1. Proportionality between the velocity V and radius r
In circular motion with a constant centripetal force.
This ppt is to derive the proportionality between V and r for the geosynchronous orbit laboratory.

3. Applying Newton’s 2nd Law
𝐹 =𝑚 𝑎
In this case the only force on the body is the force of gravity and the acceleration of the body is centripetal.
𝐹 𝐺 = 𝐹 𝐶

4. A centripetal force is required for circular motion
A centripetal force is required for circular motion. This is given by the equation:
The centripetal force is supplied by the force of gravity between the two bodies which is given by the equation:
m1 is the mass of the satellite, m2 is the mass of the object being orbited, G is the universal gravitational constant, and r is the radius of the circular motion. Fc represents the centripetal force and FG is the gravitational force, both measured in newtons, N.

5. Set the two forces equal to one another.
m1 = mass of the satellite
m2 = mass of object being orbited
V = velocity of the satellite
G = universal gravitational constant
r = radius of the circular motion
The equation for v shows that the velocity of the satellite DOES NOT depend on the mass of the satellite (as stated in the objective). The velocity v depends on the mass of the object being orbited and the radius r of the circular motion.
Mass of the satellite cancels out. The speed of the satellite depends on radius r and mass m of the object being orbited.

6. Orbital Velocity Solving for v Where: v = orbital velocity (m/s)
G = universal gravitational constant = 6.67 x 10-9
m = mass of the object being orbited (kg)
r = radius of circular motion (m)
As the mass of the object being orbited increases, the velocity also increases. Inversely, as the radius r increases, velocity v decreases. The farther away a satellite from the earth is, the slower it’s speed is.

7. Period of Satellite We know that
For a circular motion, the distance x travelled by the satellite in one revolution is the circumference of the circle. That would be
Symbol d can be used instead of x. Both represents distance in meters. The symbol t is the time it takes to complete one revolution measured in seconds (s).

8. Solving for t or But we also know that This gives us
For the teacher, this could be an exercise too before you show the next slide. Have the students simplify the equation for t.
This gives us

9. Squaring both sides
We end up with orbital period t of a satellite. This is the time it takes for a satellite to complete one orbit.
All measurements use KMS system. Mass m is in kilogram (kg), distance x in meters (m), time t in seconds (s). G is the universal gravitational constant; G = 6.67 x N.m2/kg2. Other Physics books uses the symbol T for period.
Again, m here is the mass of the object being orbited NOT the mass of the satellite.

10. Quick Check # 1:
What is orbital velocity of the earth around the sun? The sun has a mass of 1.99 x 1030kg, the mean distance from the earth to the sun is 1.50 x 1011 m.
This one can be used as an example, classwork, or homework. Work on the problem before showing them the answer on the next slide.

11. Quick Check #1 Solution
If needed, this is the solution. Most kids in your class know how to plug and chug numbers to any equation.

12. Quick Check #2
A satellite is in a low earth orbit, some 250 km above the earth's surface. rearth is 6.37 x 106 m and mearth = x 1024 kg. Find the period of the satellite in minutes.
Again, this could an example or classwork. Solution on the next slide. Have the kids try it before showing the answer.

13. Quick Check #2 Solution
r is the distance from the center of the earth to the satellite; thus
r = rearth km = 6.37 x 106 m x 103 m