Newton's Law of Gravitation and Orbital Velocity Calculator

The International Space Station is falling toward Earth every second — and moving sideways at roughly 7.7 kilometers per second, fast enough that the curve of its fall matches the curve of the Earth dropping away beneath it, so it never gets any closer. That's the entire secret of orbital motion in one sentence: an orbit isn't the absence of falling, it's falling that never quite lands. The same law that pulls an apple off a tree also keeps the Moon looping around Earth and Earth looping around the Sun. It might be the single most consequential equation Isaac Newton ever wrote down.

This page works two ways. It's an orbital velocity calculator you can drag around — set a central mass and an orbital radius, and read off speed, period, perihelion and aphelion in real time on a live elliptical orbit. And it's a full walkthrough of the physics behind those numbers: Newton's law of gravitation, $F = GMm/r^2$, and Kepler's third law, $T^2 = 4\pi^2r^3/(GM)$. Between them, those two equations describe every orbit that exists, from a satellite skimming low Earth orbit to Neptune's 165-year lap around the Sun.

What Is Newton's Law of Universal Gravitation?

Newton's law of universal gravitation says that any two objects with mass pull on each other, and the strength of that pull follows one compact rule:

$F = \frac{GMm}{r^2}$

$M$ and $m$ are the two masses, $r$ is the distance between their centers, and $G$ is the gravitational constant, $6.674 \times 10^{-11}\ \text{N·m}^2/\text{kg}^2$ — a number so small that gravity between everyday objects is utterly negligible, and so universal that it applies equally to a dropped pen and two orbiting galaxies.

A few things about that formula are easy to skim past, but worth slowing down for. The force is always attractive — gravity only pulls, never pushes. It acts along the straight line connecting the two centers of mass, no matter the shape or size of either object. And by Newton's third law, it's exactly mutual: Earth pulls the Moon with precisely the same magnitude of force that the Moon pulls Earth, just in opposite directions. So the Moon doesn't really orbit Earth in the passive sense most people picture — both bodies orbit their shared center of mass, which for the Earth-Moon system happens to sit about 1,700 km beneath Earth's surface. Inside the planet, but nowhere close to its actual center.

Why Is Gravity Called an Inverse-Square Law?

The $r^2$ in the denominator means gravity thins out fast. Double the distance between two masses and the force drops to a quarter of what it was; triple it and the force falls to a ninth. This isn't an arbitrary exponent someone picked to fit the data — picture the gravitational influence of a mass spreading outward over the surface of an ever-expanding sphere. A sphere's surface area grows with $r^2$, so the same total pull gets spread thinner and thinner as that sphere grows, and the force per unit area falls exactly as $1/r^2$. Coulomb's law for electric charges has the identical mathematical shape, for exactly the same geometric reason.

Here's that falloff drawn out, so you can see it rather than take it on faith:

Why Don't Planets Fall Into the Sun?

They are falling. That's not a simplification — it's literally true, and it's the cleanest way to understand an orbit. Earth is in constant free fall toward the Sun, accelerating inward the entire time. What keeps it from actually hitting the Sun is that it also has an enormous sideways velocity, about 29.8 km/s, and the combination of falling inward plus moving sideways traces out a curve that matches the Sun's own curvature. Earth keeps missing.

This is exactly the same situation as circular motion under centripetal force, just with gravity standing in for whatever usually plays that role — a string, friction, the walls of a curve in the road. Set the gravitational force equal to the centripetal force required to hold something on a circular path, and orbital mechanics falls straight out of it, which is exactly what the next section does.

Interactive Orbit Simulator

Drag the sliders below and watch a real elliptical orbit respond. Central Mass sets how heavy the body at the focus is, in solar masses. Orbital Radius sets how far out the orbiting body sits, in astronomical units — 1 AU is the Earth-Sun distance. Eccentricity stretches the circle into an ellipse. Watch the green trail dots closely: they're sampled at equal time intervals, not equal angle intervals, and they spread out near perihelion (closest approach, where the body moves fastest) while bunching up near aphelion (farthest point, where it crawls). That uneven spacing is Kepler's second law playing out in front of you — equal areas swept in equal times. Try dragging eccentricity down near 0.02, which is how close Earth's real orbit actually is to a perfect circle. The readout underneath doubles as a live calculator for orbital speed and period at whatever configuration you've dialed in.

What Is the Formula for Orbital Velocity?

For a circular orbit, the speed needed to stay in orbit comes directly from setting gravitational force equal to the centripetal force requirement:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

The orbiting object's own mass $m$ appears on both sides and cancels completely, leaving:

$v = \sqrt{\frac{GM}{r}}$

Look at what's missing from that formula: nothing about the orbiting object at all, beyond how far away it is. Only the central body's mass $M$ and the orbital radius $r$ matter.

Does a Satellite's Own Mass Affect Its Orbital Speed?

No — and this is one of the stranger results in mechanics once it actually sinks in. A stray bolt and a fully fueled space station at the same altitude orbit Earth at exactly the same speed, because $m$ cancelled out of the derivation above. A heavier satellite needs proportionally more gravitational force to hold it on the same path, but gravity conveniently supplies proportionally more force on a heavier object too — the two effects cancel. It's the same reason a bowling ball and a feather fall at the same rate in a vacuum near Earth's surface.

What Is Kepler's Third Law and How Do You Calculate It?

Johannes Kepler worked this relationship out empirically, decades before Newton explained why it was true. Combine $v = \sqrt{GM/r}$ with the fact that orbital period is circumference divided by speed, $T = 2\pi r / v$, and you get:

$T^2 = \frac{4\pi^2 r^3}{GM}$

The square of the orbital period is proportional to the cube of the orbital radius. Double the radius and the period doesn't just double — it multiplies by $2^{3/2} \approx 2.83$.

There's a shortcut worth knowing if you're dealing with anything orbiting the Sun specifically. Measure radius in astronomical units (AU), mass in solar masses, and period in years, and the constant $4\pi^2/(GM)$ conveniently works out to exactly 1 for $M = 1\,M_\odot$. The formula collapses to:

$T^2 = \frac{a^3}{M}$

No calculator needed for a planet at, say, 4 AU from a Sun-like star: $T = \sqrt{4^3} = \sqrt{64} = 8$ years.

How Do You Find Orbital Period from Orbital Radius?

Plug the radius straight into Kepler's third law and solve for $T$. Using the Sun-relative shortcut, a body at $a$ AU around a star of mass $M$ solar masses has period $T = \sqrt{a^3/M}$ years. Run it the other direction — given a measured period, solve for radius — and you get $a = (T^2 M)^{1/3}$, which is exactly how astronomers pin down the orbital radius of exoplanets they can only detect indirectly, from the wobble or dimming they cause in their host star. Figuring out that radius is the easy half of the problem, too; knowing how far away the host star itself sits in the first place is a much harder question, and it's the subject of its own ladder of techniques for measuring stellar distances.

Kepler's Third Law Explorer: Every Planet's Orbit, One Chart

The explorer below plots orbital period against orbital radius for every planet in the Solar System, on a log–log scale. On axes like these, $T^2 = a^3/M$ becomes a dead straight line with a slope of 3/2 — and every planet, from Mercury to Neptune, sits right on it. Real astronomical data lining up with an equation Kepler wrote down roughly 400 years ago, to within a fraction of a percent. Drag the Central Star Mass slider and watch the line itself slide. The eight grey planet markers don't move, because they really do orbit the real Sun, but the diamond marker tracking your hypothetical planet always stays glued to wherever the line currently sits.

How Strong Is Gravity on Other Planets and Moons?

Surface gravity follows from the same law. Set $r$ to a body's own radius $R$ and gravitational acceleration becomes:

$g = \frac{GM}{R^2}$

That's the constant more familiar as $9.8\ \text{m/s}^2$ on Earth's surface. It changes from world to world because both $M$ and $R$ change:

Mass and radius pull in opposite directions here, so the table isn't just a size ranking in disguise. Mars has roughly half of Earth's radius but only about a tenth of Earth's mass, so its surface gravity ends up well under half of Earth's. Jupiter is so much more massive than Earth that its gravity is over twice as strong at the "surface" — really just the cloud tops, since Jupiter has no solid ground to speak of — even though that surface sits eleven times farther from the planet's center.

Worked Examples for Physics Exams

Example 1: Gravitational force between Earth and the Moon

Find the gravitational force between Earth ($M = 5.972 \times 10^{24}$ kg) and the Moon ($m = 7.342 \times 10^{22}$ kg), separated by their average distance of $3.844 \times 10^8$ m.

$F = \frac{GMm}{r^2} = \frac{(6.674\times10^{-11})(5.972\times10^{24})(7.342\times10^{22})}{(3.844\times10^8)^2} \approx 1.98 \times 10^{20}\ \text{N}$

That's roughly 5.7 trillion times the thrust of a Saturn V rocket, and it's what keeps 384,400 km of empty space from ever letting the Moon wander off.

Example 2: Orbital velocity and period of the ISS

The International Space Station orbits at roughly 400 km altitude. Find its orbital speed and period, using $GM_{\text{Earth}} \approx 3.986 \times 10^{14}\ \text{m}^3/\text{s}^2$ and Earth's radius $R = 6.371\times10^6$ m.

Orbital radius: $r = R + 400\,\text{km} = 6.771 \times 10^6$ m.

$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.986\times10^{14}}{6.771\times10^6}} \approx 7.67\ \text{km/s}$

$T = 2\pi\sqrt{\frac{r^3}{GM}} \approx 92.4\ \text{minutes}$

A 92-minute period means the ISS laps the planet about 15.6 times a day. Astronauts on board see roughly that many sunrises every 24 hours.

Example 3: Finding orbital radius from Kepler's third law

An asteroid orbits the Sun with a measured period of 500 days. How far out is it?

Convert to years: $T = 500/365.25 \approx 1.369$ years. Using $a^3 = T^2 M$ with $M = 1$ solar mass:

$a = (T^2)^{1/3} = (1.369^2)^{1/3} \approx 1.23\ \text{AU}$

That places it just beyond Earth's orbit — without a single value of $G$ ever entering the calculation.