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Opus in profectus

# Gravity of Extended Bodies

## Practice

### practice problem 1

Almost all stars will die as hot, dense lumps, but some of them decide to get fat and cold for awhile first. In 5 billion years our sun will expand into a red giant (red because it will be much cooler) swallowing up Mercury, Venus, Earth, and (possibly) Mars. A few hundred million years after that, it will begin shrinking until it becomes a white dwarf (white because it will be much, much hotter) about as small as the Earth but still retaining much of it mass. Pairs of stars in orbit around one another are common in our galaxy — more common than lone stars like our sun. The stars never have the same mass and one will always age faster than the other. The diagram above shows a binary system where one star is a red giant and the other is a white dwarf. If the two stars are close enough, the white dwarf may begin "feeding" on the red giant. The matter torn off the surface of the red giant spirals into the white dwarf forming what is known as an accretion disk. Determine the separation between the two stars shown in this diagram. Let…
 r = separation between the stars a = radius of red giant b = radius of white dwarf ma = mass of red giant mb = mass of white dwarf
Ignore any centrifugal effects and state your answer using the given symbols only.

#### solution

Matter will not be pulled off the red giant until the gravitational field of the white dwarf at some distance is at least equal to the gravitational field of the red giant on its surface.

gb = ga

At this point, the distance to the center of the red giant is a and the distance to the white dwarf is (r − a).

 Gmb = Gma (r − a)2 a2

Now for the dirty work. Solve this apparently simple equation for the separation, r. Begin by canceling out the gravitational constant G and cross multiplying.

mba2 = ma(r − a)2

Expand the right side.

mba2 = mar2 − 2maar + maa2

The variable r is quadratic in this equation. Rearrange it until it looks something like the prototypical quadratic…

ax2 + bx + c = 0

like this…

mar2 − 2maar + (ma − mb)a2 = 0

 x = −b ± √(b2 − 4ac) 2a

like this…

 r = −(−2maa) ± √[(−2maa)2 − 4ma(ma − mb)a2] 2ma

Believe it or not, this simplifies into something simple. I have no desire to show all the steps. Here's what you get.

 r = a ⎛⎜⎝ 1 ± √ mb ⎞⎟⎠ ma

The equation has two solutions, but only one of them makes sense. The one with the plus sign makes the separation bigger than the radius of the red giant by some amount. The one with the minus sign makes it smaller, which means the white dwarf is already inside the red giant. If it's already inside, the two objects have already merged. That's what you call an extraneous or spurious solution. The valid solution is then…

 r = a ⎛⎜⎝ 1 + √ mb ⎞⎟⎠ ma

### practice problem 2

Suppose that the Earth was an infinite flat slab of thickness t with the same mean density as the Earth. Calculate t in order that this infinite flat Earth has the same acceleration due to gravity on its surface as is found on the actual spherical Earth.
1. Solve it once the easy way, using Gauss's law for gravity.

∯ g · dA = −4πGm

2. Solve it once the hard way, using Newton's law of universal gravitation for continuous distributions of matter.
 g(r) = − G ⌠⌠⌠⌡⌡⌡ r̂ dm r2
3. How does this number compare to the radius of the actual spherical Earth?

#### solution

We'll derive the equation twice, but leave it in symbolic form. At the end we'll come back and put the numbers in.

1. Encase the slab in an imaginary rectangular box with two faces entirely outside the slab and parallel to it. Let the box have an area A and a height that is greater than the thickness of the slab by an amount h on either side. This will be our Gaussian surface for Gauss's law for gravity.

∯ g · dA = −4πGm

Given the geometric simplicity of the situation, the gravitational field vector must strike either side of the slab at a right angle. The two faces of the imaginary box outside the slab are parallel to the sides of the slab. These two faces are the only ones that can capture any gravitational field. The area vector on each of these faces is pointing outward. The angle between these two vectors is 180°, which makes the dot product negative. Field in. Area out. Repeat. The integral on the left side of Gauss's law simplifies to…

−2gA = −4πGm

The mass contained within the imaginary box has a volume equal to the cross sectional area of the box times the thickness of the slab. Multiply density by volume and you get the mass of the slab contained within the imaginary box.

−2gA = −4πGρAt

Stuff cancels.

g = 2πGρt

Solve for thickness, t.

 t = g 2πρG

Note how the height, h, of the imaginary box does not appear in this equation. Gravity is independent of altitude on this world. If gravity remained constant with altitude in our world, we'd have evidence that we lived on an infinite flat Earth. It doesn't, so we don't.

2. Start from the basic concept that the gravitational field in any location near a large object (spherical, flat, or any other shape) is due to contributions from an infinite number of infinitesimal masses (dm) at some distance (r) summed up (∫∫∫) over the whole volume of the object.

 g(r) = − G ⌠⌠⌠⌡⌡⌡ r̂ dm r2 whole object

This problem is best solved with rectangular coordinates, where x and y are the horizontal axes and z is the vertical axis. Since our model flat Earth has a uniform density, we can replace the infinitesimal masses with infinitesimal volumes. The easiest shape to work with is a box.

dm = ρ dV = ρ dx dy dz

The distance to any one of these boxes can be found using Pythagorean theorem.

r2 = x2 + y2 + z2

If z is the vertical coordinate with up being positive, then the infinitesimal boxes will be at horizontal positions ranging from −∞ to +∞ and vertical positions ranging from −t to 0.

The gravitational field is a vector field, which means we need to consider the direction of the gravitational field from all these little boxes. No matter where you lie on an infinite plane, you are effectively at its center. It has a kind of symmetry — planar symmetry. What you see to your left is the same as what you see to your right (a horizon that is infinitely far away) no matter where you are located. This is also true if you look forward or backward, or for that matter, in any horizontal direction. On the infinite flat Earth, all the little masses pulling you to the right are balanced by an identical arrangement of little masses pulling you to the left. The same is true forward and backward, and in all horizontal directions. The only component of the gravitational field we need to consider then is the vertical one, gz. Thus, each infinitesimal needs to be multiplied by…

 sin θ = z √(x2 + y2 + z2)

where θ is the angle of depression from our imagined location to one of the infinitesimal boxes. A box directly under our feet (θ = 90°) would only pull us down (gz = max). A box located very far away (θ ≈ 0°) would mostly be pulling us sideways and hardy be pulling us down at all (gz ≈ 0).

Combine all these thoughts into one monster triple integral.

 0 +∞ +∞ g = −G ⌠⎮⌡ ⌠⎮⌡ ⌠⎮⌡ z ρ dx dy dz ẑ √(x2 + y2 + z2) x2 + y2 + z2 −t −∞ −∞

Simplify a bit.

 0 +∞ +∞ g = −ρG ⌠⎮⌡ ⌠⎮⌡ ⌠⎮⌡ z dx dy dz ẑ (x2 + y2 + z2)3/2 −t −∞ −∞

Let's do one integral at a time starting with x as the variable and y and z as constants.

 +∞ ⌠⎮⌡ z dx (x2 + y2 + z2)3/2 −∞
 +∞ ⎡⎢⎣ xz ⎤⎥⎦ (y2 + z2)√(x2 + y2 + z2) −∞

When evaluated over the limits ±∞ we get…

 +z − −z = 2z y2 + z2 y2 + z2 y2 + z2

Now evaluate this expression over y and let z be constant.

 +∞ +∞ ⌠⎮⌡ 2z dy = ⎡⎢⎣ 2 tan−1 y ⎤⎥⎦ y2 + z2 z −∞ −∞

The limits of the inverse tangent function at ±∞ are ±½π, respectively. This makes the y integral equal to…

2(+½π) − 2(−½π) = +2π

That giant beast of a triple integral reduced to almost nothing in two steps. Here's what we're left with — a trivial integral. (Watch the signs. When you work it out, however, there are three negatives in succession. As a result, the end product is negative, i.e., downward.)

 0 g = −2πρG ⌠⎮⌡ dz ẑ −t 0 g = −2πρG ⎡⎢⎣ z ⎤⎥⎦ ẑ −t g = −2πρG ⎡⎢⎣ 0 − (−t) ⎤⎥⎦ ẑ g = −2πρGt ẑ

Thickness is our goal. Solve for thickness.

 t = g 2πρG

Ten times the work for the same outcome. Brilliant!

Once again we can show that gravity on an infinite flat Earth does not change with altitude. Recall that we treated z as a constant when we integrated over x and y and it vanished. It only came back when we did the last integral. If we moved the origin of our coordinate system to a height h above the surface, then our limits of integration would be ht and h. Pop them into the integral and out comes the same answer.

 −h g = −2πρG ⌠⎮⌡ dz ẑ −h−t −h g = −2πρG ⎡⎢⎣ z ⎤⎥⎦ ẑ −h−t g = −2πρG ⎡⎢⎣ −h − (−h−t) ⎤⎥⎦ ẑ g = −2πρGt ẑ
3. Some facts about the Earth.

 g = 9.798 m/s2 r = 6.37 × 106 m m = 5.97 × 1024 kg ρ = 5510 kg/m3

t =
 g 2πρG

t =  9.798 m/s2
2π(5510 kg/m3)(6.67 × 10−11 N m2/kg2)
t = 4240 km

Interestingly, this is appears to be two thirds the radius of the actual Earth.

 4250 km = 0.666 6370 km

More interestingly perhaps, is that it's exactly the radius of a perfectly spherical Earth.

Recall that the gravity on a spherical Earth is…

 g = Gm r2

and the density of a spherical Earth is…

 ρ = m 43πr3

Substitute these two expressions into the equation for gravity on the infinite flat Earth and watch nearly everything cancel…

t =
 g 2πρG

t =
 1 Gm 43πr3 2πG r2 m

t = ⅔r

### practice problem 3

Dark matter
1. The orbital speed of the planets decreases with distance from the Sun. Why does this happen? Derive a formula that shows the relationship.

Magnify

2. 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.

Magnify

3. Calculate the mass of the Milky Way given an orbital speed of 230 km/s and a radius of 50,000 light years. Give your answer in solar masses (m = 1.99 × 1030 kg) and compare it to the number of stars in the Milky Way (~1011).
4. 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?

#### 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 mv2 = Gm1m2 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.

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.

 m = rv2 ⇐ v = √ Gm G r

This shows us that the mass around which a star orbits is directly proportional to its distance from the center of the galaxy when orbital velocity is constant. 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
r3/3
ρ =  3v2
Gr2

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(r) = 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 r2 and thickness dr from the center 0 out to a distance r we get back the expression we derived earlier for mass.

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

And all is well again.

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

 m = rv2 G

 m = (50,000 light year)(9.46 × 1015 m/light year)(230,000 m/s)2 6.67 × 10−11 N m2/kg2

m = 3.82 × 1041 kg

m = 1.92 × 1011 solar masses

This is a decent order of magnitude calculation for the number ofstars in the Milky Way.

4. If the orbital speed remains constant for bodies near but outside the Milky Way, then the mass-distance relationship…

 m = rv2 G

and density distribution…

 ρ(r) = 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 occasional 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 (83%) 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 space-time from its normally flat geometry. As we all know light travels in straight lines. But when light encounters the warped space-time around a galaxy, straight lines have no choice but to bend. The result is a phenomenon 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 phenomenon 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

The word nebula (plural nebulae) means cloud in latin. In astronomy, a nebula is a diffuse collection gas and dust that looks something like a cloud. Nebulae are larger than stars, but smaller than galaxies — on the order of 10-1000 solar systems in diameter. A few representative images are shown below.

A simplified model of a nebula is a spherical collection of matter whose density varies linearly from a maximum at its center to zero at its "surface". Determine the following quantities both inside and outside such a simplified nebula in terms of its radius R, the distance from the center r, the density at the center ρ0, and fundamental constants…

Magnify

1. density
2. gravitational field strength
3. gravitational potential energy per unit mass

#### solution

1. A linear function can be written in the form…

y = a + bx

where x is the independent variable, y is the dependent variable, and a and b are constants. In our case, we should modify this into the more appropriate…

ρ(r) = a + br

At the center of the nebula…

 r = 0 & ρ(0) = ρ0

and on the surface…

 r = R & ρ(R) = 0

Substituting these values into our generic equation will give us two equations with two unknowns: a and b. This is how we will generate the equation for density inside the nebula.

ρ(0) =  a + b 0  = ρ0  ⇒  a =  ρ0

ρ(R) =  a + b R  = 0  ⇒  b =  −ρ0/R

 ρ(r) = ρ0 ⎛⎜⎝ 1− r ⎞⎟⎠ R

Everywhere outside the nebula, the density is zero.

ρ(r) = 0

When graphed, the density looks like this…

Magnify

2. Start with the basic idea that at some distance r from the center of the nebula, the only matter m(r) that contributes to the gravitational field is within the sphere defined by r. Divide the nebula up into a series of infinitely thin shells of radius r, surface area r2, and thickness dr. Multiply the volume of this spherical shell by the density function, then integrate. Use the resulting expression for mass in the gravitational field formula.
 g(r) = − Gm(r) r2
 r g(r) = − G ⌠⎮⌡ ρ(r) dV r2 0

Replace density with the function we just derived and clean it up a bit.

 r g(r) = − G ⌠⎮⌡ ρ0 ⎛⎜⎝ 1− r ⎞⎟⎠ 4πr2 dr r2 R 0
 r g(r) = − 4πρ0G ⌠⎮⌡ ⎛⎜⎝ r2 − r3 ⎞⎟⎠ dr r2 R 0

Calculate the integral from the center out to the variable distance r.

 g(r) = − 4πρ0G ⎛⎜⎝ r3 − r4 ⎞⎟⎠ r2 3 4R

Simplify and you're done. The gravitational field inside the nebula is given by the expression…

 g(r) = − 4πρ0G ⎛⎜⎝ r − r2 ⎞⎟⎠ 3 4R

Repeat this procedure changing the limits of integration. Calculate the integral from the center (r = 0) all the way out to the edge of the nebula (r = R). The gravitational field outside the nebula is given by the expression…

 R g(r) = − Gm(R) = − G ⌠⎮⌡ ρ(r) dV r2 r2 0
 R g(r) = − G ⌠⎮⌡ ρ0 ⎛⎜⎝ 1− r ⎞⎟⎠ 4πr2 dr r2 R 0
 R g(r) = − 4πρ0G ⌠⎮⌡ ⎛⎜⎝ r2 − r3 ⎞⎟⎠ dr r2 R 0
 g(r) = − 4πρ0G ⎛⎜⎝ R3 − R4 ⎞⎟⎠ r2 3 4R
 g(r) = − πρ0GR3 3r2

When graphed over the appropriate ranges, the the two expressions together look like this…

Magnify

3. Integrate the field from to a position r outside the nebula. This is not too difficult.
 r Vg(r) = − ⌠⎮⌡ g(r) dr ∞
 r Vg(r) = − ⌠⎮⌡ − πρ0GR3 dr 3r2 ∞
 Vg(r) = − πρ0GR3 3r

Integrate the field from to the surface R and then again from R to a position r inside the nebula. This is a tedious procedure.

 r Vg(r) = − ⌠⎮⌡ g(r) dr ∞
 R r Vg(r) = − ⌠⎮⌡ g(r) dr − ⌠⎮⌡ g(r) dr ∞ R
 R Vg(r) = − ⌠⎮⌡ − πρ0GR3 dr 3r2 ∞
 r − ⌠⎮⌡ − 4πρ0G ⎛⎜⎝ r − r2 ⎞⎟⎠ dr 3 4R R

 R Vg(r) = − πρ0GR3 ⎡⎢⎣ 1 ⎤⎥⎦ 3 r ∞
 r + 4πρ0G ⎡⎢⎣ r2 − r3 ⎤⎥⎦ 6 12R R

 Vg(r) = − πρ0GR3 ⎛⎜⎝ 1 − 1 ⎞⎟⎠ 3 r ∞
 + 4πρ0G ⎛⎜⎝ r2 − r3 − R2 + R3 ⎞⎟⎠ 6 12R 6 12R

Simplify at the end.

 Vg(r) = − πρ0G ⎛⎜⎝ r3 − 2r2 +2R2 ⎞⎟⎠ 3 R

When graphed, the gravitational potential looks like this…

Magnify