Satellites, Orbits & Kepler's Laws

An orbit is a fall that keeps missing the ground.

14 minUpdated 21 October 2026orbital speedorbital periodgeostationary orbitKepler's lawsweightlessness

Lesson at a glanceShow the key points

Circular orbit of radius a around mass M: v = √(GM/a) (11.25); T = 2π√(a³/GM) (11.26), so T² ∝ a³ (Kepler's third law).
Energy in a circular orbit: K = GMm/2a, U = −GMm/a, E = −GMm/2a; negative E means a bound orbit.
A geostationary satellite orbits above the equator with T = 24 h at radius ≈ 42 000 km (≈ 36 000 km altitude).
Kepler's laws: ellipses with the Sun at a focus; equal areas in equal times (angular momentum conservation); T² ∝ a³.
In orbit everything falls together, so the floor pushes on nothing: weightlessness, although gravity is still about 90% of its surface value at the ISS.

Throw a stone sideways and it curves to the ground. Throw it fast enough and the ground curves away beneath it as fast as it falls. That endless fall is an orbit.

Circular orbits

Planetary orbits are ellipses, but most are nearly circular, so circles give excellent estimates. Take a body of mass m in a circle of radius a around a much heavier body M (the Sun for planets, the Earth for satellites). Gravity supplies the centripetal force:

\frac{GMm}{a^2} = \frac{mv^2}{a}

Orbital speed and period(11.25, 11.26)

v = \sqrt{\frac{GM}{a}}, \qquad T = \frac{2\pi a}{v} = 2\pi\sqrt{\frac{a^3}{GM}}

independent of the orbiting body’s own mass

The kinetic energy is $\tfrac12mv^2 = GMm/2a$ , the potential energy $-GMm/a$ , so the total is

Energy in a circular orbit

E = K + U = -\frac{GMm}{2a}

negative, as for every bound system; K = −E and U = 2E

Worked example

The International Space Station

The ISS orbits about 400 km above the Earth. Taking R = 6400 km and M = 6.0 × 10²⁴ kg, find its speed and period.

Interactive

Orbit calculator

Choose a height above the Earth (M = 6 × 10²⁴ kg, R = 6400 km). Higher orbits are slower and take much longer: T grows as r^(3/2). At about 36 000 km the period matches one rotation of the Earth.

v = \sqrt{GM/r}

7.67km/s

T

1 h 33 min

g at that height8.65m/s²

Height above the surface

h

400km

Geostationary satellites

A satellite above the equator, moving the same way as the Earth turns, with a period of one day, stays above the same spot. Dishes can point at it without tracking. Solving $T = 2\pi\sqrt{a^3/GM}$ for a with T = 24 h:

a = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3} \approx 4.2\times10^4\ \text{km}

That is about 36 000 km above the surface. Communications and weather satellites live there.

Kepler’s laws, explained

Ellipses. An inverse-square attraction produces conic-section paths. A circle needs exactly the right speed, √(GM/a), aimed exactly sideways; anything else gives an ellipse with the attracting body at a focus (or, fast enough, an escape).
Equal areas in equal times. Gravity points straight at the Sun, so it exerts no torque about the Sun, and the planet’s angular momentum L is conserved. In a short time dt the radius sweeps a thin triangle of area $\tfrac12 r\,(v_\perp dt)$ , so $dA/dt = L/2m$ : constant. Planets speed up near the Sun and slow down far from it.
T² ∝ a³. For circles we just derived $T^2 = (4\pi^2/GM)a^3$ ; the same holds for ellipses with a the semi-major axis. The constant is the same for every planet of the Sun.

Interactive

Launch a satellite

A satellite starts 1 unit from the planet moving sideways at v, measured in units of the circular-orbit speed √(GM/r). At v = 1 it circles; faster or slower gives an ellipse with the planet at one focus; at √2 ≈ 1.41 it escapes. Turn on equal areas: each shaded wedge is swept in the same time (Kepler's second law).

v / v_{circ}

1.20

orbitellipse

eccentricity0.44

E = \tfrac12v^2 - 1

-0.28

period / circular period2.39

Launch speed

v

1.20× v_circ

Weightlessness

Gravity at the ISS is still about 89% of its surface value. So why do astronauts float? Consider a body on a surface inside an orbiting satellite. The forces on it are gravity $GMm/r^2$ and the push N from the surface. The body moves with the satellite, accelerating at $GM/r^2$ toward the Earth, so

\frac{GMm}{r^2} - N = m\frac{GM}{r^2} \quad\Rightarrow\quad N = 0

What to remember

1Circular orbit: $v = \sqrt{GM/a}$ , $T = 2\pi\sqrt{a^3/GM}$ , $E = -GMm/2a$ .
2Geostationary: T = 24 h, radius ≈ 42 000 km (≈ 36 000 km up), above the equator.
3Kepler's second law is conservation of angular momentum; the third is T² ∝ a³.
4Weightlessness: everything falls together, so supports push on nothing.

Check yourself

Satellite A orbits at 4 times the radius of satellite B. A's period is…

To move to a higher circular orbit, a satellite's total energy must…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchOrbital motionKhan Academy WatchGravity for astronauts in orbitKhan Academy WatchKepler's laws, elliptical orbits, satellitesWalter Lewin · MIT 8.01x Lecture 22 PlayKepler's LawsPhET · CU Boulder PlayGravity and OrbitsPhET · CU Boulder ReadSatellite orbits and energyOpenStax University Physics §13.4 ReadKepler's laws of planetary motionOpenStax University Physics §13.5 ReadKepler's three lawsThe Physics Classroom ReadMathematics of satellite motionThe Physics Classroom ReadWeightlessness in orbitThe Physics Classroom ReadCircular orbit velocityHyperPhysics ReadKepler's lawsHyperPhysics ReadKepler's laws of planetary motionWikipedia ReadGeostationary orbitWikipedia

PreviousHow g Changes Next Escape Velocity & Black Holes

Satellites, Orbits & Kepler's Laws

An orbit is a fall that keeps missing the ground.

14 minUpdated 21 October 2026orbital speedorbital periodgeostationary orbitKepler's lawsweightlessness

Lesson at a glanceShow the key points

Circular orbit of radius a around mass M: v = √(GM/a) (11.25); T = 2π√(a³/GM) (11.26), so T² ∝ a³ (Kepler's third law).
Energy in a circular orbit: K = GMm/2a, U = −GMm/a, E = −GMm/2a; negative E means a bound orbit.
A geostationary satellite orbits above the equator with T = 24 h at radius ≈ 42 000 km (≈ 36 000 km altitude).
Kepler's laws: ellipses with the Sun at a focus; equal areas in equal times (angular momentum conservation); T² ∝ a³.
In orbit everything falls together, so the floor pushes on nothing: weightlessness, although gravity is still about 90% of its surface value at the ISS.

Throw a stone sideways and it curves to the ground. Throw it fast enough and the ground curves away beneath it as fast as it falls. That endless fall is an orbit.

Circular orbits

\frac{GMm}{a^2} = \frac{mv^2}{a}

Orbital speed and period(11.25, 11.26)

v = \sqrt{\frac{GM}{a}}, \qquad T = \frac{2\pi a}{v} = 2\pi\sqrt{\frac{a^3}{GM}}

independent of the orbiting body’s own mass

The kinetic energy is $\tfrac12mv^2 = GMm/2a$ , the potential energy $-GMm/a$ , so the total is

Energy in a circular orbit

E = K + U = -\frac{GMm}{2a}

negative, as for every bound system; K = −E and U = 2E

Worked example

The International Space Station

The ISS orbits about 400 km above the Earth. Taking R = 6400 km and M = 6.0 × 10²⁴ kg, find its speed and period.

Interactive

Orbit calculator

Choose a height above the Earth (M = 6 × 10²⁴ kg, R = 6400 km). Higher orbits are slower and take much longer: T grows as r^(3/2). At about 36 000 km the period matches one rotation of the Earth.

v = \sqrt{GM/r}

7.67km/s

T

1 h 33 min

g at that height8.65m/s²

Height above the surface

h

400km

Geostationary satellites

a = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3} \approx 4.2\times10^4\ \text{km}

That is about 36 000 km above the surface. Communications and weather satellites live there.

Kepler’s laws, explained

Ellipses. An inverse-square attraction produces conic-section paths. A circle needs exactly the right speed, √(GM/a), aimed exactly sideways; anything else gives an ellipse with the attracting body at a focus (or, fast enough, an escape).
Equal areas in equal times. Gravity points straight at the Sun, so it exerts no torque about the Sun, and the planet’s angular momentum L is conserved. In a short time dt the radius sweeps a thin triangle of area $\tfrac12 r\,(v_\perp dt)$ , so $dA/dt = L/2m$ : constant. Planets speed up near the Sun and slow down far from it.
T² ∝ a³. For circles we just derived $T^2 = (4\pi^2/GM)a^3$ ; the same holds for ellipses with a the semi-major axis. The constant is the same for every planet of the Sun.

Interactive

Launch a satellite

v / v_{circ}

1.20

orbitellipse

eccentricity0.44

E = \tfrac12v^2 - 1

-0.28

period / circular period2.39

Launch speed

v

1.20× v_circ

Weightlessness

\frac{GMm}{r^2} - N = m\frac{GM}{r^2} \quad\Rightarrow\quad N = 0

What to remember

1Circular orbit: $v = \sqrt{GM/a}$ , $T = 2\pi\sqrt{a^3/GM}$ , $E = -GMm/2a$ .
2Geostationary: T = 24 h, radius ≈ 42 000 km (≈ 36 000 km up), above the equator.
3Kepler's second law is conservation of angular momentum; the third is T² ∝ a³.
4Weightlessness: everything falls together, so supports push on nothing.

Check yourself

Satellite A orbits at 4 times the radius of satellite B. A's period is…

To move to a higher circular orbit, a satellite's total energy must…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PreviousHow g Changes Next Escape Velocity & Black Holes

Circular orbits#

Geostationary satellites

Kepler’s laws, explained#

Weightlessness#

Check yourself

Keep exploring

Circular orbits#

Geostationary satellites

Kepler’s laws, explained#

Weightlessness#

Check yourself

Keep exploring

Circular orbits

Kepler’s laws, explained

Weightlessness

Circular orbits

Kepler’s laws, explained

Weightlessness