Motion of Orbiting Satellites
Introduced| English | Meaning |
|---|---|
| orbital period/ˈɔːbɪtl ˈpɪərɪəd/ | orbital period |
The endless fall
- The Moon is always falling toward Earth, yet it never lands.
- A space station circles the planet once every ninety minutes.
- Gravity pulls it inward, but its sideways speed keeps missing the ground.
- One pull, one speed, and a perfect endless curve.
Gravity is the centripetal pull
- For a circular orbit, gravity supplies exactly the centripetal force:
- Solving for the speed: $v = \sqrt{\dfrac{GM}{r}}$.
- Lower orbits are faster; higher ones are slower.
Gravity provides the centripetal force that keeps a satellite in orbit
A satellite moves to a lower circular orbit. Its orbital speed...
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$v = \sqrt{GM/r}$, so a smaller $r$ gives a larger $v$.
What provides the centripetal force that holds a satellite in a circular orbit?
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$GMm/r^2 = mv^2/r$ -- gravity is the centripetal force.
Compute the orbital speed where $GM/r = 4.0\times10^7\ \text{m}^2/\text{s}^2$ (in m/s).
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$v = \sqrt{4.0\times10^7} \approx 6325\ \text{m/s}$.
Period grows with radius
- Cancelling and rearranging gives the orbital period 轨道周期:
- The period squared grows with the radius cubed — Kepler's third law.
- A bigger orbit takes much longer to go around.
A satellite in a larger orbit has a longer period.
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$T^2 \propto r^3$, so a bigger $r$ means a bigger $T$.
Faster when closer
- In an elliptical orbit, gravity points along $r$, so it makes no torque.
- With no torque, the planet's angular momentum is conserved.
- It speeds up near the star and sweeps equal areas in equal times.
Low orbit or high orbit?
Gravity provides the centripetal force for a satellite. Sort each fact by orbit height.
In an elliptical orbit, a planet moves fastest...
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Angular momentum is conserved, so a smaller $r$ means a larger speed -- fastest when closest.
A satellite orbits where $GM/r = 5.0\times10^7\ \text{m}^2/\text{s}^2$.
- Orbital speed: $v = \sqrt{GM/r} = \sqrt{5.0\times10^7} \approx 7100\ \text{m/s}$.
- That is about $7.1\ \text{km/s}$ — typical for a low orbit.
Doubling the satellite's own mass changes its orbital speed at the same radius.
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The mass $m$ cancels in $GMm/r^2 = mv^2/r$, so $v$ is independent of $m$.
Measure $r$ from the centre of the planet, not the altitude above the surface — add the planet's radius. The speed formula assumes a circular orbit; an ellipse has a changing speed (fastest at the closest point). And the satellite's own mass $m$ cancels out, so orbital speed and period do not depend on it.
Gravity is the centripetal force of a circular orbit, giving $v = \sqrt{GM/r}$ (lower means faster) and the orbital period $T^2 = \tfrac{4\pi^2}{GM}r^3$ (Kepler's third law). Because gravity makes no torque, a planet's angular momentum is conserved, so it moves fastest at its closest approach. Measure $r$ from the centre; the orbiting mass cancels.
Write from an orbital speed of 3000 m/s
- GM over r equals 9.0 million m squared per s squared.
- Find the circular orbital speed.
For a circular orbit, GM divided by r equals 9.0 times 10 to the 6, in m squared per s squared. What is the orbital speed, in m/s?
The square root of 9 million is 3000.
What is the orbital speed?
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Speed is the square root of GM over r. Other options — 6325 m/s (the lesson speed); 9.0 million m/s (the value under the square root).