Orbital velocity and period

A satellite never fires engines to stay up — it simply falls, forever, missing the Earth because it also moves sideways fast enough that the ground curves away beneath it exactly as fast as the satellite drops. This article pins down that balance mathematically: the exact speed a satellite needs at a given orbital radius (orbital velocity, $v_o$ ) and the exact time it takes to complete one full circuit (time period, $T$ ). Both fall straight out of one idea — gravity alone supplies the centripetal force that keeps the satellite curving around the planet instead of flying off in a straight line or spiralling in. The same balance also explains why an astronaut orbiting Earth floats freely: not because gravity has switched off, but because everything in the cabin is falling together at the same rate. This is core ISC Class 11 Gravitation, and it sets up the geostationary-satellite and Kepler's-law topics that follow.

The core insight

Orbital velocity $v_o=\sqrt{GM/r}$ and time period $T=2\pi\sqrt{r^3/GM}$ both follow from setting gravitational force equal to the centripetal force needed for a satellite of any mass to move in a circle of radius $r$ around a planet of mass $M$ .

The core idea

Setting up the balance. A satellite of mass $m$ orbits a planet of mass $M$ in a circle of radius $r$ , measured from the planet's centre. Gravity is the only force acting on it in orbit, and it must supply exactly the centripetal force needed to keep the path circular:

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

where: $G$ is the universal gravitational constant, $M$ is the planet's mass, $m$ is the satellite's mass, $r$ is the orbital radius, and $v_o$ is the orbital (tangential) speed. Cancelling one factor each of $m$ and $r$ from both sides gives the orbital velocity formula, worked out step by step in the derivation below.

Two forms worth knowing. In terms of $G$ and $M$ , $v_o=\sqrt{GM/r}$ . For a satellite skimming just above a planet's surface, $r\approx R$ (the planet's radius) and $GM=gR^2$ , so the formula simplifies to $v_o=\sqrt{gR}$ — Earth's value is about $7.9\ \text{km/s}$ , the "first cosmic velocity." Neither form contains the satellite's own mass $m$ : orbital speed depends only on the central body being orbited and the radius of the orbit, never on what is doing the orbiting.

From speed to period. The time period $T$ is how long one full revolution takes, so it is simply the orbit's circumference divided by the speed: $T=2\pi r/v_o$ . Substituting $v_o=\sqrt{GM/r}$ gives $T=2\pi\sqrt{r^3/GM}$ — a relation showing $T^2\propto r^3$ , the same proportionality you will meet again, generalised to elliptical orbits, as Kepler's third law.

Figure

A satellite in circular orbit with gravity arrow pointing to the planet centre and velocity arrow tangent to the path

Weightlessness in orbit. Inside an orbiting satellite, the satellite and everything in it — including an astronaut — are all falling toward the planet at exactly the same rate, $g_r=GM/r^2$ , while also moving sideways fast enough to continuously "miss" the surface. Since nothing accelerates faster or slower than anything else, there is no relative force between the astronaut and the cabin floor: the astronaut experiences apparent weightlessness, even though gravity is fully present and is precisely what keeps the satellite in orbit at all. It is the same effect as standing on a scale inside a lift in free fall — the scale reads zero not because gravity vanished, but because scale and passenger accelerate downward together.

Derivation ADeriving orbital velocity and time period from centripetal balance

The goal is to derive both $v_o=\sqrt{GM/r}$ and $T=2\pi\sqrt{r^3/GM}$ for a satellite of mass $m$ moving in a circular orbit of radius $r$ around a planet of mass $M$ , starting from nothing but Newton's law of gravitation and the formula for centripetal force, and to show along the way that neither result depends on the satellite's own mass.

Step 1. Identify the force supplying circular motion. For the satellite to move on a circle of radius $r$ instead of flying off in a straight line, a net inward (centripetal) force of magnitude $mv_o^2/r$ must act on it. In orbit, gravity is the only force present, and it has magnitude $GMm/r^2$ by Newton's law of gravitation. Since gravity is the sole source of that centripetal force, the two must be equal: $\dfrac{GMm}{r^2}=\dfrac{mv_o^2}{r}$ .

Step 2. Cancel the satellite's mass. Both sides contain a factor of $m$ , and it cancels completely, leaving $\dfrac{GM}{r^2}=\dfrac{v_o^2}{r}$ . This is the algebraic origin of the fact that orbital speed never depends on the mass of the orbiting body — a feather-light probe and a massive space station at the same radius $r$ around the same planet move at identical speed.

Step 3. Solve for $v_o$ . Multiplying both sides by $r$ gives $v_o^2=\dfrac{GM}{r}$ , so $v_o=\sqrt{\dfrac{GM}{r}}$ . Near a planet's surface, $r\approx R$ and $GM=gR^2$ , giving the equivalent surface form $v_o=\sqrt{gR^2/R}=\sqrt{gR}$ .

Step 4. Convert speed into time period. One full orbit covers the circle's circumference, $2\pi r$ , at constant speed $v_o$ , so $T=\dfrac{2\pi r}{v_o}$ . Substituting $v_o=\sqrt{GM/r}$ : $T=\dfrac{2\pi r}{\sqrt{GM/r}}=2\pi r\sqrt{\dfrac{r}{GM}}=2\pi\sqrt{\dfrac{r^3}{GM}}$ , which is the required period formula, showing $T^2\propto r^3$ .

Worked example

A satellite orbits Earth at a distance $r=7.0\times10^6\ \text{m}$ from Earth's centre. Take $GM=4.0\times10^{14}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}$ . Find its orbital velocity and time period, and check the period against a typical low-Earth-orbit satellite.

Step 1. Apply $v_o=\sqrt{GM/r}$ : $v_o=\sqrt{\dfrac{4.0\times10^{14}}{7.0\times10^6}}=\sqrt{5.714\times10^7}\approx7559\ \text{m/s}\approx7.56\ \text{km/s}$ .

Step 2. Apply $T=2\pi\sqrt{r^3/GM}$ . First find $r^3=(7.0\times10^6)^3=343\times10^{18}=3.43\times10^{20}\ \text{m}^3$ .

Step 3. Divide and take the square root: $\dfrac{r^3}{GM}=\dfrac{3.43\times10^{20}}{4.0\times10^{14}}=8.575\times10^5$ , so $\sqrt{8.575\times10^5}\approx926.0$ .

Step 4. Multiply by $2\pi$ : $T=2\pi\times926.0\approx5818\ \text{s}\approx97\ \text{minutes}$ . The satellite's orbital velocity is about $7.56\ \text{km/s}$ and its time period is about $97$ minutes — close to the roughly $90$ – $100$ minute period of a real low-Earth-orbit satellite, a useful sanity check.

Competitive stretch

For a satellite orbiting close to Earth's surface, using $g=9.8\ \text{m/s}^2$ and $R=6.4\times10^6\ \text{m}$ , the surface-form check gives $v_o=\sqrt{gR}=\sqrt{9.8\times6.4\times10^6}=\sqrt{6.272\times10^7}\approx7920\ \text{m/s}\approx7.92\ \text{km/s}$ — the familiar first cosmic velocity. It is slightly higher than the $7.56\ \text{km/s}$ found above because that satellite orbits at $r=7.0\times10^6\ \text{m}$ , noticeably farther out than Earth's surface radius $R=6.4\times10^6\ \text{m}$ , and $v_o=\sqrt{GM/r}$ decreases as $r$ grows.

Common traps

Plugging the satellite's own mass $m$ into $v_o$ or $T$ — it always cancels in Step 2 of the derivation; both formulas depend only on the central mass $M$ (or $g$ , $R$ ) and the orbital radius $r$ .
Confusing orbital radius $r$ , measured from the planet's centre, with height above the surface $h$ — remember $r=R+h$ , and both formulas must use $r$ , never $h$ , directly.
Treating weightlessness in orbit as "gravity switched off" — gravity is fully present and is exactly what curves the path into an orbit; the astronaut is in continuous free fall, not in a gravity-free zone.
Applying $v_o=\sqrt{gR}$ at an orbital radius that is not close to the surface — that shortcut only holds when $r\approx R$ ; for any other $r$ , use $v_o=\sqrt{GM/r}$ directly.

Ask the tutor: Want a walkthrough of why gravity acts as the centripetal force, or practice converting between orbital radius and height above the surface in numericals? Ask the AI tutor about Orbital velocity and period.

Why does a satellite's own mass not affect its orbital velocity?

Because in the balance equation $GMm/r^2 = mv_o^2/r$ , the satellite's mass $m$ appears identically on both sides — once inside the gravitational force and once inside the centripetal force requirement — so it cancels completely. Physically, a heavier satellite feels a proportionally larger gravitational pull, but it also needs a proportionally larger centripetal force to keep moving on the same curved path, and the two scale together. That is why a communications satellite and a tiny cubesat at the same orbital radius move at exactly the same speed.

Is a satellite in orbit really in free fall the whole time?

Yes. At every instant, gravity is the only force acting on it, pulling it straight toward the planet's centre — exactly what free fall means. What keeps it from crashing is that it also has a large sideways (tangential) velocity, so as it falls, the curved surface of the planet keeps dropping away beneath it at the same rate. It never stops falling; it simply never catches up to the ground, tracing a closed circular path instead of a straight vertical drop.

How is time period T related to Kepler's third law that I will study next?

The relation $T=2\pi\sqrt{r^3/GM}$ derived here, rearranged, gives $T^2=\dfrac{4\pi^2}{GM}r^3$ , so $T^2\propto r^3$ for any circular orbit around a given planet. Kepler's third law states the same proportionality for planets orbiting the Sun (or, more generally, for elliptical orbits using the semi-major axis in place of $r$ ). So this derivation is not a separate fact to memorise later — it is Kepler's third law appearing early, in the simplest circular-orbit case, straight out of Newton's law of gravitation.

Orbital velocity and period

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads