Symbolic Expression of Universal Gravitation: The Flux Picture and Newton's Formulation
We have investigated how the inverse-square nature of gravitation influences gravitational field strength in the vicinity of Earth, orbital and escape velocities for Earth, and kinetic and potential energy changes between circular orbits of different orbital radii.
We will now generalize our picture in order to find expressions for the gravitational field strength in the vicinity of any mass, and for the mutual attraction of any two masses.
We will use two equivalent formulations.
The second is Newton's original formulation of the law of universal gravitation, which was obtained as a mathematical model explaining the observed orbits of planets and the Moon, while the first is a consequence of Newton's formulation.
The two pictures are equivalent, in that either can be derived from the other, as we will see.
If we have a mass m, then we can think of that mass having a total gravitational flux equal to 4 `pi G m, where G is the universal gravitational constant, equal to approximately 6.7 * 10^-11 N m^2 / kg^2 (check the table in your text, inside the front cover, to verify this unvalidated value).
- The flux of a light source can be expressed in watts of power.
- For example, from a 100-watt point source of light, the energy produced at any instant spreads out uniformly in all directions, so that the energy will at any instant be uniformly distributed over a sphere centered at the source.
- The radius of the sphere increases at the speed of light.
- The entire 100 watts will therefore pass continuously through any such sphere which the light has had time to reach.
- However, for an increasingly large sphere the 100-watt flux of the light is spread out over the increasingly large surface area of the sphere, and hence the light provides less and less illumination.
- We see that the illumination of the source is therefore inversely proportional to the area, and hence to the square of the radius, of the sphere.
- More specifically, we define illumination as the number of watts per square meter falling on the surface of a sphere
- For a sphere of radius r, its area will be 4 `pi r^2 and the illumination at the surface of the sphere will be 100 watts / 4 `pi r^2
- This illumination will be measured in watts / m^2.
- It also spreads outward at the speed of light, and the resulting effect at any point is equal to the flux per unit area.
- Since the units of G * m are N m^2 / kg^2 * m = N m^2 / kg = (kg m / s^2) * m^2 / kg = m^3 / s^2, the units of flux per unit area are (m^3 / s^2) / (m^2) = m / s^2.
- These are the units of gravitational field strength, or gravitational acceleration.
- This gravitational acceleration is always directed toward the center of mass of the planet (or of whatever the object with mass m might happen to be).
The figure below depicts a mass m and a sphere of radius r around the mass (the arrow indicating the radius should extend from the center of the mass, not from its surface as it appears in the figure below). The definition of gravitational flux is
In the previous example of light flux, light is always traveling along straight lines from the source, and is hence always perpendicular to the surface of the sphere.
Newton's formulation states that if masses m1 and m2 are positioned so that their centers of mass are separated by distance r, then there is a mutual gravitational force between them, with the following characteristics:
This formulation is completely equivalent to the flux formulation.
We will use the flux formulation to find the force exerted by a 100 gram mass on a 70 gram mass, when the centers of mass are separated by 3 cm.
We find therefore that the gravitational field strength, equal to the total flux divided by the area of the sphere, is 7.4 * 10^-9 m/s^2, so that the acceleration of the 70 gram mass toward the 100 gram mass is a = 7.4 * 10^-9 m / s^2.
We could have calculated the force exerted on the 100 gram mass by the 70 gram mass in a similar manner, using the acceleration corresponding to the flux of the 70 gram mass spread out over a sphere of radius 3 cm to obtain this force.
(We should note also that the total momentum of the two masses, as they approach each other under the influence of their mutual gravitational attraction, must be conserved.
- This follows from the fact that equal and opposite forces are acting on the object.
- It follows that whenever the time interval over which these forces act, the two forces will deliver equal and opposite impulses, which will result in equal and opposite momentum changes.
- The total momentum of the system will therefore remain unchanged.)
We can calculate the force F between the masses Newton's formulation F = G m1 m2 / r^2.
The calculation is straightforward. We merely substitute the masses m1 = .07 kg and m2 = .1 kg, and the separation r = .03 m of the centers of mass into the formula and we obtain the same force as before, 5.2 * 10 ^ -10 Newtons.
Note carefully that in the figure below, it is not necessary for the 100 gram mass to be fixed in order for the statement to be true.
The preceding series of calculations showed us, among other things, that the gravitational field strength at a distance of three centimeters from the 100 gram mass must be 7.4 * 10 ^-9 m/s^2, since this is the acceleration experienced by an object at this distance.
note error in figure below: 7.4 * 10^-11 should read 7.4 * 10^-9 and 8.4 * 10^-9 should read 8.4 * 10^-11.
Deriving the flux picture from Newton's formulation
We now derive the flux formulation from Newton's formulation.
Since by Newton's formulation, the force of attraction is F = G m1 m2 / r^2, the acceleration of the mass m2 has magnitude
This gravitational acceleration is the gravitational field strength on the surface of the sphere of radius r centered at the center of mass of mass m1.
- total flux = gravitational field strength * area = G m1 / r^2 * 4 `pi r^2 = 4 `pi G m1.
This is the gravitational flux of m1, according to the flux picture. It has been derived without using the flux picture, using only Newton's Law of Universal Gravitation.
Orbital Velocity and Escape Velocity
We can now use either formulation we choose to obtain a formula for the velocity of a small object in a circular orbit around a planet of mass M.
- v^2 / r = G M / r^2.
If we solve this equation for v, we obtain v = `sqrt(G M / r).
We can also obtain the expression for the work required to 'raise' an object from the surface of Earth to distance r from the center.
- In the limit as the intervals `dr approach zero we get the integral shown below, integrated from the planetary radius rE to distance r from the center of the planet.
- The integral is straightforward for calculus students. Non-calculus students can simply use the result, which is
- work required from Earth's surface to distance r from center = G M m (1/rE - 1/r).
We generalize this idea easily to an arbitrary planet with radius R.
It follows that if we have KE equal to G M m / R at the surface of the planet (ignoring air resistance), we have the energy necessary to do the work required to 'raise' an object of mass m to infinite distance.
The associated velocity is the 'escape velocity'.
We note that for an orbit just above the surface of the planet, ignoring air resistance, orbital velocity must be vOrb = `sqrt(G M / R).
