Www.ciese.org



[pic]

Content Materials

Linear Motion – Velocity vs. Time

Real World Data Collection and Analysis

Your next task is to locate a satellite orbiting the earth, collect its motion data, and model its motion using a spread sheet in addition to the other representational tools you’ve been exploring.

Go to one of these sites and record the satellite’s latitude, longitude, altitude, and time for a period of 5 minutes (both pages automatically update every minute)

Hubble Space Telescope

International Space Station

Chandra X-Ray Observatory

Here’s an example of what the Space Station page looks like:

| |

|Tracking |Sighting |Other |

By the way, positive latitudes are North and positive longitudes are East. So, extracting the important (for us) information from these data, here are some representative data:

|Time (hr:min) |Latitude ( º ) |Longitude ( º ) |Altitude (km) |

|17:33 |33.9 S |11.5 E |367.4 |

|17:34 |36.5 S |14.9 E |368.4 |

|17:35 |39.0 S |18.5 E |369.5 |

|17:36 |41.2 S |22.4 E |370.5 |

|17:37 |43.1 S |26.4 E |371.4 |

Now, go to this website to figure out how far the space station traveled during each interval. By the way, these measurements are ground-based. The actual distance traveled by the station would actually be roughly 6% further. How Far Is It?

|Time |Latitude |Longitude |Interval |Total |

|(min) |(degrees) |(degrees) |Distance (km) |Distance (km) |

|0 |33.9 S |11.5 E |0 |0 |

|1 |36.5 S |14.9 E |423 |423 |

|2 |39.0 S |18.5 E |421 |844 |

|3 |41.2 S |22.4 E |413 |1254 |

|4 |43.1 S |26.4 E |392 |1646 |

Using Microsoft Excel, here is an X-Y graph of the time and total distance data:

[pic]

As you can see, the slope of this line is fairly constant, indicating that the space station moved at fairly constant speed. If we select a data point on the graph, add a trendline, and display the equation for the line, we get:

[pic]

The slope of this line indicates the average speed of the International Space Station is 415 km/min. This is equivalent to roughly 7000 m/s, 15,500 mph or 4.3 mi/s of ground speed. At 370 km, the speed would be closer to 7500 m/s. We are ignoring the fact that the space station (and the Hubble) change altitude somewhat during these few minutes. However, it only changes by roughly 1%, so that’s a pretty safe assumption. If you choose to use the Chandra X-Ray Observatory, you’ll notice the altitude is much greater than the space station (~50x) and the altitude changes a great deal (more than 50%) during each orbit.

As an interesting side note, we can calculate the speed required for the space station to maintain an orbit at that altitude using the relationship [pic]where G = 6.67x10-11 N*m2/kg2, Me=6.0x1024kg, Re=6.4x106m and the altitude h=3.7x105m. This gives us a value closer to 7700 m/s, but that’s pretty close (less than 3% difference).

Use the site, How Far Is It?, to look at the average direction the station travels during this 4 minute interval:

|Time (min) |Latitude (degrees) |Longitude (degrees) |Direction (degrees) |

|0 |33.9 S |11.5 E | |

|4 |43.1 S |26.4 E |133 |

Due north is 0º so 133º is almost due South-East (43º South of East). For this data, the Space Station was flying over the coast near Cape Town:

The space station doesn’t always travel South East, does it? On the “other side” of the world, it would appear to be traveling North West. Furthermore, the Earth rotates at a rate of 15º/hr. So, we use the concept of inclination to describe the orbit of satellites. The space station has an inclination of ~52º (measured from the equator when the satellite is “rising.”)

Velocity is a vector that combines the speed and direction something is traveling. So, the average velocity of the space station during this interval could be described as 7700 m/s South East. There are a variety of other ways to describe velocity vectors, but this will suffice for our purposes.

Now, let’s figure out how long it would take the space station to get to the opposite side of the earth. To determine the new latitude, subtract the current latitude from 90º and change the direction. The new longitude is determined similarly except you subtract from 180º.

| |Old |New |How? |

|Latitude |33.9º S |56.1º N |90º -33.9º = 56.1º, S → N |

|Longitude |11.5º E |168.5º W |180º-11.5º = 168.5º, E → W |

Use this site to figure out where in the world you are:

The next page to appear states “This confluence has not been visited, or indexed.

Type: Water: 111 km (69 mi.) from land.” There is a map link on that page that takes you to MapQuest (click on “street” or “world” in the menu that shows:

([?] maps: street world confnav)

Antipode: 56°S 12°E.

By playing with the zoom, you can see that the point in question is just off the coast of Alaska.

Another visit to How Far Is It? shows that the two points are 17,600 km from each other.

At a ground velocity of 415 km/min, it would take around 42 minutes and 20 seconds to make half a trip around the world. The space station actually has a period of 92 minutes (roughly 8% error).

Pretty cool, huh?

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download