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Lecture Notes
Updated Friday, 27-Jan-2006 10:42:31 EST



       Isaac Newton (ca 1665) discovered the Law of Universal Gravitation

       Every two objects in the universe attract each other with a force (gravity) proportional to the product of their masses and inversely proportional to the square of the distance between them.


       r = distance between the two masses (if they are particles)

       r = distance between their centers (if they have significant size, i.e., one mass is a planet)

       G = universal gravitational constant

= 6.67 10-11 N m2/kg2

       Example (gravitational force acting on mass m2)

RE = radius of Earth = 6378 km

h = altitude (distance from Earth's surface)

r = RE + h

mE = mass of Earth

       Weight (w) of an object is the gravitational force acting on it, so w = FGRAV

       At altitude h, an object with mass m2 has weight

       At Earth's surface, the same object has weight

so by simple algebra

       If an object weighs 1000 N. at Earth's surface, then at an altitude of 10 km, it will weigh

       Even at very high altitudes, this formula still applies. So if h = 200 km (typical altitude for Space Shuttle operations) and the object is an orbiting satellite, its weight is

A satellite that weighs 1000 lb on Earth will then weigh 910 lb at that 200 miles altitude. So why do people say that objects and people are weightless in space?

       Objects (including people) are not really weightless; they're falling.

       But if they're falling, why don't they crash into the Earth? (same question Newton asked about the Moon)

       An object in orbit is both falling and moving forward, and thus travelling on a curved path. But Earth's surface is also curved. So if object moving forward fast enough, it never gets any closer to Earth because the ground curves away from the falling path. (Both the textbook and the film The Apple and the Moon explain how Newton reasoned this out in terms of firing a series of progressively faster cannonballs.)

       Gravity plays a key role in orbits. If there were no gravity, then Newton's 1st law says that a satellite would just move on a straight path (i.e., it would just shoot out into deep space)


Same effect when you swing a ball on the end of a string . As long as you provide the tension in the string, the ball continues to travel in a circular path; when you release the string, the ball flies away.

Replace the tension in the string with gravity, replace yourself with the Earth and the ball with a satellite, and you have an orbiting spacecraft.



       If satellite has just the right speed at a particular altitude, it will move in a perfectly circular orbit.

       Gravity provides the centripetal force necessary for the satellite to move on a circular orbit with radius r



       So the satellite must have the exact speed

in order to maintain a circular orbit with radius r.

       Example -- if a satellite is in a circular orbit with altitude h = 1000 km, how fast is it moving?

       First, calculate the radius of the orbit. r = h + RE = 1000 + 6378 km = 7378 km

       Then calculate the speed

       Since G and Earth's mass mE are constants, we usually multiply them together first,



       If its speed is less than or greater than the necessary circular speed, then it will move on an elliptical orbit.

       Ellipse is an elongated circle. Its eccentricity (e) is a measure of the amount of elongation. If e = 0, the ellipse becomes a circle. If e = 0.75 (as in the figure above), the elongation is quite evident.

       Perigee = point on elliptical orbit closes to Earth

       rp = distance from Earth's center to perigee

       hp = altitude of perigee (height above Earth's surface)

       Apogee = point on elliptical orbit farthest from Earth

       ra = distance from Earth's center to apogee

       ha = altitude of apogee (height above Earth's surface)

       Eccentricity determined by

       NASA usually reports Space Shuttle orbits in terms of ha and hp. So a Shuttle orbit of 230 390 km has hp = 230 km, ha = 390 km, which means that e = 0.012 (nearly circular!)

       Orbital period = time for satellite to make one complete trip (one revolution) around the orbit. Calculate the period using

       for circular orbits: (for Earth, GM = 3.986 105 km3/s2)

       for elliptical orbits:


       Additional examples for orbit problems.


       NOTE: There is a substantial amount of additional material on orbits in the nine-page handout provided in class. Those handouts are available here in two parts (pdf files):   part 1   part 2

Copyright 1998, Robert G. Melton

Updated Friday, 27-Jan-2006 10:42:31 EST