Practice

practice problem 1

solution

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

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

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

Believe it or not, this simplifies to something simple.

Additional discussion needed.

practice problem 2

solution

Answer it.

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.

Helix Nebula (Source: STScI)	Rosette Nebula (Source: Unknown)	Horsehead Nebula (Source: NOAO)

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 generic equation …

   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) = ρ0 ⎛ ⎝ 1−  r ⎞ ⎠ R
   ρ(R) =	a + b R	= 0	⇒	b =	-ρ0/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 4πr², 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.

   				r
   g(r) = −	Gm(r)	= −	G	⌠ ⌡	ρ(r) dV
   	r2		r2
   				0

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

   r

   r

   g(r) = −

   G

   ⌠ ⌡

   ρ0

   ⎛ ⎝

   1−

   r

   ⎞ ⎠

   4πr2 dr = −

   4πρ0G

   ⌠ ⌡

   ⎛ ⎝

   r2 −

   r3

   ⎞ ⎠

   dr

   r2

   R

   r2

   R

   0

   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

   R

   g(r) = −

   Gm(R)

   = −

   G

   ⌠ ⌡

   ρ(r) dV = −

   G

   ⌠ ⌡

   ρ0

   ⎛ ⎝

   1−

   r

   ⎞ ⎠

   4πr2 dr

   r2

   r2

   r2

   R

   0

   0

   R

   g(r) = −

   4πρ0G

   ⌠ ⌡

   ⎛ ⎝

   r2 −

   r3

   ⎞ ⎠

   dr = −

   4πρ0G

   ⎛ ⎝

   R3

   −

   R4

   ⎞ ⎠

   r2

   R

   r2

   3

   4R

   0

   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		r
   Vg(r) = −	⌠ ⌡	g(r) dr = −	⌠ ⌡	−	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

   Rr

   Vg(r) = −

   ⌠ ⌡

   −

   πρ0GR3

   dr

   −

   ⌠ ⌡

   − 4πρ0G

   ⎛ ⎝

   r

   −

   r2

   ⎞ ⎠

   dr

   3r2

   3

   4R

   ∞

   R

   R

   r

   Vg(r) = −

   πρ0GR3

   ⎡ ⎣

   1

   ⎤ ⎦

   +

   4πρ0G

   ⎡ ⎣

   r2

   −

   r3

   ⎤ ⎦

   3

   r

   6

   12R

   ∞

   R

   Vg(r) = −

   πρ0GR3

   ⎛ ⎝

   1

   −

   1

   ⎞ ⎠

   +

   4πρ0G

   ⎛ ⎝

   r2

   −

   r3

   −

   R2

   +

   R3

   ⎞ ⎠

   3

   r

   ∞

   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]

practice problem 4

solution

Answer it.