Practice

practice problem 1

solution

1. Start with the basic principle behind all circular orbits. Set the centripetal force equal to the gravitational force.

   Fg = Fc

   Replace the speed with the circumference divided by the period.

   Gm1m2	=	mv2	=	m		⎛ ⎝	2πr	⎞2 ⎠	=	4π2mr
   r2		r		r			T			T2

   Solve for radius to arrive at a general formula

   r =	⎛ ⎝	GmT2	⎞⅓ ⎠
   		4π2

   (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 entitled Heliocentrism.)
   1. Now substitute in the appropriate values. The period of the earth's rotation is approximately equal to the mean solar day (24 x 3600 s = 86,400 s), but for best results use the sidereal day (86,164 s).

      r =	⎛ ⎝	(6.67 × 10−11 Nm2/kg2)(5.974 × 1024 kg)(86,164 s)2	⎞⅓ ⎠
      		4π2
      r =	4.216 × 107 m ≈ 42,000 km

   2. Convert to earth radii

      r =	4.216 × 107 m	= 6.610 ≈ 7 earth radii
      	6,378,140 m

   3. Convert to earth-moon distances

      r =	4.216 × 107 m	= 0.1097 ≈ 1/9 distance from earth to moon
      	384,400,000 m

2. Answer the remaining parts.
3. Answer the remaining parts.

practice problem 2

solution

1. 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	⇒	m2v2	=	Gm1m2	⇒	v = √	Gm
   		r		r2			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) can drag 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.

2. 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, ut 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	=	rv2/g	=	3v2
   	V		4πr3/3		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 workspow 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.

   			r				r
   m =	⌠ ⌡	ρ dV =	⌠ ⌡	v2	(4πr2 dr) =	v2	⌠ ⌡	dr =	rv2
   				4πGr2		g			g
   			0				0

   And all is well again.

3. Numbers in. Answer out. Here we go …

   m =	rv2
   	g
   m =	(5 × 107 light years)(3 × 108 m/s)(365.25 × 24 × 3600 s)(2.2 × 105 m/s)2
   	(6.67 × 10−11 Nm2/kg2)(2 × 1030 kg/solar mass)
   m =	170 billion solar masses

   Although a bit high, this is in the ballpark for the number of stars in the Milky Way.

4. If the orbital speed remains constant for bodies near but outside the milky way, then the mass and density distributions we derived earlier …

   m =	rv2	&	ρ =	v2
   	g			4πGr2

   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 to them — on one side there are stars and on the other side an empty void populated only occasionally by a small cluster of stars or a cold cloud of radio waves emitting 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 haloes 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 3

solution

• Answer it

practice problem 4

solution

The first three Lagrange points lie on the line connecting the earth and sun.

For this problem let …

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.

Solving this pair of equations is difficult for two reasons.

1. 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.
2. 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.

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 …

Yields these solutions …

or in terms of "natural" units …

where an AU (astronomical unit) is the distance from the sun to the earth (1.4959787 × 10¹¹ m) and r_e-m is the distance from the earth to the moon (3.844 × 10⁸ m).