KAHIBARO
Table of Contents
Circular orbit idea
Orbital velocity is the speed an object must have to move around a massive body in a circular orbit. A planet around a star, or a satellite around Earth, does not fall straight down because its sideways motion is just right for gravity to continually bend its path into a circle.
In a circular orbit, gravity provides the centripetal force. This is the key idea behind orbital velocity.
Deriving the orbital velocity formula
Consider a body of mass $m$ orbiting a much larger body of mass $M$ at a distance $r$ from its center. The gravitational force is
$$
F_g = \frac{G M m}{r^2}
$$
For circular motion, the required centripetal force is
$$
F_c = \frac{m v^2}{r}
$$
For a circular orbit, these must be equal:
$$
\frac{G M m}{r^2} = \frac{m v^2}{r}
$$
The mass $m$ of the orbiting object cancels, giving
$$
v^2 = \frac{G M}{r}
$$
So the orbital velocity is
$$
v = \sqrt{\frac{G M}{r}}
$$
For a circular orbit around a body of mass $M$ at orbital radius $r$,
$$
v = \sqrt{\frac{G M}{r}}
$$
This speed does not depend on the mass of the orbiting object.
Meaning of the formula
The formula shows two very important facts. If the central mass $M$ is larger, the orbital speed must be larger. If the orbital radius $r$ is larger, the orbital speed is smaller.
So objects in low orbit move faster than objects in high orbit. This is why satellites close to Earth travel faster than the Moon, which is much farther away.
Orbit near a planet
For an object orbiting a planet of radius $R$ at height $h$ above the surface, the distance from the planet's center is
$$
r = R + h
$$
So the orbital velocity becomes
$$
v = \sqrt{\frac{G M}{R + h}}
$$
For a very low orbit near the surface, where $h$ is small compared with $R$,
$$
v \approx \sqrt{\frac{G M}{R}}
$$
For Earth, this gives a speed of about
$$
v \approx 7.9 \times 10^3 \ \text{m/s}
$$
which is about $7.9 \ \text{km/s}$ for a low Earth orbit.
Relation to gravitational acceleration
Using the gravitational field strength at radius $r$,
$$
g = \frac{G M}{r^2}
$$
we can also write the circular orbit condition as
$$
\frac{v^2}{r} = g
$$
so
$$
v = \sqrt{gr}
$$
This form is useful only when $g$ means the gravitational acceleration at that orbital distance, not necessarily the value at the surface.
Do not automatically use $g = 9.8 \ \text{m/s}^2$ in orbital problems.
That value is only approximately correct near Earth's surface. In orbit, use
$$
g = \frac{G M}{r^2}
$$
with the correct distance $r$ from Earth's center.
Comparison of orbital speeds
The table below shows how orbital speed changes with distance from the central body.
| Orbital radius $r$ | Orbital velocity $v = \sqrt{GM/r}$ |
|---|---|
| Small $r$ | Large $v$ |
| Large $r$ | Small $v$ |
For the same central mass, doubling the orbital radius does not halve the speed. Instead,
$$
v \propto \frac{1}{\sqrt{r}}
$$
So if $r$ becomes four times larger, the speed becomes half as large.
Example with Earth
Let a satellite orbit Earth at radius $r = 7.0 \times 10^6 \ \text{m}$. Using
$$
G = 6.67 \times 10^{-11} \ \text{N m}^2/\text{kg}^2
$$
and
$$
M_{\text{Earth}} = 5.97 \times 10^{24} \ \text{kg}
$$
we find
$$
v = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.0 \times 10^6}}
$$
$$
v \approx 7.5 \times 10^3 \ \text{m/s}
$$
So the satellite must move at about $7.5 \ \text{km/s}$.
Physical picture
An orbit can be understood as continuous falling. The object is always being pulled inward by gravity, but its forward speed is such that it keeps missing the central body. The direction of velocity changes continuously, even if the speed stays constant in a circular orbit.
Special case of surface orbit
If there were no atmosphere, an object moving horizontally fast enough near a planet's surface could orbit just above the surface. This ideal speed is found from
$$
v = \sqrt{\frac{G M}{R}}
$$
For Earth this is about $7.9 \ \text{km/s}$. In reality, an orbit too close to the surface is impossible because air resistance would slow the object down.
Common mistakes
A very common mistake is using the height above the surface instead of the distance from the center. In the formula
$$
v = \sqrt{\frac{G M}{r}}
$$
the quantity $r$ must always be measured from the center of the planet or star.
Another common mistake is to think that a heavier satellite needs a larger orbital speed. It does not. The required circular orbital speed is the same for all masses at the same radius.
In orbital velocity problems, always check these points:
$$
r = \text{distance from the center of the central body}
$$
$$
v = \sqrt{\frac{G M}{r}}
$$
The orbiting mass cancels out, so orbital speed is independent of the satellite's mass.
Connection to orbital motion
Orbital velocity is specifically the speed for circular orbit at a given radius. If the speed is different from this value, the path may no longer be circular. It may become another kind of orbit, or the object may fall inward or move farther outward.
So orbital velocity gives the exact balance between gravitational attraction and the inward acceleration needed for circular motion.