Module 5—Circular Motion

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Understanding the Acceleration of a Satellite

 

To maintain periodic, circular motion—such as that of an orbiting satellite—an inward acceleration must act at all times. The nature and origin of this acceleration is found in Newton’s universal law of gravitation. Recall the two definitions of weight derived in Module 4: Lesson 2: is as follows

 

 

Equating these two definitions of weight will yield a general expression for the magnitude of the acceleration acting on a satellite in terms of

Use the simulation to explore the nature of the acceleration acting on a satellite at various distances from Earth’s surface.

 

trajectory: the path of a moving body through space

This applet simulates the trajectory Weight and Orbits and gravitational force acting on a particle in Earth's gravitational field by varying its initial position and velocity. You can learn more about the simulation and how to use it by reading Show Me found at the top of the simulation screen.

 

Self-Check

 

SC 1. The distance between the centre of Earth and the satellite (r) is 6.37 × 106 m + 6.00 × 105 m = 6.97 × 106 m. Using the equations above, calculate the acceleration of a satellite that is orbiting Earth at an altitude of 600 km. Use the simulation to verify your answer. (Set the x position to 0 km and the y position to 600 km; then press “Enter.” Click “Data,” and look for the acceleration value in the data box.)

 

SC 2. What happens to the magnitude of the acceleration as the satellite moves farther away from Earth? Review this by clicking the “Display Vectors Selection Panel” button () and turning on the acceleration vector and then de-selecting the velocity vector.

 

 

Then click off the “Display Vectors Selection Panel.” Change the scale to 100 px = 5000 km to get a better view. Drag the satellite to various positions around Earth. What do you notice about the magnitude and direction of the acceleration vector?

 

SC 3. On the simulation, set the scale to 100 pix = 10 km. This is the highest zoom setting. (The satellite will be outside of the bounds of the window.) Set x = 0 km and y = 0 km to move the satellite back to the North Pole. Now drag the satellite around the display area, and observe the changes in the magnitude of the acceleration vector. Are the changes in magnitude larger or smaller near Earth's surface?

 

Check your work.
Self-Check Answers

 

SC 1.

 

Given

 

r = 6.97 × 106 m        
G = 6.67 × 10–11 N·m2/kg2 
mE = 5.98 × 1024 kg

 

Required

 

the acceleration of the satellite (g)

 

Analysis and Solution

 

The acceleration of the satellite is due to the force of gravity from Earth.

 

 

Paraphrase

 

The acceleration of the satellite is 8.19 m/s2.

 

SC 2. The magnitude of the acceleration vector decreases as it gets farther from Earth and increases as it gets nearer. The direction is always towards the centre of Earth.

 

SC 3. There is no noticeable change in the direction of the force and very little change in its magnitude.

 

 

How can the fact that there is so little change be explained? At this scale setting, the changes in distance that are possible within the window are only tiny fractions of Earth’s radius. Significant changes in the weight only occur when the distance from Earth’s surface changes by a considerable amount. In Newton’s universal law of gravitation, the distance of separation, r, is always measured from the centre of the mass. Therefore, Earth’s radius must always be added to a satellite’s altitude when applying Newton’s equations to describe the motion.