Earth satellites

Look up on a clear night and, if you watch patiently, you may see a steady point of light glide across the stars in a few minutes flat, unlike any planet or star — a satellite catching sunlight as it races overhead. It is not hovering, and it is not fighting gravity with some hidden engine. It is falling. Every second, Earth's gravity pulls it earthward exactly as it pulls an apple from a tree, but the satellite is also moving sideways so fast that the curved surface of the Earth drops away beneath it just as quickly as it falls. The result is a satellite that falls forever without ever landing — a closed, endless free fall we call an orbit. This single idea, first sketched by Newton with his famous cannon-on-a-mountain thought experiment, is the whole of orbital mechanics. Balance the pull of gravity against the sideways momentum needed to keep missing the ground, and out drops a precise formula for how fast a satellite must move at a given height, and how long it takes to complete one full circuit of the planet. The same balance explains why the International Space Station tears around Earth every ninety minutes while a weather satellite parked over the equator seems to hang motionlessly above one spot for its entire working life — and it explains why nobody on board those satellites ever feels their own weight, since they and their spacecraft are falling at exactly the same rate. This article builds the orbital velocity and period formulas from Newton's law of gravitation and simple circular motion, applies them to a low-Earth-orbit satellite and to the special case of a geostationary satellite, and clears up the two ideas students most often get backwards: what keeps a satellite up, and why astronauts float.

Key takeaways

Gravity supplies the centripetal force. For a satellite of mass $m$ at orbital radius $r$ , equating $GMm/r^2$ to $mv_o^2/r$ gives the orbital speed directly — no separate "anti-gravity" force is needed.
Orbital speed falls as the orbit rises. $v_o = \sqrt{GM/r}$ depends only on the planet's mass $M$ and the radius $r$ , not on the satellite's own mass; closer satellites must move faster.
The period follows Kepler's third law. $T = 2\pi\sqrt{r^3/GM}$ , so $T^2 \propto r^3$ exactly as Kepler found for planets — a satellite obeys the same law as a planet, just around a smaller central mass.
A geostationary satellite matches Earth's spin. Placed in the equatorial plane with $T \approx 24\ \text{h}$ , it sits at a unique radius (about $4.2\times10^4\ \text{km}$ from Earth's centre) and appears fixed over one point on the ground — useful for communication links, though the engineering of that link is beyond this syllabus.
No engine is required to stay in orbit. Once launched at the correct speed for its chosen radius, a satellite coasts indefinitely on gravity alone, just like the Moon; only atmospheric drag in very low orbits slowly steals energy and causes decay.
Weightlessness is free fall, not zero gravity. An orbiting astronaut and the spacecraft accelerate toward Earth at the identical rate, so there is no contact force between them — that missing contact force is what we perceive as weight.
Key idea: a satellite stays in orbit because gravity alone supplies exactly the centripetal force $mv_o^2/r$ needed for its speed $v_o = \sqrt{GM/r}$ and period $T = 2\pi\sqrt{r^3/GM}$ , so no thrust is needed once it is launched at the right speed, and weightlessness aboard is simply the sensation of falling together with the spacecraft.

The core insight

A satellite orbits because Earth's gravity $GMm/r^2$ is exactly the centripetal force $mv_o^2/r$ its circular motion needs, which fixes its speed at $v_o=\sqrt{GM/r}$ and period at $T=2\pi\sqrt{r^3/GM}$ with no engine required, and makes everyone aboard feel weightless because they and the craft fall together.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Physics chapter on Gravitation, building directly on the universal law of gravitation $F = GMm/r^2$ that opens the chapter. The prescribed expectations are to derive the orbital velocity $v_o = \sqrt{GM/r}$ and the orbital time period $T = 2\pi\sqrt{r^3/GM}$ from the centripetal-force condition, to connect this to Kepler's third law, to understand qualitatively what makes an orbit geostationary, and to explain weightlessness in an orbiting satellite as a state of continuous free fall. The detailed engineering of satellite communication links, and any treatment of orbits as vector fields or with perturbations, is explicitly outside the prescribed scope and is left for later study.

The physical picture

Consider a satellite of mass $m$ moving in a circular path of radius $r$ around Earth's centre, where Earth has mass $M$ . For the orbit to remain circular rather than spiral inward or fly off outward, the net inward (centripetal) force must exactly equal the gravitational pull, since gravity is the only force acting once the satellite is above the atmosphere:

\frac{GMm}{r^2} = \frac{mv_o^2}{r}

where: $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ is the universal gravitational constant, $M$ is Earth's mass, $m$ is the satellite's mass, $r$ is the orbital radius measured from Earth's centre (not from the surface), and $v_o$ is the orbital speed. The physical reading is that the satellite's own mass $m$ cancels from both sides — a heavy satellite and a light one at the same radius need exactly the same speed to stay in orbit, exactly as a heavy ball and a light ball fall at the same rate near Earth's surface. Solving for the speed gives the central result of this topic:

v_o = \sqrt{\frac{GM}{r}}

where: $v_o$ is the orbital speed at radius $r$ . A satellite skimming just above the atmosphere ( $r \approx R_E$ ) moves fastest; push it out to a larger $r$ and $v_o$ shrinks, since gravity is weaker there and less centripetal force — hence less speed — is needed to balance it. Since the satellite sweeps out the full circumference $2\pi r$ once every period $T$ while moving at speed $v_o$ , the period follows immediately:

T = \frac{2\pi r}{v_o} = 2\pi\sqrt{\frac{r^3}{GM}}

where: $T$ is the orbital period, the time for one complete revolution. Squaring both sides shows $T^2 = \dfrac{4\pi^2}{GM}r^3$ , so $T^2 \propto r^3$ — precisely Kepler's third law, now derived rather than merely observed, because a satellite bound by gravity to Earth obeys the same mathematics as a planet bound by gravity to the Sun.

Figure

A satellite moving on a circular path around Earth, with a red arrow showing gravity pulling inward toward the centre and a blue arrow showing the satellite velocity tangent to the circle, labelled with the orbital radius r

Geostationary satellites. If a satellite is placed in the equatorial plane and its period is made to equal Earth's own rotation period (one sidereal day, close to $24\ \text{h}$ ), it appears to hang stationary over a single point on the ground, because it turns around Earth's axis at exactly the same rate the ground does. Solving $T = 2\pi\sqrt{r^3/GM}$ for $r$ with $T \approx 86400\ \text{s}$ gives a unique orbital radius of about $4.2\times10^7\ \text{m}$ from Earth's centre — roughly $3.6\times10^4\ \text{km}$ above the surface. This single, very large radius is why only one ring of space around the equator can host geostationary satellites, and why designing the communication links that use this orbit is left for further study beyond this syllabus.

Weightlessness in orbit. Inside an orbiting spacecraft, the astronaut and the spacecraft are both in free fall toward Earth, accelerating at exactly the same rate. Weight, as we normally feel it, is the contact (normal) force the floor pushes up on us to stop us from falling — when both floor and astronaut are falling together at the same rate, that contact force drops to zero, and the astronaut floats. Gravity has not vanished; it is, in fact, what is keeping the whole system in its curved orbital path. What has vanished is the sensation of weight, because nothing is being held back from falling.

Derivation AOrbital speed and period of a satellite from the centripetal-force condition

The goal is to find the exact speed a satellite must be launched at, and the exact time it will then take to complete one orbit, for any given orbital radius $r$ around a planet of mass $M$ — using nothing beyond Newton's law of gravitation and the elementary formula for centripetal acceleration in uniform circular motion, with no calculus required.

Step 1. Identify the forces on the orbiting satellite. Once above the atmosphere, the only significant force acting on the satellite is Earth's gravitational pull, directed along the line joining the satellite to Earth's centre, of magnitude $F = GMm/r^2$ by the universal law of gravitation. There is no other force to balance, because there is no surface to push back and no air resistance in the idealised orbit considered here.

Step 2. Recognise that this force must be centripetal. For the satellite to keep moving on a circle of radius $r$ rather than travel in a straight line, some force must continuously pull it toward the centre; the required magnitude of that force, from elementary circular motion, is $F_c = mv_o^2/r$ , where $v_o$ is the (constant) orbital speed. Since gravity is the only force present, it alone must supply this required centripetal force, giving the governing equation $GMm/r^2 = mv_o^2/r$ .

Step 3. Cancel and solve for the orbital speed. Both sides carry one factor of the satellite's mass $m$ and one factor of $r$ ; cancelling these (the $m$ cancels completely, one power of $r$ cancels from $r^2$ leaving $r$ in the denominator) leaves $GM/r = v_o^2$ , so $v_o = \sqrt{GM/r}$ . This is the orbital speed formula: it depends only on the central mass $M$ and the chosen radius $r$ , never on the mass of the satellite itself, which is why a heavy communication satellite and a light nanosatellite at the same altitude move at the same speed.

Step 4. Convert speed into a period using the geometry of one full revolution. In one orbital period $T$ , the satellite travels the full circumference of its circular path, a distance $2\pi r$ , at the constant speed $v_o$ found above, so $T = 2\pi r / v_o$ . Substituting $v_o = \sqrt{GM/r}$ and simplifying the fraction gives $T = 2\pi r \big/ \sqrt{GM/r} = 2\pi\sqrt{r^3/GM}$ , the orbital period formula, which reproduces $T^2 \propto r^3$ , Kepler's third law, as a direct consequence of Newtonian gravity rather than an independently observed rule.

Worked examples

Board level - orbital speed and period of a low-Earth-orbit satellite

Find the orbital speed and time period of a satellite revolving at a height of $600\ \text{km}$ above the Earth's surface. Take $M_E = 6.0\times10^{24}\ \text{kg}$ , $R_E = 6.4\times10^6\ \text{m}$ , $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ .

Step 1. Convert the given height into an orbital radius measured from Earth's centre, since the formulas use $r$ , not $h$ : $r = R_E + h = 6.4\times10^6 + 0.6\times10^6 = 7.0\times10^6\ \text{m}$ .

Step 2. Substitute into the orbital speed formula. $v_o = \sqrt{GM_E/r} = \sqrt{(6.67\times10^{-11})(6.0\times10^{24})/(7.0\times10^6)}$ . The numerator $GM_E = 6.67\times10^{-11}\times6.0\times10^{24} = 4.0\times10^{14}\ \text{m}^3/\text{s}^2$ .

Step 3. Complete the arithmetic. $v_o = \sqrt{4.0\times10^{14}/7.0\times10^6} = \sqrt{5.71\times10^{7}} \approx 7560\ \text{m/s}$ .

Step 4. Find the period from $T = 2\pi r/v_o$ . $T = 2\pi(7.0\times10^6)/7560 \approx 4.398\times10^7/7560 \approx 5818\ \text{s}$ , which is about $97\ \text{minutes}$ .

The orbital speed is about $7.56\ \text{km/s}$ and the orbital period is about $97\ \text{minutes}$ — this is typical of a low-Earth-orbit satellite such as the International Space Station, which circles Earth roughly every hour and a half.

Competitive level - radius and altitude of a geostationary orbit, compared with the low-Earth-orbit satellite above

A satellite is to be placed in a geostationary orbit, so its period matches Earth's rotation period, $T = 24\ \text{h} = 86400\ \text{s}$ . Using the same $M_E$ and $G$ as above, find the orbital radius, the altitude above Earth's surface, and the ratio of its orbital speed to the low-Earth-orbit satellite's speed found above.

Step 1. Start from $T = 2\pi\sqrt{r^3/GM_E}$ and rearrange for $r$ : squaring both sides gives $T^2 = 4\pi^2 r^3/GM_E$ , so $r^3 = GM_E T^2/(4\pi^2)$ .

Step 2. Substitute the numbers. $GM_E = 4.0\times10^{14}\ \text{m}^3/\text{s}^2$ (from the previous example) and $T^2 = (86400)^2 \approx 7.465\times10^9\ \text{s}^2$ , so $r^3 = (4.0\times10^{14})(7.465\times10^9)/(4\pi^2) \approx 2.986\times10^{24}/39.48 \approx 7.56\times10^{22}\ \text{m}^3$ .

Step 3. Take the cube root: $r \approx (7.56\times10^{22})^{1/3} \approx 4.23\times10^7\ \text{m}$ . Subtracting Earth's radius gives the altitude, $h = r - R_E \approx 4.23\times10^7 - 0.064\times10^7 \approx 4.17\times10^7\ \text{m} \approx 3.6\times10^4\ \text{km}$ , matching the well-known figure for geostationary altitude.

Step 4. Find the orbital speed at this radius and compare it with the low-Earth-orbit case. $v_{geo} = \sqrt{GM_E/r} = \sqrt{4.0\times10^{14}/4.23\times10^7} \approx \sqrt{9.46\times10^{6}} \approx 3075\ \text{m/s} \approx 3.07\ \text{km/s}$ . Comparing with the earlier $v_o \approx 7.56\ \text{km/s}$ , the ratio is $v_{LEO}/v_{geo} \approx 7.56/3.07 \approx 2.46$ : the geostationary satellite orbits about six times farther out and moves about $2.5$ times more slowly than the low-Earth-orbit satellite, illustrating $v_o \propto 1/\sqrt r$ directly. Timing note: In a competitive-exam setting, do not re-derive $T = 2\pi\sqrt{r^3/GM}$ from scratch under time pressure — memorise the rearranged form $r^3 = GMT^2/4\pi^2$ for "period given, radius wanted" problems, keep $GM_E = 4.0\times10^{14}\ \text{m}^3/\text{s}^2$ as a ready anchor number, and always double-check whether the question wants the radius $r$ (from Earth's centre) or the altitude $h$ (from the surface) before writing the final answer, since this radius-versus-altitude mix-up is the single most common way marks are dropped on this topic.

Common traps

Believing a satellite needs its engines firing continuously to stay up. A stable orbit needs no continuous thrust at all; gravity alone supplies the centripetal force once the satellite is launched at the correct speed for its chosen radius.
Mixing up orbital radius $r$ (measured from Earth's centre) with orbital altitude $h$ (measured from Earth's surface). Every formula in this topic uses $r$ , so always convert with $r = R_E + h$ before substituting.
Treating weightlessness as though gravity has switched off in orbit. Gravity is very much present — it is precisely what curves the satellite's path into an orbit — and weightlessness is only the sensation of falling together with the spacecraft, not the absence of gravitational force.
Assuming every satellite is geostationary. A geostationary orbit is one very specific radius and an equatorial plane matched to Earth's $24$ -hour rotation; most satellites, including the ISS and most GPS satellites, orbit at other radii and are not geostationary at all.
Forgetting that orbital speed decreases as radius increases. Because $v_o \propto 1/\sqrt r$ , a common error is to assume farther-out satellites move faster; in fact they move more slowly, and it is precisely this slower motion at a larger radius that lets a geostationary satellite complete only one revolution per day.

Common trapA stable orbit is not a tug-of-war between gravity and a separate outward force

The most persistent misconception in this topic is imagining that a satellite orbits because some outward "centrifugal force" exactly cancels gravity, leaving it suspended in balance. There is no such outward force acting on the satellite in an inertial (non-rotating) frame — only one real force, gravity, acts on it at all, and that single force is the entire reason the satellite's path curves into a circle instead of continuing in a straight line. The correct picture, from Step 2 of the derivation above, is that gravity equals the centripetal force the circular motion requires — not that it is cancelled by anything. If gravity were somehow switched off mid-orbit, the satellite would not float in place; it would fly off in a straight line tangent to its former orbit, because that straight-line motion, not staying in a circle, is what a satellite does in the total absence of any force.

You should be able to…

Derive the orbital velocity formula $v_o = \sqrt{GM/r}$ from the condition that gravity supplies the centripetal force for a circular orbit.
Derive the orbital period formula $T = 2\pi\sqrt{r^3/GM}$ and show that it reproduces Kepler's third law, $T^2 \propto r^3$ .
Convert correctly between orbital radius $r$ (from Earth's centre) and altitude $h$ (from the surface) in numerical problems.
Explain qualitatively why a geostationary orbit must be equatorial and have a period of about $24$ hours, and estimate its radius given $T$ .
Explain weightlessness aboard an orbiting satellite as a consequence of free fall, not the absence of gravity.
Solve for orbital speed, period, or radius given any one of $r$ , $T$ , or $h$ , along with $M_E$ and $G$ .

Ask the tutor: Want to work through how the energy of an orbiting satellite, $E = -GMm/2r$ , connects to this same orbital-speed picture, or compare a satellite's orbit with the escape-speed condition you studied earlier?

If gravity is still acting on the ISS, why doesn't it just fall down to Earth?

It is falling — that is exactly what an orbit is. The ISS is pulled toward Earth's centre every instant, and if it had zero sideways speed it would indeed crash straight down. But it also moves sideways at about $7.7\ \text{km/s}$ , fast enough that by the time gravity has pulled it down a little, the curved Earth's surface has already dropped away by the same amount underneath it. The net effect is that the distance between the ISS and the ground never changes: it is falling continuously, but always missing. This is the same idea Newton illustrated with a cannonball fired fast enough from a very tall mountain that it would circle the entire Earth instead of landing.

Why do GPS satellites, the ISS, and geostationary TV satellites all orbit at such different heights?

Every one of these is a legitimate stable orbit — there is no single "correct" satellite altitude, only a trade-off fixed by $v_o = \sqrt{GM/r}$ and $T = 2\pi\sqrt{r^3/GM}$ . The ISS sits low (around $400\ \text{km}$ ) so astronauts and cargo need less fuel to reach it, completing an orbit in about ninety minutes. GPS satellites orbit at roughly $2\times10^4\ \text{km}$ , a compromise chosen so a modest number of satellites can cover the whole Earth with good geometry for triangulating position, taking about twelve hours per orbit. Geostationary satellites sit at the one special radius, about $3.6\times10^4\ \text{km}$ up, where the $24$ -hour period matches Earth's spin, which is invaluable for a satellite dish that must point at a fixed spot in the sky without tracking motion.

Does a heavier satellite need to go faster to stay in the same orbit as a lighter one?

No, and this often surprises students the first time they see the derivation. In Step 3 of the derivation, the satellite's mass $m$ cancels completely out of the governing equation $GMm/r^2 = mv_o^2/r$ , leaving $v_o = \sqrt{GM/r}$ , a formula with no $m$ in it at all. Physically this happens for the same reason a heavy ball and a light ball fall at the same rate near Earth's surface — a bigger mass feels a proportionally bigger gravitational force, but it also needs a proportionally bigger force to produce the same acceleration, so the two effects cancel exactly. A one-tonne satellite and a ten-kilogram cubesat at the same altitude must move at exactly the same orbital speed.

Could a satellite ever be geostationary over, say, Delhi instead of the equator?

No — a geostationary orbit must lie in Earth's equatorial plane, because the satellite has to keep pace with a specific point on the ground as Earth spins about its polar axis, and only a circle in the equatorial plane, centred on that same axis, can stay above one fixed longitude indefinitely. An orbit tilted away from the equator would still have a $24$ -hour period at the geostationary radius, but the satellite would trace a figure resembling a figure-eight above the ground each day instead of hovering over one point, since its path would carry it north and south of the equator as Earth turns beneath it. That is why every true geostationary satellite you see quoted in the news sits somewhere along the equatorial ring, never over a city at higher latitude like Delhi.