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:
solution
$$F = \frac{mv^{2}}{r}$$ $$F = \frac{(20)(3)^{2}}{5}$$ $$F = 36N$$pointed towards the center of the carousel
solution
$$d = 2r \quad \quad r = 25m$$ $$a = \frac{v^{2}}{r}$$ $$\sqrt{ar} = v$$ $$\sqrt{(9.8)(25)} = v$$ $$15.7 \small \frac{m}{s} = \normalsize v$$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 |
solution
$$r = 800,000 + 1,737,000 = 2,537,000m$$ $$v^{2} = \frac{GM}{r}$$ $$v^{2} = \frac{(6.67 \times 10^{-11})(7.346 \times 10^{22})}{2,537,000}$$ $$v^{2} = \frac{(6.67 \times 10^{-11})(7.346 \times 10^{22})}{2.537 \times 10^{6}}$$ $$\sqrt{v^{2}} = \sqrt{19.31 \times 10^5}$$ $$v = 1389.6 \small \frac{m}{s}$$solution
$$v^{2} = \frac{GM}{r}$$ $$r = \frac{GM}{v^{2}}$$ $$r = \frac{(6.67 \times 10^{-11})(5.972 \times 10^{24})}{(3,070)^{2}}$$ $$r = 4.22 \times 10^{7} m = 42,200 km$$altitude = orbital radius - Earth's surface radius
$$a = 42,200 km - 6,371 km$$ $$a = 35,829 km $$solution
They will both have the same velocity. The mass of the satellite doesn't matter, only the mass of the Earth.