Class Notes Physics I, 11/06/98

Newton's Law of Universal Gravitation; Orbital and Escape Velocities

Symbolic Expression of Universal Gravitation:  The Flux Picture and Newton's Formulation

• The first shows how the total gravitational flux, or total gravitational influence of an object, is spread over larger and larger concentric spheres, with the result that the gravitational field strength decreases according to the inverse-square law we have already seen in the context of the gravitational field of Earth.
• The second relates the mutual attractive force between two masses to the masses and the distance between their centers of mass.

• This flux is very much analogous to the flux of a light source.

• 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.

• For a gravitational mass, the total flux, 4 `pi G m, is analogous to the 100-watt flux of the light source.

• 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).

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.

• Gravitational influence can also be thought of as traveling along straight lines from the source.
• If the source is spherically symmetric, the symmetry of the situation ensures that these lines are radial from the center of the source.

• The force acts in the direction of the straight line between the centers of mass.
• The magnitude of the force is F = G m1 m2 / r^2, as depicted below.

• The total flux of the 100 gram mass is found to be approximately 8.4 * 10 ^ -11 N m^2 / kg, as shown below.
• In the figure we note also that the area of the sphere centered at the 100 gram mass and containing the 70 gram mass on its surface has radius 3 cm and therefore total surface area .0113 m^2.

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.

• It follows from Newton's second law that the force experienced by this mass is F = m a = 5.2 * 10^-10 Newtons.

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 would have obtained the same result as above, as you should verify.

(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.)

• Using this formulation we need not think of spheres and flux.
• We think only of the line between the two centers of mass, and recall that the force must be a force of attraction for both objects.

• It is stated below that if the 100 gram mass is fixed, with a 70 gram mass free, the 70 gram mass will therefore accelerate at 7.4 * 10^-9 m/s^2, as obtained from Newton's second law.
• This is a true statement, but it is misleading.

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 statement is true whether or not the 100 gram mass is fixed.
• If it is not fixed, then the masses will approach their common center of mass and the distance between the masses will change more quickly that if one of the masses is fixed.
• The attraction between the masses and the resulting acceleration of the 70 gram mass at the instant when the masses are three centimeters apart is the same in either case.

• This means that the gravitational flux through a sphere of radius three centimeters, equal to the product of field strength and area, must be 8.4 * 10^-11 m^3 / s^2, as indicated below.
• Thus we see that from Newton's formulation we obtain the same total gravitational flux as from the flux formulation.
• This should not be surprising, since the flux formulation came from Newton's formulation.

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.

http://youtu.be/R5o-uqGXxAY

Deriving the flux picture from Newton's formulation

• We begin by considering to masses m1 and m2, with their centers of mass separated by distance r.
• We will calculate the total gravitational flux through this sphere and show that it is indeed equal to 4 `pi G m.

• total flux = gravitational field strength * area = G m1 / r^2 * 4 `pi r^2 = 4 `pi G m1.

http://youtu.be/DUu5BImvk_A

Orbital Velocity and Escape Velocity

• At orbital radius r, a satellite of mass m will be at distance r from the center of the planet and will therefore experience a gravitational acceleration g = F / m = [ G M m / r^2] / m = G M / r^2.
• Since this gravitational acceleration, as we have seen before, provides the centripetal acceleration necessary to hold the satellite in a circular orbit, it must be equal to v^2 / r.
• We express the equality of these two expressions for the gravitational acceleration by

If we solve this equation for v, we obtain v = `sqrt(G M / r). 

• This is the general relationship among planetary mass, orbital radius and orbital velocity for any object whose mass is much less than M.

http://youtu.be/0c9mDDC9CFA

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. 

• As before, we obtain the work by summing up the force * distance contributions from small distance intervals over which force doesn't vary much. 
• The force function is G M m / r^2, so we sum contributions of form F(r) `dr = G M m / r^2 * `dr.

• 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).

• This work is the potential energy change from the Earth's surface to distance r from the center. 
• It is consistent with the expression we derived before for this quantity.

We generalize this idea easily to an arbitrary planet with radius R. 

• We find that the work done is now G M m (1/R - 1/r), where r is the distance from planet center to which we 'raise' the mass m.
• As we increase the distance r, the quantity 1/r shrinks. 
• As r continues to increase, 1/r shrinks to near zero. 
• As r approaches infinity, 1/r approaches zero, and the work required approached G M m / 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. 

• This quantity G M m / R is therefore called the 'escape energy'.

The associated velocity is the 'escape velocity'. 

• If we let vEsc stand for escape velocity, we have KE = G M m / R, so 1/2 m vEsc^2 = G M m / r and we easily solve for v.  
• We obtain escape velocity vEsc = `sqrt(2 G M / r).

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). 

• Comparing this with escape velocity, we see that vEsc = `sqrt(2) * vOrb. 
• For example, for low-Earth orbit, orbital velocity is about 8000 m/s and escape velocity is about 11,600 m/s (check these figures for yourself)
• The ratio vEsc / vOrb is `sqrt(2).
• This ratio applies to any spherically symmetric planet, ignoring atmospheric resistance.

http://youtu.be/2rxyqsg1D0c