The Physics
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Opus in profectus

Gravity of Extended Bodies

Practice

practice problem 1

Find the separation at which a degenerate star would suck the surface matter off of a companion star.

solution

 r = separation between objects a, b = radius of degenerate and companion star, respectively ma, mb = mass of degenerate and companion star, respectively

Ignore any centrifugal effects and set the two gravitational forces equal to one another on the surface of the companion star.

 gdegenerate star = gcompanion star GMa = GMb (r − b)2 b2

And now for the dirty work. Solve this apparently simple formula for the separation, r.

 mab2 = mb(r − b)2 = mbr2 − 2mbrb + mbb2 0 = mbr2 − 2mbrb + (mb − ma)b2

The variable r is quadratic in this equation. Use the quadratic equation to solve it.

 r = + 2mbb ± √[4mb2b2 − 4mb(mb − ma)b2] 2mb

Believe it or not, this simplifies to something simple.

 r = b ⎛⎝ 1 ± √ ma ⎞⎠ mb

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

The gravitational field vector must point into the slab on either side of it. Given the geometric simplicity of the situation, the field 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

Some 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 a 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.

 gave = 9.81 m/s2 rave = 6.37 × 106 m m = 5.97 × 1024 kg ρave = 5510 kg/m3

t =
 g 2πρG

t =  9.81 m/s2
2π(5510 kg/m3)(6.67 × 10−11 Nm2/kg2)
t =  4,250 km

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

 4,250 km = 0.666 6,370 km

More interestingly perhaps, is that it's exactly the radius of the actual Earth.

Recall that the gravity on a spherical Earth is…

 g = Gm r2

…and the density of a spherical Earth is…

 ρ = m 4/3πr3

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 4/3πr3 2πG r2 m t = ⅔r

practice problem 3

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…

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…

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…

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 =
 − πρ0GR3 3r
3r2

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…

practice problem 4

The data in the text file earth.txt gives the density of the Earth at varous depths below the surface. Using data analysis software (preferably something that can do numerical integration) generate a data column for the the gravitational field strength at various depths below the surface. Remember that the value of the field is 9.8 N/kg on the surface of the Earth and 0 N/kg in the center.