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…
- multiples of the Earth's radius
- fractions of the moon's orbital radius
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 your 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 information-hungry 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 Twenty-first 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.
Fg = Fc Gm1m2 = mv2 r2 r
Replace speed with circumference divided by the period.
Gm1m2 = m ⎛
r2 r T Gm = 4π2r r2 T2
Solve for radius to arrive at a general equation
r = ⎛
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 × 3600 s = 86,400 s), but for best results use the sidereal day (86,164 s).
r = ⎛
(6.67 × 10−11 Nm2/kg2) (5.97 × 1024 kg) (86,164 s)2 ⎞⅓
4π2 r = 4.215 × 107 m r = 42,150 km
Convert to earth radii.
r = 4.215 × 107 m 6,371,000 m r = 6.616 earth radii r ≈ 7 earth radii
Convert to earth-moon distances.
r = 4.215 × 107 m 384,400,000 m r = 0.1097 earth-moon distance r ≈ ⅑ earth-moon 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
The first three Lagrange points lie on the line connecting the Earth and sun.
For this problem let…
|ms =||mass of sun|
|me =||mass of earth|
|re =||radius of earth's orbit|
|r =||displacement from sun to satellite|
|x =||displacement from earth to satellite|
x = re − 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.
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 one-dimensional 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.
L3 behind the sun
|−||4π2||r = +||me||+||ms|
|GT2||(re − r)2||r2|
L1 between the sun and earth
|−||4π2||r = +||me||−||ms|
|GT2||(re − r)2||r2|
L2 behind the Earth
|−||4π2||r = −||me||−||ms|
|GT2||(re − r)2||r2|
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…
|G||6.67259 × 10−11 Nm2/kg2||universal gravitational constant|
|T||365.25 × 24 × 3600 s||period of the Earth's rotation about the sun|
|re||1.4959787 × 1011 m||radius of the Earth's orbit|
|me||5.9742 × 1024 kg||mass of earth|
|ms||1.9891 × 1030 kg||mass of sun|
Yields these solutions…
|lagrange point||distance from sun, r||distance from earth, x|
|L1||1.481 × 1011 m||1.49 × 109 m|
|L2||1.511 × 1011 m||1.50 × 109 m|
|L3||1.496 × 1011 m||2.98 × 1011 m|
or in terms of "natural" units…
|lagrange point||distance from sun, r||distance from earth, x|
|L1||0.99 au||3.88 rem|
|L2||1.01 au||3.90 rem|
|L3||1.00001 au||775 rem|
Where au (astronomical unit) is the Earth-sun distance (1.4959787 × 1011 m) and rem is the Earth-moon distance (3.844 × 108 m).
practice problem 3
- The orbital speed of the planets decreases with distance from the sun. Why does this happen? Derive a formula that shows the relationship.
- The orbital speed of the stars remains roughly constant with distance from the center of the Milky Way. (This is true for other galaxies as well.) What does this tell us about the distribution of mass in galaxies? Derive a formula that shows the relationship.
- Calculate the mass of the Milky Way given a typical orbital speed of 220 km/s and a radius of 50,000 light years. Give your answer in solar masses (m☉ = 2 × 1030 kg) and compare it to the approximate number of stars in the Milky Way (1011).
- Dwarf galaxies, star clusters, and gas clouds beyond the edge of the visible galaxy have nearly the same orbital speed as the stars within visible galaxy.There is evidence that rotational speeds remain roughly constant at 220 km/s out to distances of 300,000 light years or six times the radius of the Milky Way. What is so amazing about this observation and what does it imply?
The basic principle behind all circular orbits is that the the centripetal force needed to keep the planet in orbit is supplied by an inverse-square gravitational force. When these two equations are set equal and solved for speed we get…
Fc = Fg mv2 = GMm rp rp2 v = √ Gm r
which shows that speed drops off as one over the square root of the distance from whatever it is we're orbiting — the sun in this case. This makes sense since gravity gets weaker as distance increases. At large distances, the outer planets (jupiter, saturn, uranus, and neptune) plod along and still stay in orbit. Closer in where gravity is strong, the inner planets (mercury, venus, earth, and mars) need much larger speeds to avoid being gobbled up by the sun.
Starting with the end product of our previous solution we can see that speed could be made constant if mass was allowed to increase with distance. This is exactly what happens in a distributed arrangement of mass like a galaxy. Objects that are farther from the center are orbiting around more stuff. On such vast scales, kilograms won't do to describe the mass of things. Instead we will use the mass of the sun as our standard unit. Stars in the core are orbiting only a few million solar masses of material, our sun, which is some two thirds of the way to the edge of the galaxy, is orbiting tens of billions of solar masses, and stars at the edge of the Milky Way are basically going around all one hundred billion solar masses of material that make up our galaxy.
If the observed speed of stars in the Milky way is more or less uniform, then the mass contained within the orbit of any one star must be proportional to the radius of its orbit, but it's really density that we're after — or rather, a density function. Take the equation derived above and solve for mass.
v = √ Gm ⇒ m = rv2 r G
This shows us that the mass around which a star orbits is directly proportional to its distance from the center of the galaxy. The sun is roughly two thirds of the way to the edge of the Milky Way. Its orbit should therefore encircle two thirds of the mass of the entire disk of the galaxy. Interesting, but we're not finished. Substitute this expression for mass and the volume of a sphere into the density formula and simplify. This gives us the density function for a galaxy with an observed flat rotation curve.
ρ = m V ρ = rv2/g 4πr3/3 ρ = 3v2 4πGr2
In order for the orbital velocity to remain constant in a galaxy its mass must increase linearly with radius. In order for its mass to increase linearly its density must drop off as the inverse square of its radius. I hope that this makes sense. The density of the core of a galaxy, where stars are tightly packed together, should be greater than the density of the whole thing. If the core has half the radius of the disk, then it should have four times the density of the entire galaxy. It's an interesting "conspiracy" of the natural world that this is the way it works out.
The equation above is slightly wrong. It's not really a density function, it's an average density function. It doesn't give us the local density at some distance r from the center, it gives us the average density within a sphere of radius r. To fix this, we need to drop the 3 from the numerator.
ρ = v2 4πGr2
This last bit should only be read by those who understand calculus. Everyone else can jump ahead to the last part of this problem. Now when this function is integrated over a series of spherical shells with surface area 4πr2 and thickness dr from the center 0 out to a distance r we get back the expression we derived earlier for mass.
m = ⌠
ρ dV r m = ⌠
v2 (4πr2 dr) 4πGr2 0 r m = v2 ⌠
dr g 0 m = rv2 g
And all is well again.
Numbers in. Answer out. Here we go…
m = rv2 G m = (5 × 107 light year) (9.46 × 1015 m/light year) (2.2 × 105 m/s)2 (6.67 × 10−11 Nm2/kg2) m = 3.43 × 1044 kg 1.99 × 1030 kg/solar mass m = 170 billion solar masses
This is of the right order of magnitude for the mass of the Milky Way.
If the orbital speed remains constant for bodies near but outside the milky way, then the mass distribution…
m = rv2 G
…and density distribution…
ρ = v2 4πGr2
…we derived earlier must apply to the apparently empty regions beyond the edge of the galaxy. But when we look at galaxies like the Milky Way we always see a definite edge — on one side there are stars and on the other side an empty void populated only by the ocaisional small cluster of stars or cold cloud of gas. Beyond this distance one would expect an inverse square root drop in orbital speed as is seen with the planets. But this is not the case. Rotational speeds remain roughly constant six times farther than the edge of the Milky Way. Since mass is directly proportional to radius when speed is constant, this means that the total mass of the galaxy is at least six times greater than its visible mass, or equivalently, that five-sixths (roughly 85%) of all the mass in our galaxy is invisible.
Astronomers have decided to call this stuff dark matter, but I don't particularly like this term since people have a tendency to think "dark" means "black". Dark matter does not interact with light or any other form of electromagnetic radiation. You and I are giving off plenty of infrared. Many communications devices give off microwaves. Both forms of radiation are invisible to our eyes, but we have other means of detecting them. I can feel infrared on my skin as heat and detect microwaves with a cellular phone or a satellite dish. Dark matter will have nothing to do with any of these forms of radiation. Dark matter neither emits, nor absorbs, nor reflects, refracts, diffracts, or interacts electromagnetically in any way with radio waves, microwaves, infrared, visible light, ultraviolet, x-rays, or gamma rays. The only way dark matter can be detected is through its gravitational effects — and they are significant. So much so that the dark matter halos around galaxies will bend spacetime from its normally flat geometry. As we all know light travels in straight lines. But when light encounters the warped spacetime around a galaxy, straight lines have no choice but to bend. The result is a phenomena called gravitational lensing (a more complete discussion of which is best left to another part of this book). What's important to note here is that this phenomena can be used to measure the amount of matter in moderately distant galaxies and that results always show a significantly larger amount of dark matter than ordinary matter (as high as 10:1 in some cases).
Dark matter exists in other galaxies besides the Milky Way. Flat rotation curves have been plotted for other nearby spiral galaxies and gravitational lensing has been used to measure dark matter distributions of more distant galaxies. Computer simulations of colliding galaxies don't work (that is, they don't agree with observations of actual colliding galaxies) unless they include dark matter as a variable. They need dark matter to give realistic results. In summary, dark matter exists. It exists as much as electrons or radio waves exist despite the fact that they can't be seen. The only remaining question is, unfortunately, a really big one. What is it? Let me know when you find out.
More on the dark side of the universe in the next section: Gravitational Potential Energy II.
practice problem 4
Start with some physics. Assume that every orbit is circular, so the gravitational force is the centripetal force.
Replace speed (v = ∆s/∆t) with circumference over period (2πr/T).
Eliminate the mass of the orbiting body (the planet) and leave behind the mass of the central body (the star).
Solve for the mass of the star.
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 1.4959787 × 1011 m). Recall that G = 6.67 × 10−11 Nm2/kg2.
|b||01.51087081||0.01111||1.704 × 1010||4.591 × 1027||1.594 × 1029|
|c||02.42182330||0.01521||4.378 × 1010||1.178 × 1028||1.592 × 1029|
|d||04.04961000||0.02144||1.224 × 1011||3.299 × 1028||1.595 × 1029|
|e||06.09961500||0.02817||2.777 × 1011||7.484 × 1028||1.594 × 1029|
|f||09.20669000||0.03710||6.327 × 1011||1.709 × 1029||1.599 × 1029|
|g||12.35294000||0.04510||1.139 × 1012||3.071 × 1029||1.595 × 1029|
|average →||1.595… × 1029|
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, r3 and T2.
Think of r3 and T2 as ∆y and ∆x on a pair of standard axes. The second fraction (r3/T2) in the equation above is then the slope of a best fit line (∆y/∆x). Use your favorite graphical analysis software and make a graph like the one below.
Then use the slope to compute another reasonable answer.
|m =||4π2(2.697… × 1017 m3/s2)|
|6.67 × 10−11 Nm2/kg2|
|m = 1.596… × 1029 kg|
Stellar masses aren't often reported in kilograms. Multiples of the sun's mass are more common (1 m☉ = 1.9891 × 1030 kg). From the first statistical method…
|m =||1.59538649 × 1029 kg|
|1.9891 × 1030 kg|
|m = 0.080206 m☉|
And from the second statistical method…
|m =||1.596454394 × 1029 kg|
|1.9891 × 1030 kg|
|m = 0.080260 m☉|
The authors of this paper computed a value of 0.0802±0.0073 m☉, which is totally in line with both of the numbers computed above.