# Orbits

In the frictionless environment of space two massive bodies can orbit each other because of their mutual gravitational attraction.

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 very little change.

Over enough time all orbits degrade as they lose energy. The Moon's orbit gets 38mm farther away every year because its kinetic energy changes into the tides.

Play around with some orbits in the simulation below to develop an intuition for orbital mechanics.

What happen to the path of a satellite when you turn gravity off?

How would you describe the directions of the velocity and the force?

How does the orbital path change as you increase the mass of either object?

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]

centripetal: pointing at the center (Not to be confused with centrifugal)
tangential: touching at only one point

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 20kg dog is riding on a carousel 5m away from the center. If the dog is moving at 3m/s, what magnitude and direction force will it have to exert to produce the circular motion needed to stay on the ride?
solution $$F = \frac{mv^{2}}{r}$$ $$F = \frac{(20)(3)^{2}}{5}$$ $$F = 36N$$

pointed towards the center of the carousel

Example: You want to reproduce the feeling of Earth's gravity in space. This can be done by rotating your space station to produce the feeling of acceleration. Your space station is shaped like a wheel with a diameter of 50.0m. How fast does the edge have to move to produce Earth's gravity?
solution $$d = 2r \quad \quad r = 25m$$ $$a = \frac{v^{2}}{r}$$ $$\sqrt{ar} = v$$ $$\sqrt{(9.8)(25)} = v$$ $$15.7 \small \frac{m}{s} = \normalsize v$$
Example: The second hand on a clock moves smoothly in a circle, but there is a brave 3 mg ant on the tip of the second hand. How much force must the ant apply to not fall off? The ant is 21 mm from the center of the clock. You can't directly measure the velocity, but you do know it takes 60 seconds for the second hand to complete a circle.
solution

The distance is the circumference of the circle. The time is 60s. Velocity is distance over time.

$$c = 2 \pi r$$ $$v = \frac{\Delta x}{\Delta y}$$ $$v = \frac{2 \pi (0.021)}{60}$$ $$v = 0.0022 \small \frac{m}{s}$$
$$F = \frac{mv^{2}}{r}$$ $$F = \frac{0.000003(0.0022)^{2}}{0.021}$$ $$F = 0.00000175N$$ $$F = 6.914 \times 10^{-10}N$$

Circular Orbits

Each satellite below starts with a different velocity. Simulation speed =

The orbits of objects 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. $$F = \frac{M_{1}v^{2}}{r} \quad F = \frac{GM_{1}M_{2}}{r^{2}}$$
set the force of gravity and the centripetal force equal $$\frac{M_{1}v^{2}}{r} = \frac{GM_{1}M_{2}}{r^{2}}$$
cancel terms that show up on both sides $$\frac{\color{blue}M_{\color{blue}1}v^{2}}{\color{blue}r} = \frac{G\color{blue}M_{\color{blue}1}M_{2}}{r^{\color{blue}2}}$$ $$v^{2} = \frac{GM_{2}}{r}$$

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

$$v$$ = 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]

Only true for circular orbits!
The orbiting body needs to be much smaller than the central body!

### radius vs. orbital velocity for a satellite of Earth

It seems counter-intuitive, but as you orbit farther from the central body your velocity goes down.

Take a look at the simulation below to see circular orbits at different distances.
Click and drag to move. Double click for full screen.

### Local Massive Objects Data Table

Planet mass (kg) radius (km) density (g/cm 3)
Sun 2.00×10 30 695,700 1.408
Mercury 3.301×10 23 2,440 5.427
Venus 4.867×10 24 6,052 5.243
Earth 5.972×10 24 6,371 5.515
Moon 7.346×10 22 1,737 3.344
Mars 6.417×10 23 3,390 3.933
Jupiter 1.899×10 27 70,000 1.326
Saturn 5.685×10 26 58,232 0.687
Uranus 8.68×10 25 25,362 1.270
Neptune 1.024×10 26 24,622 1.638
exoplanets data table
Example: If a satellite is 800.0 km from the surface of the moon how fast must it move to travel in a circular orbit?
solution $$r = 800,000 + 1,737,000 = 2,537,000m$$ $$v^{2} = \frac{GM}{r}$$ $$v^{2} = \frac{(6.67 \times 10^{-11})(7.346 \times 10^{22})}{2,537,000}$$ $$v^{2} = \frac{(6.67 \times 10^{-11})(7.346 \times 10^{22})}{2.537 \times 10^{6}}$$ $$\sqrt{v^{2}} = \sqrt{19.31 \times 10^5}$$ $$v = 1389.6 \small \frac{m}{s}$$
Example: A geosynchronous orbit is always above the same point on the earth so it takes one day to orbit. This orbit requires a velocity of 3,070 m/s. Calculate how far away a geosynchronous satellite needs to be from the center of the earth.
solution $$v^{2} = \frac{GM}{r}$$ $$r = \frac{GM}{v^{2}}$$ $$r = \frac{(6.67 \times 10^{-11})(5.972 \times 10^{24})}{(3,070)^{2}}$$ $$r = 4.22 \times 10^{7} m = 42,200 km$$

$$a = 42,200 km - 6,371 km$$ $$a = 35,829 km$$