Conservation of energy in orbital motion explained: HSC Physics Module 5

For a circular orbit, the total mechanical energy is:

$E = -\frac{G M m}{2 r}$.

Orbital radius: $r = R_E + h = 6.37 \times 10^6 + 6.0 \times 10^5 = 6.97 \times 10^6$ m.

$E = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500}{2 \times 6.97 \times 10^6}$
$E = -\frac{1.99 \times 10^{17}}{1.39 \times 10^7}$
$E = -1.43 \times 10^{10}$ J.

The negative sign indicates the satellite is gravitationally bound: positive work must be done to lift it to infinity (where $E = 0$).

Markers reward the formula derivation (or correct quotation), the use of $r = R_E + h$, the negative answer with units, and an explicit comment on the physical meaning of the negative sign.

For a circular orbit: $K = \frac{G M m}{2 r}$, $U = -\frac{G M m}{r}$, $E = -\frac{G M m}{2 r}$.

Moving from $r_1$ to $r_2$ (where $r_2 > r_1$):

• Gravitational potential energy increases (becomes less negative). $U_2 - U_1 = G M m \left(\frac{1}{r_1} - \frac{1}{r_2}\right) > 0$.
• Kinetic energy decreases. $K_2 - K_1 = \frac{G M m}{2} \left(\frac{1}{r_2} - \frac{1}{r_1}\right) < 0$. The satellite moves more slowly in the higher orbit.
• Total mechanical energy increases (becomes less negative). $\Delta E = \frac{G M m}{2} \left(\frac{1}{r_1} - \frac{1}{r_2}\right) > 0$.

The increase in $U$ is twice the magnitude of the decrease in $K$, so the net change in total energy is positive and equal to the work done by the rocket. Counter-intuitively, a higher orbit has more total energy but lower speed.

Markers reward the three correct comparisons with signs, the explicit reference to the work done by the rocket, and the comment on the unusual relationship between altitude and speed.

Kinetic energy. $K = \dfrac{GMm}{2r} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(800)}{2(8.5 \times 10^6)}$

$K = \dfrac{3.186 \times 10^{17}}{1.70 \times 10^7} = 1.87 \times 10^{10}\ \text{J}$.

Potential energy. $U = -\dfrac{GMm}{r} = -2K = -3.75 \times 10^{10}\ \text{J}$.

Marks: one for the correct $K$ formula with substitution, one for $K = 1.87 \times 10^{10}$ J, one for $U = -3.75 \times 10^{10}$ J (using either the direct formula or $U = -2K$).

(a) $E$ is negative for a bound circular orbit ($E = -GMm/(2r)$). This means the satellite does not have enough energy to reach infinity ($r \to \infty$, where $U \to 0$) - it is gravitationally trapped.

(b) $E = 0$ exactly at the escape condition: the object reaches infinity with zero kinetic energy left over.

Marks: one for stating $E < 0$ for the bound orbit, one for stating $E = 0$ at escape with the "trapped versus free at infinity" explanation.

For a circular orbit, $E = -K$ (since $E = K + U$, $U = -2K$, so $E = K - 2K = -K$).

$K = -E = 7.32 \times 10^9\ \text{J}$.

Marks: one for correctly quoting/using the relationship $E = -K$, one for the correct positive value with unit.

$K_1 = \dfrac{GMm}{2r_1} = 1.731 \times 10^{10}\ \text{J}$, $K_2 = \dfrac{GMm}{2r_2} = 1.298 \times 10^{10}\ \text{J}$, so $\Delta K = -4.33 \times 10^9\ \text{J}$ (kinetic energy decreases).

$U_1 = -\dfrac{GMm}{r_1} = -3.463 \times 10^{10}\ \text{J}$, $U_2 = -\dfrac{GMm}{r_2} = -2.597 \times 10^{10}\ \text{J}$, so $\Delta U = +8.66 \times 10^9\ \text{J}$ (potential energy increases).

$\Delta E = \Delta K + \Delta U = -4.33 \times 10^9 + 8.66 \times 10^9 = +4.33 \times 10^9\ \text{J}$.

Marks: one for correct $\Delta K$ (negative), one for correct $\Delta U$ (positive, double the magnitude of $\Delta K$), one for $\Delta E = +4.33 \times 10^9$ J, one for noting $\Delta E = -\Delta K$ as a consistency check.

(a) Reading the graph at $r = 2$: $K \approx 0.5$ units and $U \approx -1.0$ units. Then $|U| = 1.0 = 2 \times 0.5 = 2K$, confirming the virial relation $|U| = 2K$ for a circular gravitational orbit.

(b) At $r = 1$, $E \approx -1.0$ units; at $r = 5$, $E \approx -0.2$ units. Gradient $= \dfrac{\Delta E}{\Delta r} = \dfrac{-0.2 - (-1.0)}{5 - 1} = \dfrac{0.8}{4} = 0.2$ units per unit of $r$.

The gradient is positive because $E = -GMm/(2r)$ increases (becomes less negative) as $r$ increases - it is the rate at which total mechanical energy must be supplied per unit increase in orbital radius, i.e. it reflects how much work per metre a rocket motor must do to raise the orbit at that point on the curve.

Marks: one for correctly reading $K \approx 0.5$ and $U \approx -1.0$ at $r = 2$, one for confirming $|U| = 2K$, one for a correct gradient calculation with a reasonable pair of points, one for the physical interpretation (rate of energy input needed to raise the orbit, positive because $E$ increases with $r$).

$K = GMm/(2r)$ decreases as $r$ increases, and $K = \tfrac{1}{2}mv^2$, so a smaller $K$ at the same mass means a smaller speed $v$ - the satellite genuinely moves more slowly at larger $r$.

At the same time, $U = -GMm/r$ increases (becomes less negative) as $r$ increases, and this increase is twice the size of the decrease in $K$, so $E = K + U$ still increases overall.

So "more total energy" does not mean "more speed": the extra energy at a higher orbit is stored entirely as (less negative) potential energy, while the kinetic energy - and hence the speed - has actually fallen.

Marks: one for linking lower $K$ to lower speed via $K = \tfrac{1}{2}mv^2$, one for stating $U$ increases by twice as much as $K$ falls, one for the explicit resolution that the extra $E$ is stored as $U$, not as speed.

Band-6 plan. (1) State the three orbital-energy formulas and their signs. (2) Analyse how each changes as $r_1 \to r_2 > r_1$, with the virial relation $|U| = 2K$. (3) Identify where the added energy goes. (4) Directly address and refute the student's claim with the correct physics, giving a judgement.

Model answer. For a circular orbit of radius $r$, kinetic energy is $K = GMm/(2r)$, gravitational potential energy is $U = -GMm/r$, and total mechanical energy is $E = K + U = -GMm/(2r)$. All three depend only on $r$ (for fixed $M$ and $m$), and $|U| = 2K$ always holds.

Transferring the satellite from $r_1$ to a higher orbit $r_2 > r_1$ requires the rocket to do positive work, since $\Delta E = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right) > 0$ for $r_2 > r_1$. As $r$ increases, $K$ decreases (the satellite orbits more slowly, since $K = \tfrac{1}{2}mv^2$), while $U$ increases by exactly twice the magnitude that $K$ falls, so the net total energy $E$ rises. The energy supplied by the rocket's engines therefore ends up almost entirely as gravitational potential energy, not as kinetic energy: in fact kinetic energy is lost even as total energy is gained.

The student's claim is incorrect. It is true that the satellite gains total mechanical energy $E$ during the transfer - but "gaining energy" does not mean "gaining speed", because $E$ is the sum of $K$ and $U$, and here all of the gain (and more) shows up in $U$ while $K$ falls. The satellite in the higher orbit is measurably slower ($v = \sqrt{GM/r}$ decreases with $r$), even though it sits higher in the gravitational well and is closer to escaping (closer to $E = 0$). The correct statement is that a higher orbit has more total energy but less kinetic energy and lower speed.

Marker's note: the top band states all three formulas correctly with signs, uses the $|U| = 2K$ relation (or equivalent) to show the energy bookkeeping, and explicitly refutes the student's claim by distinguishing total energy from kinetic energy/speed. A response that only says "the student is wrong" without the $K$-versus-$E$ distinction caps in the middle band.

Band-6 plan. Thesis: energy accounting, not just distance, governs the manoeuvre. Cover (1) the total energy required using $\Delta E = \tfrac{GMm}{2}(1/r_1 - 1/r_2)$, (2) how $K$ and $U$ individually evolve throughout the slow continuous burn (not just start/end states), (3) practical consequences (fuel/engine power, duration, the counter-intuitive speed decrease), (4) a concluding assessment of what "higher orbit" costs in energy terms.

Model answer. The total energy the ion engine must ultimately supply is fixed by the start and end radii alone: $\Delta E = E_2 - E_1 = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$, which is positive for a higher final orbit ($r_2 > r_1$) and independent of the path taken or how slowly the thrust is applied. This sets a hard minimum on the work the propulsion system must do, and hence (with the engine's efficiency and exhaust velocity) the minimum propellant and total burn time required - the central design constraint for the mission.

Because the burn is continuous and low-thrust rather than an instantaneous impulse, the satellite does not jump directly between two circular orbits; it spirals slowly outward through a continuum of near-circular orbits of increasing $r$. At every stage of this spiral, kinetic energy $K = GMm/(2r)$ is falling and potential energy $U = -GMm/r$ is rising by twice as much, so $E = K + U$ rises steadily and continuously in step with the work being done by the engine at that instant - the mission planners can therefore track progress by monitoring $E$ (or equivalently orbital radius) rather than speed.

A key operational consequence is that the satellite is continuously losing speed as it climbs, which is counter-intuitive given that thrust is being applied in (broadly) the direction of motion: the extra energy delivered by the engine is going into $U$, not into speed. Planners must also budget for the fact that the required $\Delta E$, and hence fuel, scales with how much $1/r_1$ exceeds $1/r_2$ - so most of the energy cost is incurred raising the satellite out of the deep, steep part of the potential well near $r_1$, with diminishing returns (in energy terms) for further raising once $r$ is already large.

Overall, the energy analysis shows that the mission's cost is set entirely by $\Delta E$ between the two orbits, that the process trades kinetic energy for a larger gain in potential energy throughout, and that engineers must therefore size the ion engine and mission duration to supply this $\Delta E$, not simply to "increase the satellite's speed."

Marker's note: the top band uses the path-independent $\Delta E$ formula explicitly, correctly describes the continuous $K$-falls/$U$-rises-more behaviour during the slow spiral (not just endpoints), and draws out at least one genuine mission-planning consequence (minimum fuel/energy budget, or the counter-intuitive slowing). Responses that treat this as a single instantaneous transfer without addressing the continuous nature of a low-thrust burn cap below the top band.