11-4 The Gravitational Field g
As we discussed previously, the gravitational force exerted by a point particle
of mass m1 on a second particle of mass m2 at a distance of r1,2 away is given by
where the unit vector r1,2 points from particle 1 to particle 2.
IF you want to find the gravitational field at a given point, P, put a particle of
mass m at that point, and calculate the gravitational force on it die to all of the
other particles that act upon it (with respect to gravitational pull). The
gravitational force, F, divided by the mass, m, is the gravitational field, g, at
that point, P.
That point, P, is called a field point. So, the gravitational field at point P due to
a collection of point particles is the vector sum of all of the fields due to the
individual masses of the particles at that point, P.
g = ∑ gi
i
The places where the particles are located are called source points. If an object
is a continuous collection of points, we can compute dg due to a small element
of volume dm and integrate over the entire mass, just like you’d expect. ☺
g = ∫ dg
The gravitational field of the earth at a distance r ≥ RE points toward the earth,
and has a magnitude given by:
This equation can be used un-amended when g is in the same direction as the
force. If it is not, you will need to multiply the equation by the relevant
resolved portion of g in the direction of interest.
See Ex 11-7, p 354.
Two point particles, each of mass M, are fixed on the y-axis at y = +a and y = -a
as shown. Find the gravitational field of all points on the x-axis as a function
of x.
To find the field wrt the x-axis, you need to relate the gravity vectors to the xaxis. The gravity vectors, g, are described in the direction of r, where r is the
same magnitude for both masses. Thus, g is the same for both masses. The
same is true for the value of a, as depicted in the picture. In terms of r, a, x, g,
and Ө, the relationship can be described as:
x2 + a2 = r2
or r = √(x2 + a2)
as well as
cos Ө = x / r
Since both masses contribute equivalently along the x-axis, the total g along the
x axis (g points negative wrt the x-direction) = - 2g cos Ө = - 2gx / r. The
question only asks you about g in the x-direction, so the y-components are
irrelevant.
NOTE: We need to use eq 11-25, but because the g doesn’t point directly along
the x-axis, we need to multiply that by the portion that is along the x-axis.
Use eq 11-25 above to get g = -2GM x i
r2 r
= - 2GMx i
r3
If you now substitute r = √(x2 + a2), you get g = - 2GMx i
(x2 + a2)3/2
See Ex 11-8, p 355
A uniform rod of mass M and length L is centered on the origin and lies along
the x-axis as shown. Find the gravitational field due to the rod at all points
along the x-axis for x > L/2.
Because they use the magic words “uniform” and “at all points,” you are going
to use the equation g = ∫ dg.
So, we will once again need to use eq 11-25, which says that g = GM which
r2
means that dg = - Gdm Again, the value is negative because of the drawn
r2
direction of dg wrt the x-axis. Further, when you look at the drawing, you can
see that dm = M dx, and that point P in terms of x and x0 gives r = x0 - x
L
Now, if you substitute r and dm back into eq 11-25, you get dg = - G (M/L) dx
(x0 – x)2
So, g = ∫ dg = - GM ∫ __dx__ = - GM [ 1 / (x0 – x)] from x = - L/2 to x = + L/2
L (x0 – x)2
L
NOTE: The integral involves the negative power of the denominator times the
negative value of x in the denominator, so, the coefficient in front of the
integral stays put at – GM
L
Assuming x0 is some arbitrary point, you can replace it with regular ol’ x, after
the fact, and get g = – ___GM___ i
x2 – (L/2)2
g of a Spherical Shell and of a Solid Sphere
Q: Who developed calculus?
A:
One of Newton’s greatest motivations for developing calculus was to prove that
the gravitational field outside of a solid sphere is the same as if all of the mass
of the sphere were concentrated at its center. (Hey, what else did you expect
him to do while he was locked up trying to avoid the plague?)
The next section, which is optional, contains the proofs for the next few
statements. You may read them if you like, but we won’t cover section 11-5 in
class. So, you’ll just have to accept the next few statements as givens.
Figure 11-4: A uniform spherical shell of mass M and radius R
The gravitational field due to a hollow sphere (shell) at a distance r from the
center of the shell is given by:
Inside of the shell, g = 0 because the opposing g’s cancel each other out.
Figure 11-15
Figure 11-15 above shows a mass right at the center of the sphere, but even if
you move off center, so that the point is, say, twice as close to the right side as
to the left, Newton’s Law of Gravity says that the respective distance on one
side will be twice that of the other, and that gravity is proportional to distance
squared. 22 = 4. Keeping that in mind, if you then extend a cone outward from
that point, you will see that the cone sweeps out an area of the opposite side
that is 4 times greater than on the smaller side. 4 = 4 so the internal gravities
cancel each other out.
(If you need me to explain this further in class, jes’ ask! ☺)
NOTE: This is true whether or not the sphere has a constant density, as long
as that density depends only on r, and the sphere is symmetrical.
g Inside a Solid Sphere
Figure 11-16 below helps illustrate what happens to g inside a solid sphere.
Figure 11-16
As we already know, g inside of a shell = 0. The mass of that part of the sphere
outside r exerts no force at or inside of r. So, only the mass M’ within the
radius r contributes to the field at r. This mass produces a field equal to that
of a point mass M’ at the center of the sphere. The fraction of the total mass of
the sphere within r is equal to the ratio of the volume of the sphere of radius r
to that of radius R. So, for a uniform mass distribution,
M’ = M (4/3) ∏ r3 = M _r3
(4/3) ∏ R3
R3
If you substitute this back into eq 11-25, and multiply by the fractional
equivalent, you get
g = - GM r ,for r < R (as the r3 on top cancels out both of the r2 on the bottom)
R3
Try Ex’s 11-9 and 11-10 now. ☺
Do p 362 – 367, #’s 1, 3, 8, 14, 19, 21, 26, 35, 43, 48, 54, 58, 61, 67, 71