AVIATION TECHNOLOGY: Gravity

Newton's Law of Gravity

Spherical Mass Objects:

Between any two objects that have mass, there exists an attractive gravitational force acting on each object separately. These two forces have the same magnitude but are in opposite directions.
When the two objects have spherical mass distributions (or are point masses), then the gravitational force acts along a line connecting the center of mass of each of the two objects.

F_G	= Magnitude of the Gravitational Force (N)
m₁	= Mass of one body (kg)
m₂	= Mass of a second body (kg)
G	= 6.672x10^-11N.m²/kg² (Universal Gravitational Constant)
r	= Distance between the centers of mass of the two objects (m)

The gravitational force does not depend upon the size or the density of the two bodies, provided their mass distribution is spherical. Nor does it even depend upon the speed of motion of the two bodies. It only depends upon their separation and masses.
The origin is usually taken to the center of one mass or the other mass. By do so we know that the direction of the force exerted is radial towards the origin.

We call G a universal constant because we believe that it has the same value no matter where the two bodies are located either in the universe or in the age of the universe. Comparison of the Four Fandamential Forces It is interesting to note that the gravitational force is the weakest of the four forces yet it holds the universe together on its largest scale.

Non-Spherical Mass Distributions:
When a body has an irregular shape, we can use calculus to determine the gravitational force. The force between two infinitesimally small masses (dm₁ and dm₂₎ will obey the gravitational force law

We can integrate this equation over all the mass in the two bodies to determine the total gravitational force. In general, the gravitational force does not act between the center of mass of irregularly shaped objects.

Normally we do not have to worry about this because the gravitational force between to human size objects in negligible. Moreover, when objects get larger enough (like the size of a planet) gravity itself causes them to become round.

Gravitational Force inside a Spherical Mass Distribution:

Using the differential form above, Newton showed that if a mass object is located inside a spherical mass distribution then the gravitational force is only due to the mass in a sphere closer to the center than that of a small mass inside the sphere.
For example, if you dug a deep hole, then the gravity acting on a small mass at the bottom of the hole would only be due to the mass of the Earth below the bottom of the hole. The gravitational force of the mass of the Earth above the bottom of the hole cancels out, i.e. its net vector sum adds up to zero.

Acceleration of Gravity
At any distance r from a large circular body of mass M a smaller body of mass m will experience a gravitational force.

Applying Newton's Second (F_net = ma) to the small object we can determine its acceleration if no other force are acting on the small mass m but gravity (F_net = F_G).

Acceleration of Gravity Near the Surface of a Large Body
Assume that a large body of mass M and radius R has a radically uniform mass distribution. The gravitational force on a small mass m at height h above the surface of the large body can be expressed as,

If h << R, then

If we neglect air resistance, then this is the net force acting on the mass m. Applying Newton's Second Law, F_net = ma,

Because of its importance we usually give the acceleration of gravity due to the Earth a separate name,

* This equation does not depend upon the mass of the small object - the mass canceled out on both sides of the force equation. This proves that the acceleration due to gravity alone is the same on all objects independent of their mass, their size, their density, or even their speed.

* Local concentrations of mass in the Earth can change the magnitude of the acceleration of gravity in the third significant figure. In fact, it is often true that as you go up a mountain the vale of g gets larger not smaller as you would expect because mountains contain a larger concentration of denser rocks.

* Since the value of g is not a fixed constant, in this course we will normally use 9.80 m/s² when solving problems.

Acceleration of Gravity at the North Pole
The Earth's polar radius is 6356 km and its mass is 5.9736x10²⁴ kg.

Acceleration of Gravity at the Equator
The Earth's equatorial radius is 6378.1 km. Neglecting the Earth's rotation,

The Earth is also rotating so that the acceleration of gravity at the equator will be reduced by the centripetal acceleration. Since these are both along the same direction (radially) at the equator we don't have to subtract them vectorially - just numerically,

Since the Earth makes one revolution per day, w = 2p rev/day

Weight as Function of Latitude:
If you are at some latitude F above the equator then your period of rotation will still be one day, but your distance from the axis of rotation will be reduced to R cos(F). This reduces the effect of the centripetal acceleration. Moreover the Earth is not a perfect sphere, so R will also change.

Assuming a spherical Earth we can determine the weight on a mass m at a latitude F. By applying the law of cosines to the force vectors we get

Where

For Bellingham WA, F = 48.78

_^o

, and the centripetal acceleration is a_c = .02223 m/s². This gives an effective gravity of

A value that the student's here a Western Washington University have consistently found when measuring the acceleration of gravity over the pasted 30 years. There are other factors that could effect the local value of g such as elevation above sea level and the local concentrations of denser rock. Gravitational Potential Energy
The Gravitational Potential Energy due to a spherical body of mass M at a distance r from the center of the body is defined to be

* The gravitational potential energy is zero at r = infinity. This is equivalent to choosing the base level of the potential energy to be located very far away from the body.

* One could choose the surface of the body as a base point, but then the potential energy at this reference point would not be zero.

* The gravitational PE is negative for all values of r less than infinity. This is can be visualized by looking at the large mass M as creating a gravitational energy well around it that goes down as you get closer to it.

Derivation of Gravitational Potential:
The gravitational potential energy is defined as the negative of the work it takes to move an object from some reference point to a height h above the surface of the Earth. Near the surface of the earth the force of gravity is nearly constant.

If we do not neglect the force of gravity's dependence upon distance r, we get the following integral to evaluate,

If we let r = R_E + y then dr = dy and

Thus we can see that the change in potential energy is the difference between two negative numbers which turns out to be a positive number.

Let us start with the definition of gravitational potential energy and show that it gives the same equation as above. Starting with

the change in potential is

Reference Level
Had we defined the base level for the potential energy to be the surface of the body, the potential energy could be defined as

Observe that the change in potential energy DPE would still be the same, except our definition of potential energy would also depend upon the size of the body as well as its mass. Moreover it will not be zero at the base level - the surface of the body.

Altitude Correction Factor
It is sometimes useful to use the body's surface as a base point.

Here g is the normal acceleration of gravity at the surface of the body. The term (1 - h/R_E)^-1 represents an altitude correction factor. If h is small compared to the radius of the Earth then this formula goes over to familiar mgh expression for potential energy.

Kepler's Three Laws of Planetary Motion

1. Law of Ellipses

The shape of the orbit of any body (be it an asteroid, planet, comet, or space craft) orbiting the Sun is an Ellipse with the Sun at one of the foci.

This law is a consequence of the fact that Newton's law of gravity varies as one over distance squared.

Importance:
This law can be used to determine a planet's location r at any position of the planet relative to the Sun.

r	= Distance of the planet from the Star (SI: m)
q	= Angular location of the Planet from measured from Perihelion (SI: rad)
a	= Semimajor Axis "Mean Size of the Orbit" (SI: m)
c	= Distance of the Star/Focus from Center (SI: m)
e	= Eccentricity "Measures the out-of-roundness of the Orbit" (dimensionaless ratio)
r_p	= Perihelion distance "Closest Approach to Star" (SI: m)
r_a	= Apohelion distance "Maximum Separation from Star" (SI: m)

The origin is located at the center of the primairy body - the Sun - for the above equations.

2. Law of Areas
* The line joining the planet to the Sun sweeps out equal areas in equal times.

This law is a consequence of the conservation of angular momentum applied to the orbiting body. (Derivation)

Importance:
This law can be used to determine a uniform passage of time. This method (called Ephemeris time) was for many years the basis of our most accurate time keeping method before the use of atomic clocks replaced it.

Conservation of Angular Momentum:
Since the gravitational force is always directed radially through the center of mass, the gravitational force can exert no torque on either body. Thus angular momentum is conserved. This means that the angular momentum of the orbiting object will always have the same value at any point in its orbit. (Derivation)

For the velocities see Orbital Velocities.

3. Harmonic Law
* The square of the sidereal orbital period of a body is proportional to the cube of the planets mean distance from the Sun.

This law is a consequence of the fact that the object is in a bound orbit about the sun that varies as one over r².

Importance:
Of its many uses, a primary one is the determination of the mass of a star or planet directly from information about the period and distance of an orbiting object.

P	= Sidereal Period of the planet "Planet Year" (SI: sec) the time to make one complete revolution relative to the stars
a	= Semimajor Axis of the orbit ( SI: m)
M	= Mass of the Star (SI: kg)

Simplified Version (Change of Units Equation):

For objects orbiting the Sun their distances are often measured in Astronomical Units (1 AU = the mean Earth-Sun distance = 1.49x10¹¹ m = the Earth's semimajor axis). Typical orbital periods are on the order of years. In a set of units that measures distance in AU, time in years, and the mass of the central star in solar masses (1.989x10³⁰ kg), the value of 4p²/G = 1.

Orbiting Satellite
A satellite's distance, velocity, and direction from Earth can be altered by dragging the satellite's location and velocity vector. Displayed is the resulting orbit about the Earth. Also calculated are the orbital parameters of the satellite's period, semimajor axis, eccentricity, perigee and apogee distance, circular satellite velocity, escape velocity, maximum height above the Earth's surface, and the total mechanical energy for 100 kg satellite.

Binary Star Systems
All the relations for Kepler's Law's assumed that the central star was massive compared to the orbiting object. In fact, any two objects will orbit about their common center of mass. Two body problems can be reduced to one object orbiting another provided:

1. We change to a frame of reference attached to the center of mass of one of the stars. Either one will work.

2. We replace this star's mass with the total mass of both stars.

3. We replace the mass of the orbiting body with the reduced mass of the system.

Note that when one of the star's masses becomes large relative to the other, the reduced mass m is approximately equal to the mass of the smaller body, and the total mass M is approximately equal to the mass of the larger object. The problem reverts to described in Kepler's law section.

Reduced Mass Frame:
* A body of mass m orbits a larger body of mass M.

Center of Mass Frame:
* Both stars orbit about their center-of-mass with the same eccentricity.

Each star has its own orbit with a different, but related semimajor axis.
The stars are always opposite each other about the center-of-mass. Both stars will reach peristron (closest approach) at same time.

Binary Star Orbits
The stable orbits of two stars about each other is displaced by specifying their masses, their initial separation, and the eccentricity of their orbits. The resulting action can also be observed from different frames of reference: star 1, star 2, center of mass, and a rocket moving at a constant speed. (IP 3.0 Simulation of Binary Star Orbits - The center of mass frame does work correctly in IP2.5)

Orbital Velocities and Conservation of Energy
When the conservation of energy is applied to an orbiting body we have

Bound Elliptical Orbits:
When an object is in a periodic orbit about the central body, it is said to be bound. In this case the mechanical energy will always be negative, but constant. The total ME is equal to (Derivation)

Thus

This is the general relationship for the velocity of any orbiting body as a function of its distance from the central body.
It is the total mechanical energy which determine the shape a bodies orbit - in particular the orbit's semimajor axis a.

Unbound Hyperbolic Orbits:
If the velocity of an object is increased enough so that the total ME becomes positive, then the object will move along a hyperbolic orbit. Such orbits are unbound in that the orbiting body will never return, and will have an eccentricity greater than one. The equations describing hyperbolic orbits are basically the same except for a few signs associated with a. See Conic Orbits Table)

Parabolic Escape Orbits:
If the total ME energy of a body is zero, then the object will be on an escape trajectory. Here the eccentricity of the orbit becomes one, and the semimajor axis goes to infinity. An elliptical orbit with an infinite semimajor axis of infinity is a parabola. See Escape Velocity.

Escape Velocities
If a object is launched straight up from the surface of a non-rotating body, it will rise to some maximum height, come to a halt, and then fall back. If object's speed is large enough, then it could go all the way to infinity before it comes to a halt. The speed needed to just stop at infinity is called the escape velocity.

Derivation of the Escape Velocity:
To find the velocity just needed to escape from the surface of a non-rotating spherical body, we use the conservation of energy.

* For any velocity larger than v_ethe object will also escape. It will slow down but never come to a stop. The object
will be in a hyperbolic orbit if v > v_e.

* The object could be launched at any angle and still escape from an airless body, provided there are no close by mountains in the object's path.

* If the body is rotating, then the speed needed to just escape will be less than v_esince the object already has some
speed. This assumes that the object is launched in such a way that the rotational velocity and the launch velocity are positively additive -- the object is launched in same direction as the Earth's rotation.

* The total mechanical energy of an escape orbit is always zero regardless of whether the object is launched from a rotating or non-rotating surface. Even if the object is in orbit about the body, the ME of any escape orbit still has to be zero.

Two Body Conic Orbits

The orbit of any body (be it a comet, planet, asteroid, or a space craft) about the Sun will have the same shape that one gets by the intersection of a plane with a cone. This is true in general, not just for a body orbiting the Sun.

If the plane intersects a vertical cone at an angle that is between 0^o and 90^o then the resulting line of the intersection is that of an ellipse - a bound elliptical orbit.
If the plane intersects the cone horizontally then the resulting orbits is that of a circle - a special case of an ellipse whose eccentricity is zero.
If the plane intersects the cone vertically (at 90^o) then the resulting orbit is that of a parabola - an escape velocity orbit.
If the plane intersects the cone at an angle greater than angle greater than 90^o then the resulting line of the intersection is that of a hyperbola - the orbit of a body that has a velocity greater that the escape velocity.
The shape of the orbit is determined by the Gravitational Mechanical Energy of the orbiting body. If ME <> then the orbit will be elliptical or circular. If ME = 0 then the body will will follow a parabolic escape orbit. If ME > 0 then the body will follow a hyperbolic path.