In the vast frictionless
environment of space,
gravitational attraction can cause two massive bodies to orbit each other.
If the relative velocity between two bodies is too low they will collide.
If the velocity is too high they will move away from each other.
In between these extremes exist a range of stable orbits that can repeat with almost no change.
Yet, over time orbits degrade.
The Moon's orbit gets 38 mm farther away every year because its kinetic energy changes into tidal
friction.
Circular Motion
Any object moving in a circular path has a net force pointed at the center of the circle.
The velocity of the object will always be perpendicular to the force.
When you need the distance traveled in one full trip around the circle, use the circumference, \(C = 2\pi r\).
$$F = \frac{mv^{2}}{r} \quad \quad a = \frac{v^{2}}{r} $$
\(F\) = centripetal force [N]
vector \(a\) = centripetal acceleration [m/s²]
vector \(v\) = tangential velocity [m/s]
vector \(r\) = radius of the circular path [m]
vector \(m\) = mass [kg]
The term centripetal means pointing at the center.
In all circular motion force and acceleration always points at the center of the circle.
This is often confused with centrifugal, which means away from the center.
Tangential means touching at only one point.
Since the velocity is perpendicular to the centripetal force, it doesn't enter or exit the circle.
Nearly circular motion can occur in many situations:
swinging a rope attached to a mass
a rotating wheel or sphere, (like a fidget spinner)
many orbits are nearly circular, but technically elliptical
some roads have circular curves
merry go round
charged particles in a uniform magnetic field
Example: A 20 kg box is placed 5 m from the center of a rotating carousel.
If the box has a 3 m/s tangential velocity, what magnitude and direction friction force will it have to
exert to stay on the ride?
solution
$$F = \frac{mv^{2}}{r}$$
$$F = \frac{(20)(3)^{2}}{5}$$
$$F = 36\,\mathrm{N}$$
$$\text{towards the center of the carousel}$$
Question: The ball above maintains a constant speed while moving in a circle. Is it
accelerating?
answer
Yes, the ball is accelerating towards the center of the circle.
It's possible to produce an acceleration
similar to gravity without a gravitational field. It's done by rotating a room to produce a centripetal acceleration.
Example: You are designing a rotating spaceship with artificial gravity. In your plans, the
ship is shaped like a wheel with a diameter of 50.0 m.
How fast does the outer edge have to move to produce a centripetal acceleration equal to the gravity felt on
Earth's surface?
solution
$$d = 2r \quad \quad r = 25\,\mathrm{m}$$ $$a = \frac{v^{2}}{r}$$ $$\sqrt{ar} = v$$ $$\sqrt{(9.8)(25)} =
v$$
$$15.7 \,\mathrm{\tfrac{m}{s}} =v$$
Example: Use the circumference of a circle and the tangential velocity of the ship to
calculate how long one rotation will take.
hint
$$C = 2 \pi r$$
$$v=\frac{\Delta x}{\Delta t}$$
$$v=\frac{2 \pi r}{\Delta t}$$
solution
Your 50 meter spaceship design was built.
It produced the illusion of 9.8 m/s² of gravity, but people are complaining they feel strange when
standing up.
Example: Use the circumference of a circle at a person's head and the time for one rotation
to calculate the velocity at the top of a 2 meter tall person's head.
strategy
A person on the rotating spaceship will have their feet on the outer edge, and their heads pointed
towards the center.
This means the head will have a shorter radius and a slower speed.
We also know that the total time for one rotation must be the same at any distance from the center.
We calculated the time for one rotation in the previous example.
solution
We already calculated that one rotation takes 10 seconds.
$$d = 2r \quad d = 50 \, \mathrm{m}$$
$$r = 25-2 = 23\, \mathrm{m}$$
$$v=\frac{\Delta x}{\Delta t} \quad C = 2 \pi r$$
$$v=\frac{2 \pi r}{\Delta t}$$
$$v=\frac{2 \pi (23)}{(10)}$$
$$v=14.45\, \mathrm{\tfrac{m}{s}}$$
Example: What acceleration would a 2 meter tall person feel at their head?
Compare that to the acceleration felt at their feet.
solution
$$a = \frac{v^{2}}{r}$$ $$a = \frac{14.45^{2}}{23}$$
$$a = 9.07\,\mathrm{\tfrac{m}{s^2}}$$
$$\frac{9.07}{9.8} = 0.92$$
A person's head only feels 92% of the gravity at their feet.
Question: What changes could be made to the spaceship design to reduce the difference in
acceleration between a person's head and feet?
answer
The ship's radius could be increased, but that would increase the costs.
The ship could produce a lower acceleration, maybe half of Earth's gravity.
Circular Orbits
Each satellite below starts with a different velocity. Simulation speed =
The orbits of bodies in space are elliptical. They are never perfectly circular, but many
orbits come close enough to circular motion to make a rough approximation useful.
derivation of orbital velocity
If we assume that the central mass is much larger and doesn't move, we can combine the equations for
circular motion with the universal gravitation equations.
\(v\) = orbital tangential velocity [m/s]
\(G\) = 6.67408 × 10-11 = universal gravitation constant [N m²/kg²]
\(r\) = radius of the circular orbit [m]
\(M\) = mass of the central body being orbited [kg]
The orbiting body needs to be much smaller than the central body!
Only true for circular orbits!
Situations where there is circular motion and a much larger central body include:
planets orbiting the Sun, moons orbiting large planets, and satellites.
Planet
mass (kg)
radius (km)
Sun
2.00 × 1030
695 700
Mercury
3.301 × 1023
2440
Venus
4.867 × 1024
6052
Earth
5.972 × 1024
6371
Moon
7.346 × 1022
1737
Mars
6.417 × 1023
3390
Jupiter
1.899 × 1027
70 000
Saturn
5.685 × 1026
58 232
Uranus
8.68 × 1025
25 362
Neptune
1.024 × 1026
24 622
Example: If a satellite is 800.0 km above the surface of the moon how fast must it move to
travel in a circular orbit?
solution
$$r = 800\,000\,\mathrm{m} + 1\,737\,000\,\mathrm{m} = 2\,537\,000\,\mathrm{m}$$
$$v^{2} = \frac{GM}{r}$$
$$v^{2} = \frac{(6.67 \times 10^{-11})(7.346 \times 10^{22})}{2\,537\,000}$$
$$\sqrt{v^{2}} = \sqrt{19.31 \times 10^5}$$
$$v = 1389.6 \,\mathrm{\tfrac{m}{s}}$$
Question: A 100 kg satellite and a 200 kg satellite are both in a circular orbit
around the Earth at an orbital radius of 10 000 000 m. Compare the velocities of the two satellites. Which is faster?
answer
They both have the same velocity. The mass of the satellite doesn't matter, only the mass of the Earth.
Radius Vs. Orbital Velocity
(for satellites in circular motion around Earth)
Example: A geosynchronous orbit
can stay above the same point on the Earth.
To be able to do this, the orbit must equal one Earth day, which requires a velocity of 3070 m/s.
Calculate how far away a geosynchronous satellite needs to be from the center of the Earth.
solution
$$v^{2} = \frac{GM}{r}$$
$$r = \frac{GM}{v^{2}}$$
$$r = \frac{(6.67 \times 10^{-11})(5.972 \times 10^{24})}{(3070)^{2}}$$
$$r = 4.22 \times 10^{7} \,\mathrm{m} = 42\,200 \,\mathrm{km}$$
$$\text{altitude = orbital radius - Earth's surface radius}$$
$$\text{altitude} = 42\,200 \,\mathrm{km} - 6371 \,\mathrm{km}$$
$$\text{altitude} = 35\,829 \,\mathrm{km}$$
You can get a feeling for orbits with this PhET gravitational orbit simulation.
Click to Run
What happens to the path of a satellite when you turn off gravity?
How would you describe the direction of the velocity compared to the force?
How does the orbital path change as you increase the mass of either object?
In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!
In orbit problems, gravity is the inward force, and the radius is measured from the center of the planet or moon, not from the surface.
Example: A 40 kg rider is on a circular ride with radius 8 m. The ride has bright lights around the edge, but the circular motion depends on radius and speed. If the rider moves at 6 m/s, what centripetal force is required, and which direction does it point?
solution
$$F = \frac{mv^2}{r}$$
$$F = \frac{(40)(6)^2}{8}$$
$$F = 180\,\mathrm{N}$$
The force points toward the center of the circle.
Example: A car moves through a circular highway curve with radius 75 m at a speed of 15 m/s. The road sign gives an advisory speed, but use the given speed here. What is the car's centripetal acceleration?
solution
$$a = \frac{v^2}{r}$$
$$a = \frac{(15)^2}{75}$$
$$a = 3.0\,\mathrm{m/s^2}$$
The acceleration points toward the center of the curve, even if the car's speed is constant.
Example: A rotating space station design has radius 40 m, about the length of a large classroom hallway. How fast does the outer edge need to move to create an artificial gravity acceleration of 9.8 m/s²?
solution
$$a = \frac{v^2}{r}$$
$$v^2 = ar$$
$$v = \sqrt{ar}$$
$$v = \sqrt{(9.8)(40)}$$
$$v = 19.8\,\mathrm{m/s}$$
A larger station can produce the same acceleration with a less extreme rotation rate.
Example: The outer edge of the 40 m radius station moves at 19.8 m/s. How long does one rotation take?
solution
One rotation covers one circumference.
$$C = 2\pi r$$
$$C = 2\pi(40)$$
$$C = 251\,\mathrm{m}$$
$$v = \frac{\Delta x}{\Delta t}$$
$$\Delta t = \frac{\Delta x}{v}$$
$$\Delta t = \frac{251}{19.8}$$
$$\Delta t = 12.7\,\mathrm{s}$$
Example: A 1.8 m tall person stands with their feet on the outer edge of the same 40 m radius station. Their head is 38.2 m from the center. If one rotation takes 12.7 s, what acceleration does the person's head feel?
solution
The head has the same rotation time, but a smaller circular path.
The head feels slightly less artificial gravity than the feet.
Example: A 65 kg person in the rotating station has a head-level circular speed of 18.9 m/s at a radius of 38.2 m. What inward force would be needed if their whole mass were moving at that radius?
solution
$$F = \frac{mv^2}{r}$$
$$F = \frac{(65)(18.9)^2}{38.2}$$
$$F = 608\,\mathrm{N}$$
This is smaller than the force needed for 9.8 m/s² because the acceleration is smaller.
Example: A small satellite is 100 km above the Moon's surface. Use the Moon's mass as 7.346 × 1022 kg and radius as 1.737 × 106 m. What circular orbital speed is needed?
solution
The orbital radius is measured from the Moon's center.
Example: A 500 kg satellite orbits 300 km above a planet's surface. What is its orbital speed?
answer
This cannot be solved from the information given. Orbital speed needs the planet's mass and the orbital radius from the planet's center, so the planet's radius is also missing.
Example: A satellite is 400 km above Earth's surface. Use Earth's mass as 5.972 × 1024 kg and radius as 6.371 × 106 m. What circular orbital speed is needed?
solution
$$r = 6.371 \times 10^6 + 400\,000$$
$$r = 6.771 \times 10^6\,\mathrm{m}$$
$$v = \sqrt{\frac{GM}{r}}$$
$$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.972 \times 10^{24})}{6.771 \times 10^6}}$$
$$v = 7.67 \times 10^3\,\mathrm{m/s}$$
This is about 7.67 km/s.
Example: A very low circular orbit around Mars has radius 3.390 × 106 m from Mars's center. Use Mars's mass as 6.417 × 1023 kg. What orbital speed is needed?
solution
$$v = \sqrt{\frac{GM}{r}}$$
$$v = \sqrt{\frac{(6.67 \times 10^{-11})(6.417 \times 10^{23})}{3.390 \times 10^6}}$$
$$v = 3.55 \times 10^3\,\mathrm{m/s}$$
Mars needs a lower orbital speed than Earth because Mars has less mass.
Example: A 100 kg satellite and a 400 kg satellite both orbit Earth at a radius of 1.00 × 107 m. First find the orbital speed, then find the centripetal force needed for each satellite. Use Earth's mass as 5.972 × 1024 kg.
solution
$$v = \sqrt{\frac{GM}{r}}$$
$$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.972 \times 10^{24})}{1.00 \times 10^7}}$$
$$v = 6.31 \times 10^3\,\mathrm{m/s}$$
$$F = \frac{mv^2}{r}$$
$$F_{100} = \frac{(100)(6.31 \times 10^3)^2}{1.00 \times 10^7}$$
$$F_{100} = 398\,\mathrm{N}$$
$$F_{400} = \frac{(400)(6.31 \times 10^3)^2}{1.00 \times 10^7}$$
$$F_{400} = 1590\,\mathrm{N}$$
The satellites need the same speed, but the more massive satellite needs more force.
Example: A satellite around Earth has a circular orbital speed of 6000 m/s. Use Earth's mass as 5.972 × 1024 kg. What is the orbital radius, and what is the altitude above Earth's surface?
solution
$$v^2 = \frac{GM}{r}$$
$$r = \frac{GM}{v^2}$$
$$r = \frac{(6.67 \times 10^{-11})(5.972 \times 10^{24})}{(6000)^2}$$
$$r = 1.11 \times 10^7\,\mathrm{m}$$
$$altitude = r - R_E$$
$$altitude = 1.11 \times 10^7 - 6.371 \times 10^6$$
$$altitude = 4.69 \times 10^6\,\mathrm{m}$$
The altitude is about 4690 km.
Example: A GPS satellite has an orbital speed of about 3870 m/s. Use Earth's mass as 5.972 × 1024 kg and Earth's radius as 6.371 × 106 m. Estimate its orbital radius and altitude.
solution
$$r = \frac{GM}{v^2}$$
$$r = \frac{(6.67 \times 10^{-11})(5.972 \times 10^{24})}{(3870)^2}$$
$$r = 2.66 \times 10^7\,\mathrm{m}$$
$$altitude = r - R_E$$
$$altitude = 2.66 \times 10^7 - 6.371 \times 10^6$$
$$altitude = 2.02 \times 10^7\,\mathrm{m}$$
This is about 20 200 km above Earth's surface.
Example: Use the 400 km altitude Earth satellite from earlier. Its orbital radius is 6.771 × 106 m and its speed is 7670 m/s. How long does one orbit take?
solution
$$C = 2\pi r$$
$$C = 2\pi(6.771 \times 10^6)$$
$$C = 4.25 \times 10^7\,\mathrm{m}$$
$$v = \frac{\Delta x}{\Delta t}$$
$$\Delta t = \frac{\Delta x}{v}$$
$$\Delta t = \frac{4.25 \times 10^7}{7670}$$
$$\Delta t = 5.55 \times 10^3\,\mathrm{s}$$
That is about 92 minutes.
Example: Around the same planet, satellite B orbits at 4 times the orbital radius of satellite A. How does satellite B's circular orbital speed compare to satellite A's?
solution
$$v = \sqrt{\frac{GM}{r}}$$
The central mass is the same, so only the radius changes.
Satellite B moves at half the speed of satellite A.
Example: A satellite around Earth has orbital radius 4.22 × 107 m and speed 3070 m/s. If another circular orbit around Earth has one-fourth that radius, what speed would it need?
solution
Both satellites orbit Earth, so the value of GM is the same for both orbits.
$$v = \sqrt{\frac{GM}{r}}$$
If the new radius is one-fourth as large, the fraction inside the square root becomes 4 times as large.
Example: A 900 kg satellite is in a 400 km altitude circular orbit around Earth. Its orbital radius is 6.771 × 106 m and its speed is 7670 m/s. What centripetal force is required?
solution
$$F = \frac{mv^2}{r}$$
$$F = \frac{(900)(7670)^2}{6.771 \times 10^6}$$
$$F = 7.82 \times 10^3\,\mathrm{N}$$
In orbit, this inward force is supplied by gravity.
Example: A satellite is at an orbital radius of 1.00 × 107 m from Earth's center and is moving at 5500 m/s. Use Earth's mass as 5.972 × 1024 kg. Is that fast enough for a circular orbit at that radius?
solution
$$v = \sqrt{\frac{GM}{r}}$$
$$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.972 \times 10^{24})}{1.00 \times 10^7}}$$
$$v = 6.31 \times 10^3\,\mathrm{m/s}$$
The satellite needs about 6310 m/s for a circular orbit. Since 5500 m/s is too slow, it will not maintain that circular orbit.
Example: A satellite orbits 800 km above the Moon's surface while mapping crater shadows. Its orbital radius is 2.537 × 106 m and its circular speed is 1390 m/s. Estimate the time for one orbit.
solution
$$C = 2\pi r$$
$$C = 2\pi(2.537 \times 10^6)$$
$$C = 1.59 \times 10^7\,\mathrm{m}$$
$$\Delta t = \frac{\Delta x}{v}$$
$$\Delta t = \frac{1.59 \times 10^7}{1390}$$
$$\Delta t = 1.15 \times 10^4\,\mathrm{s}$$
That is about 3.2 hours.
Reading (10 minutes): Read Explainer: All about orbits from Science News Explores. Then answer these questions.
Why does a planet move fastest when it is closest to the Sun?
answer
Gravity is stronger when the planet is closer to the Sun. The stronger pull changes the planet's motion so that it moves faster along that part of its orbit.
How can a spacecraft be continuously falling toward Earth without crashing into it?
answer
It is moving sideways fast enough that, as it falls, Earth's surface curves away beneath it. Gravity continually bends its path into an orbit.
Why must a spacecraft change its path around the Sun when traveling from Earth toward Mars?
answer
It begins with Earth's orbital motion around the Sun. Its engines must change that orbit into an ellipse that reaches Mars at the right time.
