The three energies of a circular orbit
Place a satellite of mass $m$ in a circular orbit of radius $r$ around the Earth (mass $M$). Here $r = R_E + h$ is measured from the centre of the Earth, not the surface — a height $h$ above the ground is an orbital radius $R_E + h$. The gravity supplies the centripetal force, which fixes the orbital speed at $v = \sqrt{GM/r}$. Three energies follow.
Kinetic energy. Using the orbital speed $v^2 = GM/r$,
$$ KE = \tfrac{1}{2}mv^2 = \frac{1}{2}m\cdot\frac{GM}{r} = +\frac{GMm}{2r}. $$It is positive, as kinetic energy always is.
Potential energy. Taking the zero of gravitational potential energy at infinity (the standard NCERT convention), the potential energy at distance $r$ from the Earth's centre is
$$ PE = -\frac{GMm}{r}. $$It is negative because gravity is attractive — work must be done against gravity to carry the mass out to infinity where the energy is defined to be zero.
Total energy. Adding the two,
$$ E = KE + PE = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}. $$NCERT puts it plainly: "The K.E. is positive whereas the P.E. is negative. However, in magnitude the K.E. is half the P.E., so that the total $E$ is negative." That single sentence contains the three relations NEET tests most often.
The virial relations: E = −KE = PE/2
Lined up side by side, the three energies are in fixed proportion $KE : PE : E = 1 : -2 : -1$. Two compact relations fall out, and they are worth memorising because they let you skip arithmetic in an exam.
- $E = -KE$ — the total energy equals the negative of the kinetic energy. Know the speed, and you know the total energy without touching the potential energy.
- $E = \tfrac{1}{2}\,PE$ — the total energy is exactly half the potential energy.
- $PE = -2\,KE$ — equivalently, the potential energy is minus twice the kinetic energy.
These are the gravitational form of the virial theorem for an inverse-square force. They hold for any circular orbit — a planet about the Sun, the Moon about the Earth, an electron picture of the atom — provided the force goes as $1/r^2$.
Why total energy is negative
The sign of the total energy is not a quirk of algebra — it diagnoses whether the orbit is closed. NCERT spells out the logic in the summary: "The total energy is negative for any bound system, that is, one in which the orbit is closed." Read the three cases.
- $E < 0$ (negative): the satellite is bound. It cannot reach infinity, where it would need at least zero energy. The orbit is closed — a circle or an ellipse.
- $E = 0$: the borderline. The satellite has just enough energy to crawl to infinity and arrive with zero speed. This corresponds to a parabolic escape trajectory at exactly the escape speed.
- $E > 0$ (positive): the satellite is unbound. It escapes to infinity and still has kinetic energy left over — a hyperbolic path.
NCERT states the consequence directly: "Satellites are always at a finite distance from the earth and hence their energies cannot be positive or zero." A real, orbiting satellite therefore must have $E < 0$. If a problem hands you a satellite and you compute a positive total energy, you have made a sign error.
Energy versus orbital radius
The single most useful picture in this topic plots all three energies against the orbital radius $r$. As $r$ grows, $KE = GMm/2r$ falls toward zero from above, $PE = -GMm/r$ rises toward zero from below, and $E = -GMm/2r$ rises toward zero from below — but the total always stays sandwiched between zero and the potential-energy curve.
Three readings to take from the diagram. First, the total-energy curve is always below the $E=0$ line: every orbit is bound. Second, as the orbit widens the total energy rises toward zero — a larger orbit has more (less negative) total energy. Third, the kinetic-energy curve falls as $r$ grows: the satellite in the outer orbit moves slower. These last two points look contradictory until you separate "energy" from "speed", which is exactly where NEET sets its traps.
Binding energy of a satellite
The binding energy is the minimum energy that must be supplied to a bound satellite to just free it from the Earth's gravity — to raise its total energy from $-\dfrac{GMm}{2r}$ all the way up to zero (the energy of a body at rest at infinity). Since you are climbing from $E$ to $0$,
$$ BE = E_{\text{final}} - E_{\text{initial}} = 0 - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{2r}. $$So the binding energy is simply the negative of the total energy, $BE = -E = +\dfrac{GMm}{2r}$. Notice it is also numerically equal to the orbital kinetic energy, because $E = -KE$. To liberate an orbiting satellite you must hand it an amount of energy equal to the kinetic energy it already carries.
Raising a satellite to a higher orbit
To move a satellite from a circular orbit of radius $r_1$ to a larger circular orbit $r_2 > r_1$, you must change its total energy from $-\dfrac{GMm}{2r_1}$ to $-\dfrac{GMm}{2r_2}$. The energy you have to supply is
$$ \Delta E = E_2 - E_1 = -\frac{GMm}{2r_2} - \left(-\frac{GMm}{2r_1}\right) = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right). $$Because $r_2 > r_1$, the bracket is positive: $\Delta E > 0$, so energy must be added to climb to a higher orbit. Now look at what happens to each piece, using $KE \propto 1/r$ and $PE \propto -1/r$:
| Quantity | Formula | Moving to higher orbit (r ↑) |
|---|---|---|
| Orbital speed $v$ | $\sqrt{GM/r}$ | Decreases — satellite moves slower |
| Kinetic energy $KE$ | $+GMm/2r$ | Decreases |
| Potential energy $PE$ | $-GMm/r$ | Increases (rises toward 0) |
| Total energy $E$ | $-GMm/2r$ | Increases (rises toward 0) |
| Binding energy $-E$ | $+GMm/2r$ | Decreases — easier to free |
The bookkeeping is clean once you realise the change in potential energy is twice the change in total energy, while the change in kinetic energy is minus the change in total energy: $\Delta PE = 2\,\Delta E$ and $\Delta KE = -\Delta E$. The added energy splits so that the potential energy rises by $2\Delta E$ and the kinetic energy falls by $\Delta E$, netting $+\Delta E$. This is the relationship NCERT Example 7.8 demonstrates with numbers.
Every energy here rides on the orbital speed $v=\sqrt{GM/r}$. If that result is shaky, revisit earth satellites & orbital velocity before drilling the energy problems.
Worked Example 1 — energy to free a satellite at 400 km (NCERT Example 7.19 style)
A satellite of mass $200~\text{kg}$ orbits the Earth at a height of $400~\text{km}$ above the surface. How much energy must be expended to rocket it out of the Earth's gravitational influence? Take $M = 6.0\times10^{24}~\text{kg}$, $R_E = 6.4\times10^{6}~\text{m}$, $G = 6.67\times10^{-11}~\text{N m}^2\text{kg}^{-2}$.
Step 1 — orbital radius. $r = R_E + h = 6.4\times10^{6} + 4.0\times10^{5} = 6.8\times10^{6}~\text{m}$. Note the height is added to the Earth's radius, not used by itself.
Step 2 — total energy in orbit. $E = -\dfrac{GMm}{2r} = -\dfrac{(6.67\times10^{-11})(6.0\times10^{24})(200)}{2(6.8\times10^{6})}$.
The numerator $GMm = (4.002\times10^{14})(200) = 8.004\times10^{16}$, and $2r = 1.36\times10^{7}$, giving $E = -5.89\times10^{9}~\text{J}$ (about $-5.9\times10^{9}~\text{J}$).
Step 3 — energy to free it. To rocket the satellite out of the Earth's influence its total energy must be raised to zero, so the energy to expend is $-E = +5.9\times10^{9}~\text{J}$.
Answer: about $5.9\times10^{9}~\text{J}$ must be supplied. This equals the satellite's binding energy, which is also numerically equal to its orbital kinetic energy.
Worked Example 2 — shifting an orbit from 2Rᴇ to 4Rᴇ (NCERT Example 7.8)
A $400~\text{kg}$ 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 kinetic and potential energies?
Initial and final total energies. $E_i = -\dfrac{GMm}{2(2R_E)} = -\dfrac{GMm}{4R_E}$ and $E_f = -\dfrac{GMm}{2(4R_E)} = -\dfrac{GMm}{8R_E}$.
Energy to supply. $\Delta E = E_f - E_i = -\dfrac{GMm}{8R_E} + \dfrac{GMm}{4R_E} = +\dfrac{GMm}{8R_E}$. Writing $GM = gR_E^2$, this is $\dfrac{g m R_E}{8}$. With $g = 9.81~\text{m s}^{-2}$, $m = 400~\text{kg}$, $R_E = 6.37\times10^{6}~\text{m}$, NCERT obtains $\Delta E = +3.13\times10^{9}~\text{J}$.
Change in kinetic energy. Since $\Delta KE = -\Delta E$, the kinetic energy drops by $\Delta K = -3.13\times10^{9}~\text{J}$ — the satellite is slower in the wider orbit.
Change in potential energy. Since $\Delta PE = 2\,\Delta E$, the potential energy rises by $\Delta V = +2(3.13\times10^{9}) = +6.25\times10^{9}~\text{J}$.
Check: $\Delta KE + \Delta PE = -3.13\times10^{9} + 6.25\times10^{9} = +3.13\times10^{9}~\text{J} = \Delta E$. The energy budget balances.
Worked Example 3 — energies in an elliptical orbit
A satellite is in an elliptical orbit. State how its kinetic, potential and total energies behave as it moves from perigee (closest approach) to apogee (farthest point).
Total energy. Constant throughout. NCERT notes that for an elliptical orbit "both the K.E. and P.E. vary from point to point. The total energy which remains constant is negative as in the circular orbit case." For an ellipse $E = -\dfrac{GMm}{2a}$, where $a$ is the semi-major axis.
Potential energy. At perigee the satellite is nearest ($r$ small), so $PE = -GMm/r$ is most negative (lowest). At apogee ($r$ large) the potential energy is highest (least negative).
Kinetic energy. Because $E = KE + PE$ is fixed, $KE = E - PE$ is largest where $PE$ is most negative — at perigee. The satellite moves fastest at perigee and slowest at apogee, exactly as Kepler's law of areas demands.
Take-away: in an ellipse, KE and PE trade off point by point, but their sum — the total energy — never changes and stays negative.
Satellite energy in one breath
- For a circular orbit of radius $r$: $KE = +\dfrac{GMm}{2r}$, $PE = -\dfrac{GMm}{r}$, $E = -\dfrac{GMm}{2r}$.
- Virial relations: $E = -KE = \tfrac12 PE$, and $PE = -2\,KE$. Ratio $KE:PE:E = 1:-2:-1$.
- Total energy is negative for any bound orbit; zero means just-escaping, positive means unbound.
- Binding energy $= -E = +\dfrac{GMm}{2r}$ — the energy to free the satellite, equal to its orbital KE.
- Higher orbit ⇒ larger total energy, but smaller KE and slower speed; you must supply $\Delta E = \dfrac{GMm}{2}\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$.
- Orbit-change splits: $\Delta KE = -\Delta E$, $\Delta PE = +2\,\Delta E$.
- Use $r = R_E + h$ (from the centre), and $GM = gR_E^2$ to convert between forms.