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.

$$F = \frac{mv^{2}}{r} \quad \quad a = \frac{v^{2}}{r} $$

$F$ = 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

$$C = 2 \pi r$$ $$C = 2 \pi (25)$$ $$C = 157 \, \mathrm{m}$$

$$v=\frac{\Delta x}{\Delta t}$$ $$\Delta t=\frac{\Delta x}{v}$$ $$\Delta t=\frac{157}{15.7}$$ $$\Delta t=10\, \mathrm{s}$$

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.

$$d = 2r \quad d = 50 \, \mathrm{m}$$ $$r = 25-2 = 23\, \mathrm{m}$$

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.

$$F = \frac{M_{1}v^{2}}{r} \quad F = \frac{GM_{1}M_{2}}{r^{2}} $$ $$\frac{M_{1}v^{2}}{r} = \frac{GM_{1}M_{2}}{r^{2}} $$ $$\frac{ {\color{blue}M_{1} }v^{2}}{\color{blue}r} = \frac{G{\color{blue}M_1} M_{2}}{{\color{blue}r}^2}$$ $$v^{2} = \frac{GM_{2}}{r} $$

$$v^{2} = \frac{GM}{r} $$

$v$ = 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?