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]
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:
solution
$$F = \frac{mv^{2}}{r}$$ $$F = \frac{(20)(3)^{2}}{5}$$ $$F = 36\,\mathrm{N}$$ $$\text{towards the center of the carousel}$$solution
$$d = 2r \quad \quad r = 25\,\mathrm{m}$$ $$a = \frac{v^{2}}{r}$$ $$\sqrt{ar} = v$$ $$\sqrt{(9.8)(25)} = v$$ $$15.7 \,\mathrm{\tfrac{m}{s}} =v$$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. $$F = \frac{M_{1}v^{2}}{r} \quad F = \frac{GM_{1}M_{2}}{r^{2}} $$ $$\frac{M_{1}v^{2}}{r} = \frac{GM_{1}M_{2}}{r^{2}} $$ $$\frac{\color{blue}M_{\color{blue}1}v^{2}}{\color{blue}r} = \frac{G\color{blue}M_{\color{blue}1}M_{2}}{r^{\color{blue}2}}$$ $$v^{2} = \frac{GM_{2}}{r} $$
$$v^{2} = \frac{GM}{r} $$
\(v\) = 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]
The orbiting body needs to be much smaller than the central body!
Only true for circular orbits!
There are several restrictions on where we can use this equation. Most planets orbiting the sun, moons orbiting large planets, and satellites are all applicable.
This is a graph of radius vs. orbital velocity for a satellite orbiting Earth in circular motion. 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 |
solution
$$r = 800000\,\mathrm{m} + 1737000\,\mathrm{m} = 2537000\,\mathrm{m}$$ $$v^{2} = \frac{GM}{r}$$ $$v^{2} = \frac{(6.67 \times 10^{-11})(7.346 \times 10^{22})}{2,537,000}$$ $$\sqrt{v^{2}} = \sqrt{19.31 \times 10^5}$$ $$v = 1389.6 \,\mathrm{\tfrac{m}{s}}$$solution
$$v^{2} = \frac{GM}{r}$$ $$r = \frac{GM}{v^{2}}$$ $$r = \frac{(6.67 \times 10^{-11})(5.972 \times 10^{24})}{(3,070)^{2}}$$ $$r = 4.22 \times 10^{7} \,\mathrm{m} = 42,200 \,\mathrm{km}$$$$\text{altitude = orbital radius - Earth's surface radius}$$ $$\text{altitude} = 42,200 \,\mathrm{km} - 6,371 \,\mathrm{km}$$ $$\text{altitude} = 35,829 \,\mathrm{km}$$
solution
They will both have the same velocity. The mass of the satellite doesn't matter, only the mass of the Earth.