Gravitational Attractions
of Small Bodies
Calculating the gravitational
attraction of an arbitrary body
Given an elementary body with
mass mi at position ri, the associated
acceleration gi at any point r is
gi (r )  
Gmi
ri
3
(r  ri )
Forces add linearly, so the total
acceleration due to a group of n
bodies is
n
gi (r )  G
i 1
Where mi = (ri)vi.
mi
ri
3
n
(r  ri )  G
i 1
(ri )
ri
3
(r  ri )vi
For a continuous mass, let vi go to zero and the sum becomes an integral:
gi (r )  G 
(r')(r  r ')
V
(r  r ')
3
dv(r ')
Thus, given the density distribution (r’), you can calculate the
gravitational acceleration.
Two
practical comments:
1. Usually we try to take advantage of any symmetry, to make the
vector addition easier.
2. Potential is often easier to work with because it is a scalar.
Specifically:
U(r )  G 
V
(r')dv(r ')
(r  r ')
In practice, we measure |g| as opposed to the vector quantity itself.
Suppose we are measuring the gravitational anomaly caused by a small
body within the Earth.
Let ge be the acceleration of the Earth without the body, and Dg be the
perturbation due to the anomalous mass (Dg << ge).
Then, the total field is found from
g  ge  Dg
g
g  Dg  g  Dg
1/ 2
e
e
 ge  ge  2ge  Dg  Dge  Dg
1/ 2
1/ 2
2 

Dge
g

Dg
e

g  ge 1 2

2
2

ge
ge 


Expanding (as (1+x)1/2 ~ 1 + x/2) we get:
2 



 Dg cos 
Dg
g

Dg
g

Dg
e
  ge 1 e 2  ge 1
g  ge 1 e 2 
 ge  Dg cos
2
ge




ge
2 ge 
ge 






Or, since Dgcos is the z component of Dg:
g  ge  Dg cos  ge  Dg z
The residual R is therefore

R  g  ge  Dg z
Therefore, in calculating the anomaly due to a body, we only
worry about the z component.

We will now demonstrate how to calculate the gravity anomaly
due to bodies with simple shapes.
Calcuation of Dg for some simple bodies
1. Sphere of constant density . An example of using the potential U.
Recall that
U(r )  G 
(r')dv(r ')
V
(r  r ')
In this example, (r’) = constant = o.
For a sphere, dv(r’) = r’2sinddfdr’
Using the
law of cosines, |r - r’| = [r2 + r’2 - 2rr’cos]1/2. Thus
U(r )  G o 
V
r
r '2 sindr 'ddf
2
 r ' 2rr 'cos 
2

a
1/ 2
 G o  r ' dr ' 
2
0
0
r
The last integral is just 2. Note that
1/ 2
2a  bx
dx
 a  bx 1/ 2 
b


2
sinddf
2
 r ' 2rr 'cos 
2
1/ 2
 df
0
Substitute x = cos and dx = -sind in the above and we get
a
U(r )  2G o 
0


a
1/ 2
r '2 dr ' 2
r '2 dr '
2
2

r

r
'
2rr
'cos


2

G

r

r
'




o

rr '
rr
'
0
0
 
1/ 2

2 1/ 2
 r  r '
Now consider the case where r > r’ (outside the sphere):
a
U(r )  2Go 
0
a
r'2 dr'
r'2 dr' 4 Ga 3 o GM


r  r'  r  r'  4 Go 

rr'
r
3r
r
0
Which is the same as the potential for a point mass. Thus
U

GM
GMz
GMz
gz  

 2 2 2 3/ 2  3
2
2
2 1/ 2
z
z (x  y  z )
(x  y  z )
r




Now consider the case where r < r’ (inside the sphere):
r
a
r '2 dr'
r '2 dr'
U(r )  2G o 
r  r '  r  r' 
r  r '  r' r


rr '
rr '
0
r
r 2r '2 dr' a

 2 r 2 
U(r )  2G o
  2r'dr' 2Go a  
r
3 

0

r
Thus
U 4 G o r 3 GM r
g 

 2
2
r
3 r
r
Meaning that only the mass within the radius r contributes to the
acceleration.

A Buried Vertical Cylinder. An Example of Using Symmetry.
Let’s try computing the anomaly due to a vertical cylinder within the
earth at a point directly above the center of the cylinder.
Consider a cylinder of radius a buried in the earth at a depth h1. The
cylinder extends to depth h2. It has a density 2, while the surrounding
earth has a density 1 (so D = 2 - 1).
Horizontal (sin) attractions cancel because of symmetry. The
remaining vertical (cos) gravitational anomaly caused by a volume
element is
GD
gz 
R
2
For a cylinder, dv = rdrdzdf.
Also, cos = z/(r2 + z2)1/2. Therefore
GD zrdrdzdf

gz  2 2
r  z r 2  z 21/ 2
cosdv
2
h2
a
Dgz  GD  df  zdz 
0
h1
0
h2
rdr
r
2
z

2 3/ 2
a
 2GD  zdz 
h1
0
rdr
r
2
z



1
1
 2GDh  h  
Dgz  2GD  zdz 
2
1
1/
2
2
2



z
h1
 a  z  


2 1/ 2
2 1/ 2 
2
2
Dgz  2GDh2  h1  a  h2
 a  h1



h2

 
h2

h1

2 3/ 2


1/
2
a 2  z 2 
zdz

If we let a become very large, the the equation above reduces to
Dgz  2GDh2  h1  2GDDh
This will turn out to be a very useful equation later on, so
remember it!
Anomalies from Two-Dimensional Structures
Very often it is useful to treat structures in the Earth as twodimensional. It turns out there are clever ways to analyze gravity
anomalies from such structures. Here is how you can do it:
For a 2-D structure, r = r(x’, z’) (i.e., no y dependence). In what
follows, we use primed coordinates (x’, y’, z’) for the anomalous mass,
and unprimed coordinates (x, y, z) for the observing station. All
measurements are made along the y = 0 axis.
The distance R in the x-z plane is
R2 = (x-x’)2 + (z-z’)2
And the total distance r is
r2 = R2 + y’2
The gravitational potential is
U(x,y,z)  G 
(x',z')dx'dy'dz'
r
L
dy'
2
2 1/ 2
L (R  y' )
 G  (x',z')dx'dz' 
The integral over y’ is now finite. Later we will let it get big. For now



U(x,y,z)  G  (x',z ')dx'dz' ln y' (R  y' )
U(x,y,z)  G  (x',z ')dx'dz' ln
2
2
L
L
(R 2  L2 )  L
(R 2  L2 )  L
Here comes cute trick #1. Because we are really interested in g and
not U, and
because g = U, we can add anything we want to U that

is not a function of (x,y,z) without affecting g.
So, let’s define a constant:
Uo  G 
 (R 2  L2 )  L 

(x',z')dx'dz' ln o2
2

 (Ro  L )  L 

Then
U  U Uo  G 
If we let L >> R, then

logarithm becomes:
2

2
2
(Ro  L2 )  L 
(R

L
)

L

(x',z')dx'dz' ln 2 2

2
2

(Ro  L )  L 
 (R  L )  L

2 

R
(R2  L2 )  L1 2  . The argument in the
 2L 

 2 R 2 R 2 
R2
Ro 2 
2L 

Ro  o 2 
R 2 
o
2L
2L
4L




ln


ln

ln
 2ln(Ro) ln(R)

2
2
2 2
2 
Ro 
 R
 2 Ro R 
R 
2L

R



 2L



2L 
4L2 
So
U  2G  (x',z')ln(Ro) ln(R)dx'dz'

And

gz 
U
(x',z')(z  z')
 2G 
dx'dz'
2
2
z
(x  x') (z  z')
The integral for g is often easy to integrate. In many applications we
can assume (x’,z’) = constant and we set z = 0 at the surface.
Now comes cute trick #2: let’s use cylindrical coordinates instead of
Cartesian. Then
2
2
2
R  (x  x')  (z  z ')
dx'dz' RdRd
Substituting these expressions into the formula for gz gives

Noting that
(z  z')
 sin
R

gz  2G 
RdRd (z  z')
R2
we have
gz  2G  sinddR
Fora given value of , the limits of integration are from R() near to
the observer to R() from from the observer. We perform this kind of

integration from  at the top of the body (top) to  at the bottom
(bottom).
Noting that sindR = dz, we have
gz  2G  ddz  2G
 bottom
z( ) far
 bottom
 d  dz  2G  d z( )
far
 top
z( ) close
 top
 z( )close 
These two integrals are line integrals that together encompass
the body (the minus sign on the second integral reverses the
direction. Therefore
gz  2G  z( )d
Which is a line integral around the body.
This is a great way to compute
anomalies for complicated structures.

Note that the line integration must
always be done in a clockwise sense!
Let’s use this method to calculate the attractions of some simple
bodies.
The Infinite Sheet (Revisited)
In this case
2
3
4
1
1
2
3
4
 z( )d   z( )d   z( )d   z( )d   z( )d
 0 z 2(  ) 0 z 2 (  )
gz  2G  z( )d  2G (z2  z1)  2Gh

Which is what
 we obtained before (but with much more work!)

The Semi-Infinite Sheet of Thickness t
As above
2
3
4
1
1
2
3
4
 z( )d   z( )d   z( )d   z( )d   z( )d
 z1 o  0 z1  t o  0   ot



gz  2G  z( )d  2G ot
Now
o 
So
x -> ∞

Also
x
 arctan
2
zo

x 
gz  2Gt  arctan 
z o 
2

Note that as

 -> 
g -> 2Gt (i.e., infinite slab)
x -> 0
 -> /2 g -> Gt
x -> -∞
 -> 0 g -> 0
dgz
d arctanx z o
dx
 2Gt
dx
 z

o
 2Gt 2
2 
x

z

o 
2Gt

dx x 0
zo
dgz

Thus, the slope lessens as the depth of the sheet increases. In fact,
the slope can give us
 some information about the depth of the sheet.
Hitchhiker’s Guide to Gravity
First, since many situations use the infinite slab formula, it’s useful to
remember that if the density difference D is given in gr/cm3 and the
thickness of the slab Dh is given in kilometers, then g in mgals is
g = 2GDDh = 2 x 3.14159 x 6.67DDh = 41.9DDh ~ 42 DDh
So, if you can remember the answer to Life, the Universe, and
Everything, you can remember to how to calculate the infinite slab
attraction!
Practical example #1: A basin filled with sediments
Let’s suppose we are standing on top of a basin that is 2 km thick and
filled the sediments with a density that is 0.5 gr/cm3 less than the
basement below, we would expect that in the middle of the basin:
Dgmid = 42 x 2 x 0.5 = 42 mgals
At the edge of the basin (x=0), Dgx=0 = Dgmid/2 = 21 mgals
and dg/dx = (2G DDh/zo)=13.34 mgals/km (note zo = 1 km)
Practical example #2: The continental margin
In this case, we can think of a margin as two semi-infinite slabs on
either side of an infinite slab that has a thickness equal to the ocean
side of the margin. The upper semi-∞ slab make a D>0
contribution, while the lower semi-∞ slab make a D<0 contribution.
BUT, the contributions don’t cancel because they are at different
depths.
Example of Equivalence and Non-uniqueness
Convince yourself
that the 3 geometries
shown to the right
give identical relative
gravity anomalies:
Identical Response from different Mass Distributions