Predict quantitatively the orbital properties of planets and artificial satellites in a variety of situations, including near-Earth and geostationary orbits, using the relationship between orbital speed, radius, and period

Set gravity equal to the centripetal force for a circular orbit.

$\frac{G M m}{r^2} = \frac{m v^2}{r}$, giving $v = \sqrt{\frac{G M}{r}}$.

Orbital radius: $r = R_E + h = 6.37 \times 10^6 + 4.0 \times 10^5 = 6.77 \times 10^6$ m.

$v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.77 \times 10^6}}$
$v = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^6}}$
$v = \sqrt{5.88 \times 10^7}$
$v = 7670$ m/s, or about $7.67$ km/s.

Orbital period: $T = \frac{2 \pi r}{v} = \frac{2 \pi \times 6.77 \times 10^6}{7670} = 5546$ s, about $92.4$ minutes.

Markers reward the derivation from setting gravity equal to centripetal force, the substitution with $r = R_E + h$ (not just $h$), and answers in correct SI units.

Orbital speed: $v = \sqrt{G M / r}$, so speed decreases with increasing radius.

A geostationary satellite orbits at the same angular rate as Earth's rotation (one revolution per sidereal day, in the equatorial plane). To a ground observer it remains fixed in the sky, allowing ground antennas to be aimed permanently without tracking. This is ideal for television broadcast, weather monitoring, and continuous communication links.

LEO satellites move rapidly across the sky and need tracking or large constellations (such as Starlink) to provide continuous coverage. Their lower altitude gives shorter signal delay (about 5 ms versus 240 ms for geostationary), which suits high-bandwidth and low-latency applications.

Markers reward the quantitative comparison, the geostationary geometry (24-hour equatorial), and a clear application-driven contrast.

(a) Radius. Orbital radius is measured from Earth's centre, not the altitude:

$r = R_E + h = 6.37 \times 10^6 + 6.00 \times 10^5 = 6.97 \times 10^6\ \text{m}$.

(b) Speed. Gravity supplies the centripetal force, so $v = \sqrt{GM/r}$:

$v = \sqrt{\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.97 \times 10^6}} = \sqrt{5.71 \times 10^{7}} = 7.56 \times 10^{3}\ \text{m/s}$.

Marks: one for correctly forming $r = R_E + h$, one for the formula $v = \sqrt{GM/r}$ with values substituted, one for $v = 7.56 \times 10^{3}$ m/s (7.56 km/s) with the unit.

$v = \sqrt{\dfrac{GM}{r}}$.

Setting gravity equal to the centripetal force gives $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$. The satellite's mass $m$ appears on both sides and cancels, so the orbital speed at a given radius is the same for a $1\ \text{kg}$ probe or a $10\ \text{tonne}$ satellite - it depends only on the central mass $M$ and the radius $r$.

Marks: one for the correct equation, one for explaining the cancellation of $m$ from both sides of the force equation.

The only force acting on the satellite (ignoring drag) is gravity, directed toward Earth's centre. For a circular orbit this gravitational force is exactly equal to the centripetal force required to keep the satellite moving on its circular path: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$.

The satellite is continuously accelerating toward Earth (falling), but its tangential velocity carries it sideways at the same rate the curved surface of Earth falls away beneath it, so it never gets closer. If its speed were less than the orbital speed for that radius, gravity would exceed the centripetal requirement and it would spiral in; if greater, it would rise to a larger radius.

Marks: one for identifying gravity as the sole (centripetal) force, one for the "continuously falling but missing the Earth" explanation, one for linking speed too low/too high to spiralling in/rising out.

(a) Radius. Rearranging $T = 2\pi\sqrt{r^3/(GM)}$ for $r$:

$r = \left(\dfrac{GMT^2}{4\pi^2}\right)^{1/3}$.

Convert the period: $T = 6.0 \times 3600 = 2.16 \times 10^{4}\ \text{s}$.

$r = \left(\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(2.16 \times 10^{4})^2}{39.48}\right)^{1/3} = (4.70 \times 10^{21})^{1/3} = 1.68 \times 10^{7}\ \text{m}$.

(b) Altitude. $h = r - R_E = 1.68 \times 10^{7} - 6.37 \times 10^{6} = 1.04 \times 10^{7}\ \text{m}$ (about $10\,400\ \text{km}$), a medium Earth orbit.

Marks: one for the rearranged Kepler formula, one for correctly converting $T$ to seconds before substituting, one for $r = 1.68 \times 10^{7}$ m, one for $h = r - R_E$ correctly computed.

(a) Orbital speed is $v = \sqrt{GM/r} = \sqrt{GM} \times \dfrac{1}{\sqrt{r}}$. This has the form $v = (\text{constant}) \times x$ with $x = 1/\sqrt{r}$, so plotting $v$ against $1/\sqrt{r}$ gives a straight line through the origin with gradient $\sqrt{GM}$.

(b) Gradient $= \dfrac{\Delta v}{\Delta x} = \dfrac{(6.310 - 2.576)\ \text{km/s}}{(3.162 - 1.291) \times 10^{-4}\ \text{m}^{-1/2}} = \dfrac{3.734 \times 10^{3}\ \text{m/s}}{1.871 \times 10^{-4}\ \text{m}^{-1/2}} = 2.00 \times 10^{7}\ \text{m}^{1/2}\text{s}^{-1}$.

(c) The gradient equals $\sqrt{GM}$, so $M = \dfrac{(\text{gradient})^2}{G} = \dfrac{(2.00 \times 10^{7})^2}{6.67 \times 10^{-11}} = \dfrac{4.00 \times 10^{14}}{6.67 \times 10^{-11}} = 6.00 \times 10^{24}\ \text{kg}$, close to the accepted value $5.97 \times 10^{24}\ \text{kg}$.

Marks: one for identifying $v = \sqrt{GM} \times (1/\sqrt{r})$ as the reason for linearity through the origin, one for a correct gradient with unit $\text{m}^{1/2}\text{s}^{-1}$, one for equating the gradient to $\sqrt{GM}$, one for $M = (\text{gradient})^2/G$, one for the numeric answer close to $5.97 \times 10^{24}$ kg.

(a) Speed. $v = \sqrt{GM/r}$, so $v \propto r^{-1/2}$. Doubling $r$ gives a new speed $v' = v/\sqrt{2}$, a factor of $\dfrac{1}{\sqrt{2}} \approx 0.707$ (the satellite slows down).

(b) Period. $T = 2\pi\sqrt{r^3/(GM)}$, so $T \propto r^{3/2}$. Doubling $r$ gives $T' = T \times 2^{3/2} = 2.83\,T$ (the period increases by a factor of about $2.83$).

Marks: one for the correct proportionality $v \propto r^{-1/2}$ with the factor $1/\sqrt{2}$, one for stating speed decreases, one for the correct proportionality $T \propto r^{3/2}$ with the factor $2\sqrt{2} \approx 2.83$, one for stating period increases.

Band-6 plan. (1) State the governing equations $v = \sqrt{GM/r}$ and $T = 2\pi\sqrt{r^3/(GM)}$ and what they predict as $r$ increases. (2) Give quantitative LEO vs GEO figures (speed, period, altitude). (3) Explain the geostationary geometric condition (equatorial, prograde, $T =$ one sidereal day). (4) Evaluate: match each orbit's physical properties to a real application, reaching a judgement about "suited to different applications" rather than just listing facts.

Model answer. Both orbit types obey $v = \sqrt{GM/r}$ and $T = 2\pi\sqrt{r^3/(GM)}$: as orbital radius increases, speed falls and period rises. A near-Earth orbit (altitude roughly $200$-$2000\ \text{km}$, radius $\approx 6.6$-$8.4 \times 10^{6}\ \text{m}$) therefore has a high orbital speed of about $7$-$8\ \text{km/s}$ and a short period of $90$-$130$ minutes, circling Earth more than a dozen times a day. A geostationary orbit sits at the much larger radius $r = 4.22 \times 10^{7}\ \text{m}$ (altitude about $35\,800\ \text{km}$), giving a slower speed of about $3.07\ \text{km/s}$ but a period of exactly one sidereal day ($24\ \text{h}$).

The geostationary period is not a coincidence: only an equatorial, prograde orbit with $T$ equal to Earth's rotation period keeps the satellite above a fixed point on the ground, because the satellite's angular velocity then exactly matches Earth's. This makes geostationary orbits ideal for applications needing a permanently fixed ground antenna and continuous coverage of one region, such as television broadcast and weather-monitoring satellites - the trade-off is a long, roughly $240\ \text{ms}$, one-way signal delay because of the large radius.

Near-Earth orbits, by contrast, sweep rapidly across the sky and any one satellite is only overhead for a few minutes, so continuous coverage needs either constant tracking or a large constellation (as used by Earth-observation missions and the ISS). Their much shorter range gives a signal delay of only a few milliseconds, which is why low-latency applications favour LEO over GEO despite needing many more satellites. The physics of $v$ and $T$ versus $r$ therefore directly explains why mission designers choose GEO for fixed, wide-area coverage and LEO for low-latency or high-resolution imaging tasks.

Marker's note: the top band quantifies both orbits (not just "LEO is fast, GEO is slow"), explicitly links the geostationary period to the equatorial/prograde/one-sidereal-day condition, and evaluates by matching physical properties to genuine applications with a stated trade-off (delay versus coverage). A response that only defines the two orbit types without this application judgement caps in the middle band.

Band-6 plan. (1) Confirm the correct part of the claim (gravity, and hence the required centripetal force, does decrease with $r$). (2) Identify the flawed part (the conclusion that speed must increase). (3) Derive $v = \sqrt{GM/r}$ from first principles to show speed actually decreases with $r$. (4) Resolve the apparent contradiction by separating the force and speed relationships, and reach an explicit verdict on the claim.

Model answer. The first part of the claim is correct: gravitational force follows an inverse-square law, $F = GMm/r^2$, so the force required to keep any satellite in circular orbit does fall as $r$ increases. For a stable orbit this gravitational force must equal the centripetal force $mv^2/r$, so $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$.

However, the claim's conclusion does not follow. Solving the force-balance equation for speed, $v = \sqrt{GM/r}$, so as $r$ increases, $v$ decreases, not increases. The claim confuses "a smaller force is needed" with "a faster speed is needed" - but a smaller centripetal force is achieved partly because $v^2$ also becomes smaller, and partly because the same speed swept over a larger radius already requires less centripetal force ($mv^2/r$ falls as $r$ rises even before accounting for $v$'s own decrease). Both the numerator force and the required speed fall together as $r$ grows; they are not independent.

This matches the quantitative pattern seen across real orbits: a near-Earth satellite at $r \approx 6.8 \times 10^{6}\ \text{m}$ travels at about $7.7\ \text{km/s}$, while a geostationary satellite at the much larger $r = 4.22 \times 10^{7}\ \text{m}$ travels more slowly, at about $3.07\ \text{km/s}$ - exactly the opposite trend to the claim. The claim is therefore inaccurate: higher orbits do need less centripetal force, but this is achieved with a lower orbital speed, not a higher one.

Marker's note: the top band does not simply assert "the claim is wrong" - it derives $v = \sqrt{GM/r}$ from the force balance to show algebraically why speed falls, and uses real LEO/GEO figures as supporting evidence. A response that only states "speed decreases with radius" without engaging with why the claim's reasoning about force is misleading caps in the middle band.