kinetic energy goes down by exactly the amount you supplied. The satellite ends up moving more slowly than before.

Read that again. You put energy in, and the satellite slowed down.

This looks impossible, and every year it catches people out. So let us take it apart properly, because the explanation is not a trick — it is the physics.

Why adding energy slows a satellite down

Here is the thing. You did not give the satellite energy and let it stay where it was. You gave it energy and it moved outwards, and moving outwards costs potential energy at a rate that outruns what you supplied.

Suppose you supply 1 J. The satellite climbs to a bigger radius. Climbing to that radius costs 2 J of potential energy. Where does the second joule come from? There is only one other account: the kinetic energy. So the kinetic energy pays 1 J, and the satellite is slower.

$$\underbrace{+1 \text{ J}}_{\text{you supply}} \ + \ \underbrace{(-1 \text{ J})}_{\text{from } K} \ = \ \underbrace{+2 \text{ J}}_{\text{into } U}$$

The bookkeeping is forced by the fact that $|U| = 2K$ at every radius. Gravity, so to speak, takes a double share.

Key Point: Adding energy to a satellite raises its orbit and lowers its speed. Removing energy lowers its orbit and raises its speed. The total energy and the speed always move in opposite directions for a circular orbit, because $K = -E$.

The same thing running the other way: atmospheric drag

A satellite in a low orbit brushes the very thin upper atmosphere and loses energy to drag. Drag is a retarding force, so your instinct says it must slow down.

It does the opposite. Losing energy makes $E$ more negative, which makes $r$ smaller, which makes $K = -E$ bigger. A decaying satellite spirals inwards and speeds up the whole way, right until the air gets thick enough to destroy it.

This is not a paradox either, once you see the accounting: as the satellite falls, gravity does positive work on it that is roughly twice what drag takes away, and the surplus becomes speed. It is called the satellite drag paradox precisely because it fools people, and it is a favourite assertion-reason question.

The classic numbers

A 400 kg satellite is moved from a circular orbit of radius $2R_E$ to one of radius $4R_E$. With $GM_E = 3.982 \times 10^{14}$ m$^3$/s$^2$ and $R_E = 6.37 \times 10^{6}$ m:

$$\Delta E = \frac{GM_Em}{2}\left(\frac{1}{2R_E} - \frac{1}{4R_E}\right) = \frac{GM_Em}{2} \cdot \frac{1}{4R_E} = \frac{GM_Em}{8R_E}$$ $$\Delta E = \frac{3.982 \times 10^{14} \times 400}{8 \times 6.37 \times 10^{6}} = \frac{1.593 \times 10^{17}}{5.096 \times 10^{7}} = +3.13 \times 10^{9} \text{ J}$$

and therefore

$$\Delta K = -3.13 \times 10^{9} \text{ J} \qquad\qquad \Delta U = +6.25 \times 10^{9} \text{ J}$$

The speed drops from 5.59 km/s to 3.95 km/s, which is $\frac{1}{\sqrt{2}}$ of what it was — exactly what doubling the radius should do.

[JEE Tip] Whenever a question asks for "the energy required to shift a satellite from one orbit to another", it wants $\Delta E$, and $\Delta E$ alone. If it then asks for the changes in kinetic and potential energy, write them down instantly as $-\Delta E$ and $+2\Delta E$ rather than recomputing. Getting the signs right on those two is usually where the marks are.

Escape, Seen from Orbit

Section 6 derived the escape speed from the ground. The energy picture we now have lets us redo that result as a one-liner, and lets us answer the more practical question: what does it take to escape from an orbit you are already in?

Escape is just $E = 0$

Set the total energy to zero at radius $r$ and see what speed that demands:

$$\frac{1}{2}mv^{2} - \frac{GM_Em}{r} = 0 \qquad \Longrightarrow \qquad v = \sqrt{\frac{2GM_E}{r}} = v_e(r)$$

That is precisely the escape speed of Section 6, now arriving as the boundary case of the energy classification rather than as a separate derivation. Escape speed is simply the speed at which $E$ crosses zero.

Compare it with the orbital speed at the same radius:

$$\frac{v_e(r)}{v_o(r)} = \frac{\sqrt{\frac{2GM_E}{r}}}{\sqrt{\frac{GM_E}{r}}} = \sqrt{2}$$

Key Point: At any radius $r$, the escape speed is $\sqrt{2}$ times the circular orbital speed at that same radius. $$v_e = \sqrt{2}\,v_o$$ Section 6 stated this for the surface; it is in fact true everywhere, provided both speeds are evaluated at the same distance from the centre.

What escaping from orbit actually costs

A satellite in a circular orbit at radius $r$ has $E = -\frac{GM_Em}{2r}$. To escape it needs $E = 0$, so it needs an extra

$$\Delta E = \frac{GM_Em}{2r}$$

which is exactly the binding energy of Section 3 above, and exactly its current kinetic energy. Two neat consequences:

Key Point: To escape from a circular orbit, a satellite must be given

• extra energy equal to its own kinetic energy, $K = \frac{GM_Em}{2r}$;
• extra speed of $(\sqrt{2} - 1)v_o = 0.414\,v_o$, an increase of about 41%.

A 41% increase in speed doubles the kinetic energy, which is precisely what is required. The two statements are the same statement.

The numbers for a low orbit

For a satellite at 300 km, where $v_o = 7727$ m/s:

$$v_e = \sqrt{2} \times 7727 = 10\,927 \text{ m/s}$$ $$\Delta v = 10\,927 - 7727 = 3200 \text{ m/s}$$

So a probe leaving a low parking orbit for deep space needs a burn of roughly 3.2 km/s. That is a real number used in real mission planning, and it is why interplanetary spacecraft are launched into a parking orbit first and only then given their escape burn.

For a 1000 kg satellite the extra energy needed is $K = 2.985 \times 10^{10}$ J.

Putting the whole ladder together

The two steps are equal. Getting into a low orbit is exactly half the energy job of leaving the Earth entirely — the same "half" that keeps appearing, and always for the same reason: $E_{orbit} = \frac{1}{2}E_{ground}$ because $\frac{1}{2r}$ at $r = R_E$ is half of $\frac{1}{R_E}$.

(The energy to actually reach a 300 km orbit from rest on the ground is a touch more, $3.27 \times 10^{7}$ J per kilogram, because you must also climb the extra 300 km. Real rockets need far more again, because of air drag and because they must carry their own fuel — but that is engineering, not gravitation.)

[NEET Important] Two facts, asked in this exact form: the escape speed from a given point is $\sqrt{2}$ times the orbital speed at that point, and the extra energy needed by an orbiting satellite to escape equals its kinetic energy in orbit.

The Shortlist, and the Traps

Everything worth memorising

Seven traps

Trap 1 — dropping the minus sign on $U$. $U = -\frac{GM_Em}{r}$, always, with the zero at infinity. Every result in this section inherits that sign, and a positive $U$ produces a positive $E$, which would mean the satellite is not in orbit at all.

Trap 2 — confusing $U$ with $V$. $U = -\frac{GM_Em}{r}$ is an energy in joules and belongs to the satellite. $V = -\frac{GM_E}{r}$ is a potential in joules per kilogram and belongs to the field. If a question gives you a value in J/kg, it is $V$, and you must multiply by the mass before adding it to a kinetic energy.

Trap 3 — using $\frac{GM_Em}{2r}$ for the binding energy of a body on the ground. A body at rest on the surface has no kinetic energy, so its binding energy is $\frac{GM_Em}{R_E}$, twice the orbital value. The factor of $\frac{1}{2}$ belongs only to something already in orbit.

Trap 4 — thinking that adding energy speeds a satellite up. It slows it down. $K = -E$, so $E$ up means $K$ down. Say it as higher orbit, higher energy, lower speed.

Trap 5 — answering "the energy required to shift orbits" with $\Delta U$. The energy required is $\Delta E$. $\Delta U$ is twice as big, and quoting it is worth zero marks.

Trap 6 — using $r$ where the problem gave $h$, or the other way round. Every formula here uses $r$ measured from the centre. $r = R_E + h$.

Trap 7 — assuming $K$ and $U$ are separately constant on an elliptical orbit. They are not; only $E$ is. On an ellipse the satellite is fastest and most tightly bound at perigee and slowest at apogee. The formula $E = -\frac{GM_Em}{2a}$ is the one that survives.

A worked sign check you can do in your head

If someone tells you a satellite has $U = -8 \times 10^{9}$ J, you should be able to say, without writing anything: $K = +4 \times 10^{9}$ J, $E = -4 \times 10^{9}$ J, binding energy $= +4 \times 10^{9}$ J. If you can do that reliably, you have this section.

What comes next

Section 9 takes two particular orbits — the one whose period is exactly 24 hours, and the one that passes over both poles — and asks what they are good for. It also settles the question that this section has been circling: why an astronaut inside an orbiting spacecraft feels no weight at all, even though gravity is precisely what is holding the spacecraft up there.

[Board Important] The most-set long question in this topic is: "Obtain expressions for the kinetic, potential and total energy of a satellite orbiting the Earth at a height $h$. Hence define binding energy and obtain its value." Do it in that order, state the zero-of-potential convention, and finish with $\text{B.E.} = \frac{GM_Em}{2(R_E+h)}$.

Solved Examples

Constants used throughout this section, unless a problem states otherwise: $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$, $M_E = 5.97 \times 10^{24}$ kg (so $GM_E = 3.982 \times 10^{14}$ m$^3$/s$^2$), $R_E = 6.37 \times 10^{6}$ m and $g = 9.8$ m/s$^2$. The zero of potential energy is at infinity throughout, and every energy below is quoted with its sign.

Example 1: The full energy account of a low-orbit satellite

A 1000 kg satellite orbits 300 km above the Earth's surface. Find its kinetic, potential and total energy, and verify the relations $E = -K$ and $U = 2E$.

Solution:

1. The orbital radius. $$r = R_E + h = 6.37 \times 10^{6} + 3.0 \times 10^{5} = 6.67 \times 10^{6} \text{ m}$$

2. Kinetic energy. $$K = \frac{GM_Em}{2r} = \frac{3.982 \times 10^{14} \times 1000}{2 \times 6.67 \times 10^{6}} = \frac{3.982 \times 10^{17}}{1.334 \times 10^{7}} = +2.985 \times 10^{10} \text{ J}$$

3. Potential energy. $$U = -\frac{GM_Em}{r} = -\frac{3.982 \times 10^{17}}{6.67 \times 10^{6}} = -5.970 \times 10^{10} \text{ J}$$

4. Total energy. $$E = K + U = 2.985 \times 10^{10} - 5.970 \times 10^{10} = -2.985 \times 10^{10} \text{ J}$$

5. Check the relations. $-K = -2.985 \times 10^{10} = E$, and $2E = -5.970 \times 10^{10} = U$. Both hold.

6. Independent check on $K$. The orbital speed at this radius is $v_o = \sqrt{\frac{GM_E}{r}} = 7727$ m/s, and $\frac{1}{2} \times 1000 \times 7727^{2} = 2.985 \times 10^{10}$ J. Agreed.

Final Answer: $K = +2.99 \times 10^{10}$ J, $U = -5.97 \times 10^{10}$ J, $E = -2.99 \times 10^{10}$ J.

Takeaway: Write all three with their signs every time. An answer of "$2.99 \times 10^{10}$ J" for the total energy, without the minus sign, is wrong — it claims the satellite is unbound.

Example 2: One energy given, the other two wanted

The potential energy of a satellite in a circular orbit is $-8.0 \times 10^{9}$ J. Find its kinetic energy, its total energy and its binding energy. If its mass is 250 kg, find its orbital speed.

Solution:

1. Use the ratios. For a circular orbit, $K = -\frac{U}{2}$ and $E = \frac{U}{2}$. $$K = -\frac{-8.0 \times 10^{9}}{2} = +4.0 \times 10^{9} \text{ J}$$ $$E = \frac{-8.0 \times 10^{9}}{2} = -4.0 \times 10^{9} \text{ J}$$

2. Binding energy. $$\text{B.E.} = -E = +4.0 \times 10^{9} \text{ J}$$

3. Orbital speed, from $K = \frac{1}{2}mv_o^{2}$. $$v_o = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 4.0 \times 10^{9}}{250}} = \sqrt{3.2 \times 10^{7}} = 5.66 \times 10^{3} \text{ m/s}$$

Final Answer: $K = +4.0 \times 10^{9}$ J, $E = -4.0 \times 10^{9}$ J, B.E. $= +4.0 \times 10^{9}$ J, $v_o = 5.66$ km/s.

Takeaway: You never need $G$, $M_E$ or $r$ to move between $K$, $U$ and $E$. The ratios $1 : -2 : -1$ do it in one line, which is why they are worth memorising.

Example 3: Binding energy of a surface-skimming satellite

Find the binding energy per kilogram of a satellite in a circular orbit just above the Earth's surface, and compare it with the energy per kilogram needed to launch a body from rest on the ground so that it escapes.

Solution:

1. Binding energy in orbit, with $r = R_E$: $$\frac{\text{B.E.}}{m} = \frac{GM_E}{2R_E} = \frac{3.982 \times 10^{14}}{2 \times 6.37 \times 10^{6}} = \frac{3.982 \times 10^{14}}{1.274 \times 10^{7}} = 3.13 \times 10^{7} \text{ J/kg}$$

2. Escape energy from rest on the ground. A body at rest on the surface has $K = 0$ and $U = -\frac{GM_Em}{R_E}$, so $$\frac{\text{energy needed}}{m} = \frac{GM_E}{R_E} = \frac{3.982 \times 10^{14}}{6.37 \times 10^{6}} = 6.25 \times 10^{7} \text{ J/kg}$$

3. The ratio. $$\frac{3.13}{6.25} = \frac{1}{2}$$

4. Why. The orbiting body already has kinetic energy $\frac{GM_Em}{2R_E}$, which is exactly half the depth of the pit it sits in. That is the energy version of $v_o = \frac{v_e}{\sqrt{2}}$.

Final Answer: $3.13 \times 10^{7}$ J/kg in orbit, against $6.25 \times 10^{7}$ J/kg from rest on the ground — exactly half.

Takeaway: Getting into a low orbit is half the energy job of leaving the Earth. Whenever a question compares "from the ground" with "from orbit", the answer is a factor of two, and it comes from the $\frac{1}{2}$ in $\frac{GM_Em}{2r}$.

Example 4: A satellite one Earth radius up

A 500 kg satellite orbits at a height equal to the Earth's radius. Find its kinetic, potential and total energy, and its binding energy.

Solution:

1. The radius. $r = R_E + R_E = 2R_E = 1.274 \times 10^{7}$ m.

2. The common factor. $$\frac{GM_Em}{r} = \frac{3.982 \times 10^{14} \times 500}{1.274 \times 10^{7}} = \frac{1.991 \times 10^{17}}{1.274 \times 10^{7}} = 1.563 \times 10^{10} \text{ J}$$

3. The three energies. $$U = -1.563 \times 10^{10} \text{ J} \qquad K = +7.81 \times 10^{9} \text{ J} \qquad E = -7.81 \times 10^{9} \text{ J}$$

4. Binding energy. $$\text{B.E.} = -E = +7.81 \times 10^{9} \text{ J}$$

5. Per kilogram, that is $1.56 \times 10^{7}$ J/kg — half the surface-skimming value, because the radius has doubled.

Final Answer: $K = +7.81 \times 10^{9}$ J, $U = -1.56 \times 10^{10}$ J, $E = -7.81 \times 10^{9}$ J, B.E. $= 7.81 \times 10^{9}$ J.

Takeaway: Compute $\frac{GM_Em}{r}$ once and write all four answers off it. Everything in this section is that one number, halved or negated.

Example 5: Moving a satellite to a higher orbit

A 400 kg satellite is in a circular orbit of radius $2R_E$ about the Earth. How much energy is required to transfer it to a circular orbit of radius $4R_E$? What are the changes in its kinetic and potential energies?

Solution:

1. Initial and final total energies. $$E_1 = -\frac{GM_Em}{2(2R_E)} = -\frac{GM_Em}{4R_E} \qquad\qquad E_2 = -\frac{GM_Em}{2(4R_E)} = -\frac{GM_Em}{8R_E}$$

2. The energy required. $$\Delta E = E_2 - E_1 = -\frac{GM_Em}{8R_E} + \frac{GM_Em}{4R_E} = \frac{GM_Em}{8R_E}$$

3. Substitute. $$\Delta E = \frac{3.982 \times 10^{14} \times 400}{8 \times 6.37 \times 10^{6}} = \frac{1.593 \times 10^{17}}{5.096 \times 10^{7}} = +3.13 \times 10^{9} \text{ J}$$

4. The two changes. Since $\Delta K = -\Delta E$ and $\Delta U = +2\Delta E$, $$\Delta K = -3.13 \times 10^{9} \text{ J} \qquad\qquad \Delta U = +6.25 \times 10^{9} \text{ J}$$

5. Check the bookkeeping. $\Delta K + \Delta U = -3.13 + 6.25 = +3.13 \times 10^{9}$ J $= \Delta E$. Consistent.

6. What happened to the speed. $v_1 = \sqrt{\frac{GM_E}{2R_E}} = 5591$ m/s and $v_2 = \sqrt{\frac{GM_E}{4R_E}} = 3953$ m/s, a drop of exactly $\sqrt{2}$.

Final Answer: $3.13 \times 10^{9}$ J must be supplied; the kinetic energy falls by $3.13 \times 10^{9}$ J and the potential energy rises by $6.25 \times 10^{9}$ J.

Takeaway: The satellite ended up 3.13 GJ richer in total energy and 1.6 km/s slower. The potential energy took twice what you paid, and the kinetic energy made up the difference.

Example 6: The kick that frees a satellite

A satellite is in a circular orbit 300 km above the Earth. (a) By how much must its speed be increased for it to escape? (b) If its mass is 1000 kg, how much extra energy is that?

Solution:

1. Orbital speed at that radius, with $r = 6.67 \times 10^{6}$ m: $$v_o = \sqrt{\frac{3.982 \times 10^{14}}{6.67 \times 10^{6}}} = 7727 \text{ m/s}$$

2. (a) Escape speed at the same radius. $$v_e = \sqrt{2}\,v_o = 1.414 \times 7727 = 10\,927 \text{ m/s}$$ $$\Delta v = 10\,927 - 7727 = 3200 \text{ m/s}$$

3. (b) The extra energy is the binding energy, which equals the current kinetic energy: $$\Delta E = \frac{GM_Em}{2r} = \frac{3.982 \times 10^{14} \times 1000}{2 \times 6.67 \times 10^{6}} = +2.99 \times 10^{10} \text{ J}$$

4. Cross-check the energy from the speeds. $$\Delta E = \frac{1}{2}m(v_e^{2} - v_o^{2}) = \frac{1}{2} \times 1000 \times (1.194 \times 10^{8} - 5.971 \times 10^{7}) = 2.99 \times 10^{10} \text{ J}$$ The two agree.

Final Answer: an extra 3.2 km/s, which for 1000 kg is $2.99 \times 10^{10}$ J.

Takeaway: A 41% increase in speed doubles the kinetic energy, and doubling the kinetic energy is exactly what escaping needs. That is why the extra energy required equals the kinetic energy the satellite already has.

Example 7: From the launch pad to a low orbit

How much energy per kilogram is needed to take a body from rest on the Earth's surface and place it in a circular orbit 300 km up? Ignore air resistance and the Earth's rotation.

Solution:

1. Energy at the start. At rest on the ground, $K_i = 0$ and $$\frac{U_i}{m} = -\frac{GM_E}{R_E} = -6.251 \times 10^{7} \text{ J/kg} \qquad\Longrightarrow\qquad \frac{E_i}{m} = -6.251 \times 10^{7} \text{ J/kg}$$

2. Energy at the end, in a circular orbit at $r = 6.67 \times 10^{6}$ m: $$\frac{E_f}{m} = -\frac{GM_E}{2r} = -\frac{3.982 \times 10^{14}}{1.334 \times 10^{7}} = -2.985 \times 10^{7} \text{ J/kg}$$

3. The difference. $$\frac{\Delta E}{m} = -2.985 \times 10^{7} + 6.251 \times 10^{7} = +3.27 \times 10^{7} \text{ J/kg}$$

4. Split it up, for interest. Of that, $3.13 \times 10^{7}$ J/kg would have been needed even for a skimming orbit; the extra $1.4 \times 10^{6}$ J/kg is the cost of the additional 300 km of climb.

Final Answer: about $3.27 \times 10^{7}$ J for every kilogram placed in that orbit.

Takeaway: Launching is a change in TOTAL energy, from $-\frac{GM_Em}{R_E}$ to $-\frac{GM_Em}{2r}$. Do not compute the kinetic energy needed and stop there; the body also had to climb.

Example 8: The drag paradox, with numbers

A satellite in a low circular orbit loses energy slowly to atmospheric drag. Explain, with the energy relations, why its speed increases. If a 1 kg element of it starts at $r = 6.67 \times 10^{6}$ m and its total energy per kilogram falls by $5.6 \times 10^{5}$ J/kg, find the new radius and the new speed.

Solution:

1. The argument. For a circular orbit $E = -\frac{GM_Em}{2r}$ and $K = -E$. If $E$ falls (becomes more negative), then $r$ must fall, and $K = -E$ must rise. Lower orbit, faster satellite.

2. The starting values, per kilogram. $$\frac{E_1}{m} = -\frac{GM_E}{2r_1} = -\frac{3.982 \times 10^{14}}{1.334 \times 10^{7}} = -2.985 \times 10^{7} \text{ J/kg}$$

3. After the loss. $$\frac{E_2}{m} = -2.985 \times 10^{7} - 5.6 \times 10^{5} = -3.041 \times 10^{7} \text{ J/kg}$$

4. The new radius. $$r_2 = -\frac{GM_E}{2(E_2/m)} = \frac{3.982 \times 10^{14}}{2 \times 3.041 \times 10^{7}} = 6.547 \times 10^{6} \text{ m}$$ so the satellite has dropped by $6670 - 6547 = 123$ km.

5. The new speed. $$v_2 = \sqrt{\frac{GM_E}{r_2}} = \sqrt{\frac{3.982 \times 10^{14}}{6.547 \times 10^{6}}} = 7799 \text{ m/s}$$ against 7727 m/s before — an increase of 72 m/s.

Final Answer: the orbit shrinks by 123 km and the speed rises by about 72 m/s, even though the satellite has lost energy.

Takeaway: A retarding force makes an orbiting satellite go faster. Gravity does more positive work on the falling satellite than drag takes away, and the surplus becomes kinetic energy — which is exactly what $K = -E$ encodes.

Example 9: The Moon's energy account

The Moon, of mass $7.34 \times 10^{22}$ kg, orbits the Earth at a mean radius of $3.84 \times 10^{8}$ m. Find its kinetic, potential and total energy, and its binding energy to the Earth.

Solution:

1. The common factor. $$\frac{GM_Em}{r} = \frac{3.982 \times 10^{14} \times 7.34 \times 10^{22}}{3.84 \times 10^{8}} = \frac{2.923 \times 10^{37}}{3.84 \times 10^{8}} = 7.61 \times 10^{28} \text{ J}$$

2. The three energies. $$U = -7.61 \times 10^{28} \text{ J} \qquad K = +3.81 \times 10^{28} \text{ J} \qquad E = -3.81 \times 10^{28} \text{ J}$$

3. Binding energy. $$\text{B.E.} = +3.81 \times 10^{28} \text{ J}$$

4. A check on $K$. The Moon's orbital speed is $v_o = \sqrt{\frac{3.982 \times 10^{14}}{3.84 \times 10^{8}}} = 1018$ m/s, and $\frac{1}{2} \times 7.34 \times 10^{22} \times 1018^{2} = 3.80 \times 10^{28}$ J. Agreed to three figures.

Final Answer: $K = +3.81 \times 10^{28}$ J, $U = -7.61 \times 10^{28}$ J, $E = -3.81 \times 10^{28}$ J, and the Moon is bound to the Earth by $3.81 \times 10^{28}$ J.

Takeaway: The same three formulas that describe a 400 kg satellite describe the Moon. Only the numbers change, and the ratio $1 : -2 : -1$ does not change at all.

Example 10: Given the total energy, find the orbit

A satellite of mass 200 kg orbits the Earth with a total energy of $-3.0 \times 10^{9}$ J. Find the radius of its orbit, its height above the surface, its orbital speed and its period.

Solution:

1. Radius, from $E = -\frac{GM_Em}{2r}$. $$r = -\frac{GM_Em}{2E} = \frac{3.982 \times 10^{14} \times 200}{2 \times 3.0 \times 10^{9}} = \frac{7.964 \times 10^{16}}{6.0 \times 10^{9}} = 1.327 \times 10^{7} \text{ m}$$

2. Height. $$h = r - R_E = 1.327 \times 10^{7} - 6.37 \times 10^{6} = 6.90 \times 10^{6} \text{ m} = 6900 \text{ km}$$

3. Speed. Since $K = -E = 3.0 \times 10^{9}$ J, $$v_o = \sqrt{\frac{2K}{m}} = \sqrt{\frac{6.0 \times 10^{9}}{200}} = \sqrt{3.0 \times 10^{7}} = 5.48 \times 10^{3} \text{ m/s}$$

4. Period. $$T = \frac{2\pi r}{v_o} = \frac{2\pi \times 1.327 \times 10^{7}}{5.48 \times 10^{3}} = \frac{8.338 \times 10^{7}}{5.48 \times 10^{3}} = 1.522 \times 10^{4} \text{ s} = 4.23 \text{ h}$$

Final Answer: $r = 1.33 \times 10^{7}$ m, $h = 6900$ km, $v_o = 5.48$ km/s, $T = 4.23$ hours.

Takeaway: The total energy fixes the orbit completely. Give me $E$ and the satellite's mass and I can tell you its radius, its speed and its period — which is why $E$ is the single most useful number to know about an orbit.

Example 11: How the binding energy changes with height

Compare the binding energy of a satellite of mass $m$ in a surface-skimming orbit with that of the same satellite in an orbit at a height $h = R_E$. Which is more tightly bound, and which cost more to put there?

Solution:

1. The two binding energies. $$\text{B.E.}(0) = \frac{GM_Em}{2R_E} \qquad\qquad \text{B.E.}(R_E) = \frac{GM_Em}{2(2R_E)} = \frac{GM_Em}{4R_E}$$

2. The ratio. $$\frac{\text{B.E.}(R_E)}{\text{B.E.}(0)} = \frac{1}{2}$$ The higher satellite is less tightly bound: it would take only half as much extra energy to free it.

3. But which cost more to reach? Compare total energies against a body at rest on the ground, $E_{ground} = -\frac{GM_Em}{R_E}$: $$\Delta E_{\text{to skimming}} = -\frac{GM_Em}{2R_E} + \frac{GM_Em}{R_E} = \frac{GM_Em}{2R_E}$$ $$\Delta E_{\text{to } 2R_E} = -\frac{GM_Em}{4R_E} + \frac{GM_Em}{R_E} = \frac{3GM_Em}{4R_E}$$

4. Compare. $\frac{3/4}{1/2} = 1.5$, so the higher orbit cost 50% more to reach.

Final Answer: the higher satellite has half the binding energy but cost one and a half times as much energy to place.

Takeaway: "Less tightly bound" and "cheaper to reach" are opposite things. Going higher costs more and leaves you needing less to escape — which is exactly why deep-space missions leave from high orbits when they can.

Example 12: An orbit change in symbolic form

A satellite of mass $m$ orbits a planet of mass $M$ at radius $r$. Find, in terms of $\frac{GMm}{r}$, (a) the energy needed to move it to a circular orbit of radius $3r$, (b) the changes in its kinetic and potential energies, and (c) the fraction of its original kinetic energy that this represents.

Solution:

1. (a) The two total energies. $$E_1 = -\frac{GMm}{2r} \qquad\qquad E_2 = -\frac{GMm}{2(3r)} = -\frac{GMm}{6r}$$ $$\Delta E = -\frac{GMm}{6r} + \frac{GMm}{2r} = \frac{GMm}{r}\left(\frac{1}{2} - \frac{1}{6}\right) = \frac{GMm}{3r}$$

2. (b) The ledger. $$\Delta K = -\Delta E = -\frac{GMm}{3r} \qquad\qquad \Delta U = +2\Delta E = +\frac{2GMm}{3r}$$

3. (c) The original kinetic energy was $K_1 = \frac{GMm}{2r}$, so $$\frac{\Delta E}{K_1} = \frac{GMm/3r}{GMm/2r} = \frac{2}{3}$$

4. Read it back. Supplying two-thirds of the satellite's original kinetic energy triples its orbital radius. The new kinetic energy is $K_1 - \frac{GMm}{3r} = \frac{GMm}{6r}$, one third of what it was — and indeed $v \propto \frac{1}{\sqrt{r}}$ means the speed dropped by $\sqrt{3}$ and the kinetic energy by 3.

Final Answer: (a) $\frac{GMm}{3r}$; (b) $\Delta K = -\frac{GMm}{3r}$ and $\Delta U = +\frac{2GMm}{3r}$; (c) two thirds of the original kinetic energy.

Takeaway: Symbolic orbit-change questions are always $\frac{GMm}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$, followed by $-\Delta E$ and $+2\Delta E$. Do the algebra with the $\frac{1}{r}$ terms and substitute numbers only at the very end.