Orbital Mechanics I
Practice
practice problem 1
There is a special class of satellites that orbit the Earth above the equator with a period of one day.
 How will such a satellite appear to move when viewed from the surface of the Earth?
 What type of satellites use this orbit and why is it important for them to be located in this orbit? (Keep in mind that this is a relatively high orbit. Satellites not occupying this band are normally kept in much lower orbits.)
 Determine the orbital radius at which the period of a satellite's orbit will equal one day. State your answer in…
 kilometers
 multiples of the Earth's radius
 fractions of the moon's orbital radius
solution
Since they match the rotation rate of the Earth (they are geosynchronous), these satellites appear to remain fixed in the sky when viewed from the Earth (they are geostationary).
There are two major types of satellite that can be found in the geosynchronous/geostationary Earth orbit (GEO).
Telecommunications satellites: Since they maintain a fixed position in the sky, a fixed antenna can be used relay messages between the ground and a GEO satellite. Point the antenna in the right direction once and you're good to go for the lifetime of the satellite or the antenna (whichever comes first). Non GEO satellites move across the sky and must be tracked, which means building a steering mechanism for the antenna and writing computer software to drive it. GEO satellites also provide 24 hour coverage to large sections of the planet, which makes them great for broadcasting to an informationhungry world that never sleeps. Finally, a set of at least three geostationary satellites spaced evenly around the equator could be used as a relay. Just equip each satellite with one antenna pointing "forward" to the satellite ahead, one pointing "backward" to the satellite behind, and one pointing downward to the Earth. This makes it possible to send wireless messages (audio, video, text, etc.) from nearly anywhere on the Earth to nearly everywhere on the Earth. Because of their great height above the Earth, signals sent to and from GEO satellites are delayed by about a half a second each way or about a whole second round trip. This delay is known as latency and is annoying during a phone call. (Latency in broadcast and other applications isn't such a big deal.) Annoying latency is getting to be a thing of the past now that most international calls are handled by fiber optic cables laying near the surface of the Earth and true satellite phones have been developed with omnidirectional antennas capable of accessing satellites in low Earth orbit (LEO).
Meteorological satellites: Satellites in geostationary orbits stay fixed over a point on the surface of the Earth and are high enough that they can monitor the weather over nearly an entire continent or ocean. Images taken from the same location in space are easier to use than images taken from a location that changes. Images taken from weather satellites in lower orbits must be processed to correct for changes in perspective and scale and then mosaicked or stitched together. As with telecommunication, weather coverage is a 24 hour a day need in the 21st century. (Weather never sleeps, too.) Height has its drawbacks as well as its benefits. Being so far up means a loss in resolution. The information you get from a GEO satellite image may not be finely detailed enough. This is why government weather services now operate weather satellites in low Earth orbit (LEO) as well as geostationary Earth orbit (GEO).
Start with the basic principle behind all circular orbits. The gravitational force is the centripetal force.
F_{g} = F_{c} Gm_{1}m_{2} = mv^{2} r^{2} r Replace speed with circumference divided by the period.
Gm_{1}m_{2} = m ⎛
⎜
⎝2πr ⎞^{2}
⎟
⎠r^{2} r T Gm = 4π^{2}r r^{2} T^{2} Solve for radius to arrive at a general equation
r = ⎛
⎜
⎝GmT^{2} ⎞^{⅓}
⎟
⎠4π^{2} This problem can also be solved using Kepler's third law of planetary motion: the square of the period of a satellite in a circular orbit is proportional to the cube of its radius. That solution is presented in an earlier section of this book.

Now substitute in the appropriate values. The period of the Earth's rotation is approximately equal to the mean solar day (24 × 3,600 s = 86,400 s), but for best results use the sidereal day (86,164 s).
r = ⎛
⎜
⎝(6.67 × 10^{−11} N m^{2}/kg^{2}) (5.97 × 10^{24} kg) (86,164 s)^{2} ⎞^{⅓}
⎟
⎠4π^{2} r = 4.215 × 10^{7} m r = 42 150 km Convert to Earth radii.
r = 4.215 × 10^{7} m 6,371,000 m r = 6.616 earth radii r ≈ 7 earth radii Convert to Earthmoon distances.
r = 4.215 × 10^{7} m 384,400,000 m r = 0.1097 Earthmoon distance r ≈ ⅑ earthmoon distance

practice problem 2
 from the Sun and Earth in meters
 from the Earth as multiples of the Moon's orbital radius
 from the Sun as multiples of the Earth's orbital radius
solution
The first three Lagrange points lie on the line connecting the Earth and Sun.
For this problem let…
m_{s} =  mass of Sun 
m_{e} =  mass of Earth 
r_{e} =  radius of Earth's orbit 
r =  displacement from Sun to satellite 
x =  displacement from Earth to satellite 
so that…
x = r_{e} − r
All of the Lagrange points move together with the Earth about the Sun as if they were fixed on a rotating disk. In this situation, where angular velocity is constant, centripetal acceleration is directly proportional to the distance from the center of rotation.





A satellite will travel on a circular orbit wherever the required centripetal acceleration can be provided by the net gravitational field.
g_{net} =  g_{earth}  +  g_{sun}  
g_{net} =  ⎛ ⎜ ⎝ 
−  Gm_{e}  î  ⎞ ⎟ ⎠ 
+  ⎛ ⎜ ⎝ 
−  Gm_{s}  r̂  ⎞ ⎟ ⎠ 
x^{2}  r^{2} 
Solving this pair of equations is difficult for two reasons.
 It's a vector problem, which means we must contend with direction. Since the first three Lagrange points lie on the line connecting the Earth and Sun, the vector aspects of this problem are not that serious. In a onedimensional problem like this one, directions are are indicated with plus and minus signs. The nature of this problem requires that we deal with the signs in a piecewise fashion.
 It's also fifth order, which means an exact analytical solution is impossible. The way around this is to graph both equations on a calculator and let it find the points of intersection.
The next step is to set up a coordinate system. For no apparent reason, I've chosen to place the origin at the Sun and use r as the independent variable. (The Earth would have worked equally well as an origin and x as the independent variable.) This places L1, the Earth, and L2 on the positive side of the axis and leaves L3 by itself on the negative side. As is usually done, all vectors pointing to the right will be positive and those to the left will be negative.
a_{c}  =  g_{net}  
−  4π^{2}  =  ⎛ ⎜ ⎝ 
±  Gm_{e}  ⎞ ⎟ ⎠ 
+  ⎛ ⎜ ⎝ 
±  Gm_{s}  ⎞ ⎟ ⎠ 
T^{2}  x^{2}  r^{2} 
L3 behind the Sun
−  4π^{2}  r = +  m_{e}  +  m_{s}  
GT^{2}  (r_{e} − r)^{2}  r^{2} 
L1 between the Sun and Earth
−  4π^{2}  r = +  m_{e}  −  m_{s}  
GT^{2}  (r_{e} − r)^{2}  r^{2} 
L2 behind the Earth
−  4π^{2}  r = −  m_{e}  −  m_{s}  
GT^{2}  (r_{e} − r)^{2}  r^{2} 
Which when graphed looks like this.
Well, not exactly. Since the Sun is so much more massive than the Earth, the L1 and L2 points would lie so close together as to be indistinguishable at this scale. Using the following values…
symbol  value  name 

G  6.674 30 × 10^{−11} N m^{2}/kg^{2}  universal gravitational constant 
T  365.25 × 24 × 3600 s  period of the Earth's rotation about the Sun 
r_{e}  149,597,870,700 m  radius of the Earth's orbit 
m_{e}  5.9742 × 10^{24} kg  mass of Earth 
m_{s}  1.9891 × 10^{30} kg  mass of Sun 
Yields these solutions…
Lagrange point  distance from Sun, r  distance from Earth, x 

L1  1.481 × 10^{11} m  1.49 × 10^{9} m 
L2  1.511 × 10^{11} m  1.50 × 10^{9} m 
L3  1.496 × 10^{11} m  2.98 × 10^{11} m 
or in terms of "natural" units…
Lagrange point  distance from Sun, r  distance from Earth, x 

L1  0.99 au  3.88 r_{em} 
L2  1.01 au  3.90 r_{em} 
L3  1.00001 au  775 r_{em} 
where au (astronomical unit) is the EarthSun distance (1.495978707 × 10^{11} m) and r_{em} is the EarthMoon distance (3.844 × 10^{8} m).
practice problem 3
planet 
period (days) 
semimajor axis (au) 

b  01.51087081  0.01111 
c  02.42182330  0.01521 
d  04.04961000  0.02144 
e  06.09961500  0.02817 
f  09.20669000  0.03710 
g  12.35294000  0.04510 
solution
Start with some physics. Assume that every orbit is circular, so the gravitational force is the centripetal force.
F_{g}  =  F_{c}  
Gm_{1}m_{2}  =  mv^{2}  
r^{2}  r 
Replace speed (v = ∆s/∆t) with circumference over period (2πr/T).
Gm_{1}m_{2}  =  m  ⎛ ⎜ ⎝ 
2πr  ⎞^{2} ⎟ ⎠ 

r^{2}  r  T 
Eliminate the mass of the orbiting body (the planet) and leave behind the mass of the central body (the star).
Gm  =  4π^{2}r 
r^{2}  T^{2} 
Solve for the mass of the star.
m =  4π^{2}r^{3} 
GT^{2} 
There are two statistical methods that can be used to find an answer. The first is to apply the equation derived above over and over and then take an average. This method is for people that like spreadsheets. Make sure you use the right units. Convert days to seconds (multiply by 24 × 60 × 60). Convert the astronomical units to meters (multiply by 1.495978707 × 10^{11} m). Recall that G = 6.67 × 10^{−11} N m^{2}/kg^{2}.
planet 
period (days) 
semimajor axis (au) 
period (s) 
semimajor axis (m) 
stellar mass (kg) 

b  01.51087081  0.01111  1.704 × 10^{10}  4.591 × 10^{27}  1.594 × 10^{29} 
c  02.42182330  0.01521  4.378 × 10^{10}  1.178 × 10^{28}  1.592 × 10^{29} 
d  04.04961000  0.02144  1.224 × 10^{11}  3.299 × 10^{28}  1.595 × 10^{29} 
e  06.09961500  0.02817  2.777 × 10^{11}  7.484 × 10^{28}  1.594 × 10^{29} 
f  09.20669000  0.03710  6.327 × 10^{11}  1.709 × 10^{29}  1.599 × 10^{29} 
g  12.35294000  0.04510  1.139 × 10^{12}  3.071 × 10^{29}  1.595 × 10^{29} 
average →  1.595… × 10^{29} 
The second method is for people that like graphs. Take that last equation and pull out the two variable quantities — radius and period or, more precisely, r^{3} and T^{2}. Leave them on one side of the equation and put everything else on the other side.
m =  4π^{2}  r^{3}  
G  T^{2} 
Think of r^{3} and T^{2} as ∆y and ∆x on a pair of standard axes. The second fraction (r^{3}/T^{2}) in the equation above is then the slope (∆y/∆x) of a best fit line that runs through the origin. Use your favorite graphical analysis software and make a graph like the one below.
Then use the slope to compute another reasonable answer.
m =  4π^{2}(2.697… × 10^{17} m^{3}/s^{2})  
6.67 × 10^{−11} N m^{2}/kg^{2}  
m = 1.596… × 10^{29} kg  
Stellar masses aren't often reported in kilograms. Multiples of the Sun's mass are more common (1 m_{☉} = 1.9891 × 10^{30} kg). From the first statistical method…
m =  1.59538649 × 10^{29} kg  
1.9891 × 10^{30} kg  
m = 0.080206 m_{☉}  
And from the second statistical method…
m =  1.596454394 × 10^{29} kg  
1.9891 × 10^{30} kg  
m = 0.080260 m_{☉}  
The authors of the paper that inspired this problem computed a value of 0.0802±0.0073 m_{☉}, which is totally in line with both of the numbers computed above.
practice problem 4
solution
Answer it.