

Lesson 1: Motion Characteristics for Circular MotionThe Centripetal Force Requirement Mathematics of Circular Motion 
Lesson 4: Planetary and Satellite MotionMathematics of Satellite MotionThe motion of objects are governed by Newton's laws. The same simple laws which govern the motion of objects on earth also extend to the heavens to govern the motion of planets, moons, and other satellites. The mathematics which describes a satellite's motion are the same mathematics presented for circular motion in Lesson 1. In this part of Lesson 4, we will be concerned with the variety of mathematical equations which describe the motion of satellites. Consider a satellite with mass M_{sat} orbiting a central body with a mass of mass M_{Central}. The central body could be a planet, the sun or some other large mass capable of causing sufficient acceleration on a less massive nearby object. If the satellite moves in circular motion, then the net centripetal force acting upon this orbiting satellite is given by the relationship
This net centripetal force is the result of the gravitational force which attracts the satellite towards the central body and can be represented as Since F_{grav} = F_{net}, the above expressions for centripetal force and gravitational force can be set equal to each other. Thus, Observe that the mass of the satellite is present on both sides of the equation; thus it can be canceled by dividing through by M_{sat}. Then both sides of the equation can be multiplied by R, leaving the following equation. Taking the square root of each side, leaves the following equation for the velocity of a satellite moving about a central body in circular motion where G is 6.673 x 10^{11} N•m^{2}/kg^{2}, M_{central} is the mass of the central body about which the satellite orbits, and R is the radius of orbit for the satellite. Similar reasoning can be used to determine an equation for the acceleration of our satellite that is expressed in terms of masses and radius of orbit. The acceleration value of a satellite is equal to the acceleration of gravity of the satellite at whatever location which it is orbiting. In Lesson 3, the equation for the acceleration of gravity was given as Thus, the acceleration of a satellite in circular motion about some central body is given by the following equation where G is 6.673 x 10^{11} N•m^{2}/kg^{2}, M_{central} is the mass of the central body about which the satellite orbits, and R is the average radius of orbit for the satellite. The final equation which is useful in describing the motion of satellites is Newton's form of Kepler's third law. Since the logic behind the development of the equation has been presented elsewhere, only the equation will be presented here. The period of a satellite (T) and the mean distance from the central body (R) are related by the following equation: where T is the period of the satellite, R is the average radius of orbit for the satellite (distance from center of central planet), and G is 6.673 x 10^{11} N•m^{2}/kg^{2}.
There is an important concept evident in all three of these equations  the period, speed and the acceleration of an orbiting satellite are not dependent upon the mass of the satellite. None of these three equations has the variable M_{satellite} in them. The period, speed and acceleration of a satellite is only dependent upon the radius of orbit and the mass of the central body which the satellite is orbiting. Just as in the case of the motion of projectiles on earth, the mass of the projectile has no affect upon the acceleration towards the earth and the speed at any instant. When air resistance is negligible and only gravity is present, the mass of the moving object becomes a nonfactor. Such is the case of orbiting satellites.
To illustrate the usefulness of the above equations, consider the following practice problems.
Like most problems in physics, this problem begins by identifying known and unknown information and selecting the appropriate equation capable of solving for the unknown. For this problem, the knowns and unknowns are listed below.
Note that the radius of a satellite's orbit can be found from the knowledge of the earth's radius and the height of the satellite above the earth. As shown in the diagram at the right, the radius of orbit for a satellite is equal to the sum of the earth's radius and the height above the earth. These two quantities can be added to yield the orbital radius. In this problem, the 100 km must first be converted to 100 000 m before being added to the radius of the earth. The equations needed to determine the unknown are listed above. We will begin by determining the orbital speed of the satellite using the following equation: The substitution and solution are as follows: v = 7.85 x 10^{3} m/s The acceleration can be found from either one of the following equations:
Equation (1) was derived above. Equation (2) is a general equation for circular motion. Either equation can be used to calculate the acceleration. The use of equation (1) will be demonstrated here. a = (6.673 x 10^{11} N m^{2}/kg^{2}) • (5.98 x 10^{24} kg) / (6.47 x 10^{6} m)^{2} a = 9.53 m/s^{2} Observe that this acceleration is slightly less than the 9.8 m/s^{2} value expected on earth's surface. As discussed in Lesson 3, the increased distance from the center of the earth lowers the value of g. Finally, the period can be calculated using the following equation: The equation can be rearranged to the following form The substitution and solution are as follows: T = 5176 s = 1.44 hrs
Like Practice Problem #2, this problem begins by identifying known and unknown values. These are shown below.
The radius of orbit can be calculated using the following equation: The equation can be rearranged to the following form The substitution and solution are as follows: R^{3} = 5.58 x 10^{25} m^{3} By taking the cube root of 5.58 x 10^{25} m^{3}, the radius can be determined as follows:
The orbital speed of the satellite can be computed from either of the following equations:
Equation (1) was derived above. Equation (2) is a general equation for circular motion. Either equation can be used to calculate the orbital speed; the use of equation (1) will be demonstrated here. The substitution of values into this equation and solution are as follows: v = 1.02 x 10^{3} m/s
Just as in the previous problem, the solution begins by the identification of the known and unknown values. This is shown below.
The unknown in this problem is the height (h) of the satellite above the surface of the earth. Yet there is no equation with the variable h. The solution then involves first finding the radius of orbit and using this R value and the R of the earth to find the height of the satellite above the earth. As shown in the diagram at the right, the radius of orbit for a satellite is equal to the sum of the earth's radius and the height above the earth. The radius of orbit can be found using the following equation: The equation can be rearranged to the following form The substitution and solution are as follows: R^{3} = 7.54 x 10^{22} m^{3} By taking the cube root of 7.54 x 10^{22} m^{3}, the radius can be determined ro be The radius of orbit indicates the distance which the satellite is from the center of the earth. Now that the radius of orbit has been found, the height above the earth can be calculated. Since the earth's surface is 6.37 x 10^{6} m from its center (that's the radius of the earth), the satellite must be a height of above the surface of the earth. So the height of the satellite is 3.59 x 10^{7} m.

Sun 





a. T^{2}/R^{3} for planets about sunb. T^{2}/R^{3} for the moon about Earth
c. T^{2}/R^{3} for moons about Saturn
3. One of Saturn's moons is named Mimas. The mean orbital distance of Mimas is 1.87 x 10^{8} m. The mean orbital period of Mimas is approximately 23 hours (8.28x10^{4} s). Use this information to estimate a mass for the planet Saturn.
4. Consider a satellite which is in a low orbit about the Earth at an altitude of 220 km above Earth's surface. Determine the orbital speed of this satellite. Use the information given below.
M_{earth} = 5.98 x 10^{24} kg R_{earth} = 6.37 x 10^{6} m 
5. Suppose the Space Shuttle is in orbit about the earth at 400 km above its surface. Use the information given in the previous question to determine the orbital speed and the orbital period of the Space Shuttle.
Lesson 4: Planetary and Satellite Motion
 Kepler's Three Laws
 Circular Motion Principles for Satellites
 Mathematics of Satellite Motion
 Weightlessness in Orbit
 Energy Relationships for Satellites
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19962007