Gravity of Extended Bodies

The Physics Hypertextbook™
© 1998-2008 by Glenn Elert -- A Work in Progress
All Rights Reserved -- Fair Use Encouraged

Discussion

tidal forces

The tides, tidal forces, prolate spheroid, Roche limit


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Let …

r	be the separation between the objects
a, b	be the radius of the planet and moon, respectively
m_a, m_b	be the mass of the panet and moon, respectively

Derive the tidal force formula.

g_tidal =	g_front	−	g_back

g_tidal =	Gm_b	−	Gm_b
	(r − a)²		(r + a)²

Work that algebra. Worlk it!

g_tidal = Gm_b	⎛ ⎜ ⎝	(r + a)² − (r − a)²	⎞ ⎟ ⎠	= Gm_b	⎛ ⎜ ⎝	(r² + 2ra + a²) − (r² − 2ra + a²)	⎞ ⎟ ⎠
		(r − a)²(r + a)²				r⁴ − a⁴

Simplify.

g_tidal = Gm_b	⎛ ⎜ ⎝	4ra	⎞ ⎟ ⎠
		r⁴ − a⁴

Super-simplify

g_tidal ≈	4Gm_ba
	r³

Good, now derive the Roche limit.

g_tidal	≈	g_surface

4Gm_ab	≈	Gm_b
r³		b²


r ≈ b	∛	4m_a
r ≈ b	∛	m_b

flattening

oblate spheroid

Polar radius a, equatorial radius c. The flattening factor (also called oblateness) is …

ℰ =	a − c
	a

gravity inside & outside

Two ways to solve problems …

in general …


g(r) = − G	⌠⌠⌠ ⌡⌡⌡	ˆr dm	&	V_g = − G	⌠⌠⌠ ⌡⌡⌡	dm
		r²				r²
	^{all space}				^{all space}

for spherically symmetric mass distributions (something like Gauss' law)

		_r				_r
g(r) = −	G	⌠ ⌡	ρ(r) 4πr² dr ˆr	&	V_g(r) = −	⌠ ⌡	g(r) · dr
g(r) = −	r²	⌠ ⌡	ρ(r) 4πr² dr ˆr	&	V_g(r) = −	⌠ ⌡	g(r) · dr
		⁰				^∞

Prove that the gravitational field inside a uniform mass distribution is 0. This is the key.

Summary

bullet

Problems

practice

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

Solution: Let …

r	be the separation between the objects
a, b	be the radius of the degenerate and companion star, respectively
m_a, m_b	be the mass of the 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.

g_{degenerate star}	=	g_{companion star}
GM_a	=	GM_b
(r − b)²	=	b²

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

m_ab² =	m_b(r − b)² = m_br² − 2m_brb + m_bb²
0 =	m_br² − 2m_brb + (m_b − m_a)b²

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

r =	+ 2m_bb ± √[4m_b²b² − 4m_b(m_b − m_a)b²]
	2m_b

Believe it or not, this simplifies to something simple.

r = b	⎛ ⎜ ⎝	1 ± √	m_a	⎞ ⎟ ⎠
			m_b

Additional discussion needed.

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 earth has the same acceleration due to gravity at the surface.
Answer it.

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	Rosette Nebula	Horsehead Nebula
(Source: STScI)	(Source: Unknown)	(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 …

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density
gravitational field strength
gravitational potential energy per unit mass

Solutions …

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 …

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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
g(r) = −	r²	= −	r²	⌠ ⌡	ρ(r) dV
				⁰

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

g(r) = −

⌠
⌡

ρ₀

⎛
⎜
⎝

1−

⎞
⎟
⎠

4πr² dr = −

4πρ₀G

⌠
⌡

⎛
⎜
⎝

r² −

r³

⎞
⎟
⎠

r²

⁰

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


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

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


g(r) = − 4πρ₀G	⎛ ⎜ ⎝	r	−	r²	⎞ ⎟ ⎠
g(r) = − 4πρ₀G	⎛ ⎜ ⎝	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 …

g(r) = −

Gm(R)

= −

⌠
⌡

ρ(r) dV = −

⌠
⌡

ρ₀

⎛
⎜
⎝

1−

⎞
⎟
⎠

4πr² dr

r²

⁰

g(r) = −

4πρ₀G

⌠
⌡

⎛
⎜
⎝

r² −

r³

⎞
⎟
⎠

dr = −

4πρ₀G

⎛
⎜
⎝

R³

−

R⁴

⎞
⎟
⎠

r²

⁰


g(r) = −	πρ₀GR³
g(r) = −	3r²

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

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Integrate the field from ∞ to a position r outside the nebula. This is not too difficult.

V_g(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.

V_g(r) = −

⌠
⌡

g(r) · dr

^∞_R

V_g(r) = −

⌠
⌡

g(r) dr

−

⌠
⌡

g(r) dr

^∞_R

^R_r

V_g(r) = −

⌠
⌡

−

πρ₀GR³

−

⌠
⌡

− 4πρ₀G

⎛
⎜
⎝

−

r²

⎞
⎟
⎠

3r²

^∞

V_g(r) = −

πρ₀GR³

⎡
⎣

⎤
⎦

4πρ₀G

⎡
⎣

r²

−

r³

⎤
⎦

12R

^∞

V_g(r) = −

πρ₀GR³

⎛
⎜
⎝

−

⎞
⎟
⎠

4πρ₀G

⎛
⎜
⎝

r²

−

r³

−

R²

R³

⎞
⎠

∞

12R

Simplify at the end.


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

When graphed, the gravitational potential looks like this …

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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.
Answer it.

numerical

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Prolate Spheroid
The image to the right shows a typical alignment of the three most popular objects in the solar system: the sun, earth, and moon (not to scale). The letters around the earth represent various locations on the surface of the earth.


A	is closest to the moon
B	is farthest from the moon
C	is closest to the sun
D	is farthest from the sun

Complete the following table.
Use the results of your calculations to explain the effects that the moon and sun have on the shape of the earth.

quantity		moon	sun

mass (kg)		7.348 × 10²²	1.989 × 10³⁰
distance from earth (m)		3.844 × 10⁸	1.496 × 10¹¹
gravitational field (N/kg) from object
	on earth, at earth's center (g)	–	–
	on earth, nearest to object (g_near)	–	–
	on earth, farthest from object (g_far)	–	–
ratio of difference in field at opposite sides of the earth to field at center (1:x)		-	-

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Oblate Spheroid
Compare these images of earth and Jupiter not to scale.
1. Complete the following table.
2. Use the results of your calculations to explain the effect that rotation has on the shapes of earth and Jupiter.

quantity		earth	jupiter

mass (kg)		5.9742 × 10²⁴	1.8988 × 10²⁷
period of rotation (h)		23.935	9.9250
radius (m)
	polar	6,356,750	66,854,000
	equatorial	6,378,140	71,492,000
oblateness: ratio of difference in radii to polar radius (1:x)		-	-
rotational speed (m/s)
	at either pole	–	–
	on the equator	–	–
centrifugal acceleration (m/s²)
	at either pole	–	–
	on the equator	–	–
gravitational acceleration (m/s²)
	at either pole	–
	on the equator	–	–
ratio of centrifugal to gravitational accelerations at the equator (1:x)		-	-

algebraic

Determine the density function of a galaxy necessary to ensure that all of its stars orbit with the same linear velocity (that is, a galaxy with a flat rotation curve).
Determine the rotation curve for a very thin, disk-shaped galaxy with a uniform mass distribution.

calculus

Given the nebula in sample problem 3, determine the …
of the maximum gravitational field in terms of its radius R, the density at the center ρ₀, and fundamental constants.
Determine the gravitational field and gravitational potential, inside and outside the following mass distributions …
1. a sphere of mass M and uniform density ρ.
2. a simplified model of the earth, whose total mass M is evenly split between the core and the mantle — ½M for each part. (Assume the crust, oceans, and atmosphere make a negligible contribution to earth's mass.)
3. a simplified model of the earth consisting of a core with density 2ρ and mantle with density ρ. (Assume the crust, oceans, and atmosphere make a negligible contribution to earth's mass.)
4. a galaxy (including dark matter halo) with a density that decreases according to the function ρ = ρ₀/r², where r is the distance from the galactic center.

investigative

Determine the apparent acceleration due to gravity at your current location on earth. Here is the general procedure for doing this.

Start by determining your latitude and altitude. That information is probably available somewhere on the World Wide Web or you could measure it yourself with a GPS enabled device.

The earth is an oblate spheroid. Use the following fancy formula to calculate your distance (r) from the center of the earth …


r = h +	⎛ ⎜ ⎝	(a² cos φ)² + (b² sin φ)²	⎞^½ ⎟ ⎠
r = h +	⎛ ⎜ ⎝	(a cos φ)² + (b sin φ)²	⎞^½ ⎟ ⎠

where …


h	is your altitude
φ	is your latitude
a	is earth's equatorial radius (6,378,135 m)
b	is earth's polar radius (6,356,750 m)

Compute the magnitude of the gravitational acceleration at your location.
Compute your distance from the earth's axis; that is, the component of r parallel to the plane of the equator.
Compute the magnitude of the centrifugal acceleration at your location.
Combine the gravitational and centrifugal components to determine the apparent acceleration due to gravity (g') at you location. Determine both the …
1. magnitude and
2. direction (relative to r, the vector that points directly toward the center of the earth)
(Draw yourself a little picture before you try to answer this part.)
Compare your results to the values often found in textbooks and reference tables.
1. 9.8 m/s² (this value with its two significant digits of accuracy should agree with your results)
2. 9.81 m/s² (an overly accurate value that is incorrect for many places on earth)
3. 9.80665 m/s² (the defined unit acceleration due to gravity)

Resources

dark matter
- Seeing dark matter in the Andromeda galaxy, Vera Rubin, Physics Today, December 2006
- Rotation of the Andromeda Nebula from a Spectroscopic Survey of Emission Regions, Vera C. Rubin and W. Kent, Jr. Ford, Astrophysical Journal, Vol. 159 (1970): 379.

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