Posts Tagged ‘Gravitation’

Physics Friday 64

March 20, 2009

[part 3 of 3]

In part 2, we showed how if the distant point mass M orbits our oblate spheroid (or our ellipsoid orbits it) in a circular orbit, with an orbital frequency Ω much faster than the torque produced, one may time-average the torque to get a result where the rotation axis of our ellipsoid, the z’ axis of the body frame, maintains the same angle with, and preccesses about, the axis perpendicular to the plane of the orbit, the z axis of the inertial frame. In particular, we found that
, where is the angular frequency of the precession, ωz is the angular velocity of the ellipsoid’s rotation, θ0 is the angle between z’ and z, and d is the distance between our ellipsoid and the mass M.

So far we’ve kept all this in general, abstract terms. Now we will move this to a concrete example: the shape of the planet Earth may be approximated as an oblate spheroid of equatorial radius a=6378.1 km and polar radius c=6356.8. More specifically, if we use either pair of values given here for the moment of inertia of the earth, we get a value for of about 0.00323.

The other major parameters for the Earth are θ0 and ωz. The tilt of the earth’s axis, θ0, is approximately 23.44°=0.4091 radians, so that cosθ0=0.9175. The angular frequency of the earth’s rotation about its axis, ωz is found to be 7.2921×10-5 s-1, using as rotational period one sidereal day.

Now, we consider first the effect of the sun. The mass of the sun is 1.99892×1030 kg, so GM for the sun is 1.3272×1020 m3 s-2. The average distance between the earth and the sun is d=1.496×1011 m. Plugging these into our precession formula,
gives a precession frequency of about 2.417×10-12 s-1, which corresponds to a period of 2.600×1012 s, or about 82,000 years.

Now, for the moon. The mass of the moon is is 7.3477×1022 kg, so GM for the moon is 4.903×1012 m3 s-2. The average earth-moon distance is 3.844×108 m. Ignoring the tilt of the moon’s orbit relative to the ecliptic, we can use our precession formula as before, giving a precession frequency of about 5.262×10-12 s-1, which corresponds to a period of 1.194×1012 s, or about 38,000 years.

As noted before, we have made many simplifying assumptions and approximations; the actual rate of the precession of Earth’s axis, historically called the “precession of the equinoxes,” has a period of about 25,700 years. Further, there are added complications, such as nutation, due to factors we have ignored (such as eccentricity of the orbits of the moon and Earth, the tilt of the moon’s orbit relative to the ecliptic, the time-varying nature of the torque, and so on). See here for a deeper treatment.


Physics Friday 63

March 13, 2009

[Part 2 of ?]

In the previous part, we introduced a spinning oblate spheroid, and showed that a distant point mass M at displacement d will exert a torque on the spheroid that can be approximated as
. Now, suppose that d is in the xy-plane of the inertial frame. Further, suppose that our point mass is orbiting our spheroid in a circular orbit of angular frequency Ω (or that our spheroid is orbiting our point mass in a circular orbit of angular frequency Ω; the model will turn out the same). Then in the inertial frame. Suppose then at a time t the x and x’ axes coincide. Then, if we let the angle between z and z’ be θ0, we see that d has components in the body coordinates of .

Now, if the density of our spheroid is sufficiently symmetric about its axis, then we will have Ix’=Iy’, and the moment of inertia tensor in the body coordinates will be . Using our results from part one, we find the torque in this situation;
Now, we presently have and , so we can rewrite the above in a way independent or our choice of x’ and y’ axes:
giving torque:

Supposing that this torque is small enough that any precession produced is of frequency much slower than Ω, we can then average the torque over time; recalling the time average of trigonometric functions and their products, we get average torque

Recalling that our object has angular momentum along the z’ axis, we see then that our average torque is perpendicular to our angular momentum, and to our z axis (as it is along the cross product ). Thus, as in here, we have precession of our spheroid’s rotation about the z axis (so θ0 is constant), and from our previous work on torque-driven precession, we see that the precession has angular frequency
(The negative sign indicates that the direction here is opposite the sense of the rotation ωz.)
Now, supposing our spheroid has total mass ME, then Kepler’s third law for our spheroid-point mass orbit tells us that the period T of the orbit is . Since , we find , which lets us rewrite the precession frequency in terms of the orbital frequency and the ratio of the masses :

Physics Friday 62

March 6, 2009

[Part 1 of ?]

Let us consider a rotating oblate spheroid with equatorial radius a and polar semi-axis c. We choose an inertial frame (x,y,z) and body coordinate system (x’,y’,z’), both with origin at the ellipsoid’s center of mass. Suppose we have a rotation about a principal axis with angular momentum , where Iz’ is the moment of inertia along the z’ principal axis and is the z’ unit vector).

Now, let us put a point mass M at a displacement d from the center of our ellipsoid. Then a mass element of our ellipsoid at position r ( is the density at that point in the object, not necessarily uniform). experiences a gravitational force due to the mass M of

Integrating the torque element due to this force over the volume V of the ellipsoid gives a total torque of

Now, suppose our point mass is far from our body, compared to its size, so that , where and . Now using
, we see that approximating to first order in r/d,
, so we can in this situation approximate our torque by:

Now, to find the first integral on the right, we note that
For a an object in volumeV of total mass m and density , the center of mass has position (vector)

(see here); as our origin is the center of mass, we thus see that the integral , and the first integral in our approximation is the zero vector, and

Consider the moment of inertia tensior I of our object (see here):
, where E3 is the 3×3 identity matrix and is the outer product of r with itself. Applying the tensor to d, we use and to get
Next, consider the cross product of d and this vector:
which is our remaining integral in the torque approximation:

Physics Friday 56

January 23, 2009

Consider a rigid spherical body of density ρ. Now, imagine a test mass m located inside this body at a distance r from the center of the sphere. It will experience a gravitational force , with the negative sign indicating a force opposite in direction to the displacement r from the sphere center (an inward force). We see the force varies linearly with r. Thus, the gradient of the gravity is
, a constant. Here, the negative sign indicates that the gradient of the sphere’s gravitational self-attraction compresses the object.

Now, let us consider a mass M with its center of gravity a distance d from the center of our sphere. Let us consider a displacement r from the center of the sphere on the line between the centers of the sphere and M, with positive r being toward the mass M. Then the gravitational force on our test mass due to our external mass is
where the positive sign indicates the force is in the direction of positive r.
This, in turn, has a gradient

Here, the positive sign indicates that this tidal gradient puts the object under tension.

Note here that the tidal stretching increases as one approaches the mass M, while the self-attraction compression is constant. In fact, the sum of the two gradients becomes positive, and the tidal force dominates, when

Where is the radius of a sphere of density ρ with a mass 2M.

Thus, note that if d<deff, we see that the tidal forces overcome the self-attraction down to the center of the sphere, and thus a stable rigid body held together by gravitational self-attraction alone cannot exist. If our body of mass M is a sphere of radius R, it has density . Thus, writing deff in terms of ρM and R,

A real body would also have a material tensile strength to help hold it together, and also would deform from spherical into tidal bulges under the tidal force; however, there still remains a similar limit, inside which a self-gravitating celestial body (our sphere) orbitting a larger body (the mass M) will disintegrate under the tidal forces. This is called the Roche Limit.

Physics Friday 42

October 17, 2008

Last week, we considered a thin ring of radius a and mass M, and the energy of a mass m on the perpendicular axis at a distance z from the center of the ring. We noted that the equilibrium position of the mass m is the center of the ring, and that, at least with regards to motion perpendicular to the plane of the ring, the mass will be attracted toward this point.
Now, let us consider motion within the plane of the ring; in particular, is the equilibrium point at the ring center stable or unstable?

As before, we calculate the gravitational potential energy. Here, due to symmetry, the only element of the position of mass m relative to the ring that will affect the energy is the distance ρ from the center of the ring. Note that we are considering ρ<a.

Let us choose our origin to be the center of the ring, and let us choose an angular coordinate θ so that the mass is on the ray θ=0. Note that a small segment of the ring of arc would have mass . For such a portion at a coordinate angle θ, the distance between that piece and the mass is given by the law of cosines:

Thus the gravitational potential energy due to this portion of the ring is

And we integrate over all angles to find the energy:

This, however, is an elliptic integral, and cannot be done analytically.
Remember, however, we are interested in stability of the origin, and thus are considering ρa. In particular, to obtain stability, we need to know , which means we need only consider to second order in .
Our integrand is

to second order in x is
And so our integrand is, to second order,
Integrating over 0≤θ≤2π, the terms independent of theta are simply multiplied by 2π, while the and terms integrate to zero; thus, we have:

This gives at the center, as expected, and shows us that the potential energy decreases as the mass moves from the center of the ring, and thus the center is an unstable equilibrum. (The Ringworld is unstable!)

Physics Friday 25

June 20, 2008

Let us consider two small, identical masses, each of mass m, attached by a rigid, massless rod, so that their centers of mass are separated by the fixed distance 2d. We then place this “dumbbell” so that it’s center is a distance R>d from the center of mass of an object of mass M, and so that the line connecting the identical masses forms an angle θ with the line connecting the center of the dumbbell to the mass M. Treating the three masses as point masses, what is the gravitational potential energy of this arrangement, how does it vary with θ, and what does that imply physically?
The Answer: