*See National Math Standards for this Challenge.*

**Q: How close does the spacecraft have to be before Tempel 1 looks like anything more than a white dot?**

In more scientific terms, this question involves the maximum distance away from the target at which the target can be resolved.

Answering this question requires knowledge of the size of the target (comet Tempel 1 in this case), a nifty mathematical tool called the small angle formula, as well as how big a pixel is for the imager.

What is a pixel? Pixel is short for "picture element". The more pixels a picture is composed of, the more "real" it looks, and the more details you can make out - we'd say it is more resolved. As far as a computer analyzing data is concerned, a pixel is just a block of color. Put enough tiny blocks of color together in the right order, and to your eye you've got a nice picture.

If the comet is far enough away from the spacecraft, it fits inside one of the imager's pixels, and the comet is resolved as nothing more than a white dot. As the spacecraft gets closer, the comet overlaps more and more pixels, and more detail can be seen, or smaller details resolved.

The small angle formula is a mathematical formula used in astronomy for determining the angular diameter of an object that will appear as a disk when viewed from another object. It only works well for small angles - less than 10 degrees - thus its name. One way to present this formula is in this form:

a | = | diameter |

206265 arcseconds | distance |

Where "a" is the angular diameter of the object, in arcseconds*****. The 206265 is a conversion factor that converts from radians to degrees, and then from degrees to arcseconds (it is equal to [360/2p] × 3600). "Diameter" is the object's true linear diameter, and "distance" is how far away the object is. These last two need to be in the same distance unit (both in meters, for example).

In the case of Deep Impact, we are told on the Instruments page for the High Resolution Instrument (HRI) that the size of a pixel will be 1.4 meters at a distance of 700 kilometers. The Medium Resolution Instrument (MRI) has a pixel size of 7 meters at a distance of 700 kilometers.

So, the questions you need to answer are...

- What is the angular diameter, in arcseconds, of a HRI pixel?
- What is the angular diameter, in arcseconds, of a MRI pixel?
- Given that the diameter of Tempel 1 is approximately 6 kilometers, how close (in kilometers) does the spacecraft have to be before the nucleus of the comet is bigger than 1 pixel on the HRI? How about for the MRI?
- How close will the comet have to be for its nucleus to cover 3 pixels (the minimum number to be considered more than random noise) on the HRI? How about the MRI?
- At the time of impact, the flyby spacecraft will be 234,641 km away from the comet (as we figured out in last month's Challenge). How many pixels across will the nucleus of the comet be on the HRI? How about the MRI?
- If the flyby spacecraft were not in shield mode at the time of its closest approach, when it's only around 508 km away from the comet (see last month's Challenge), how many pixels across would the nucleus of the comet be on the HRI? How about the MRI?
- The impactor spacecraft has a camera identical to the one in the flyby spacecraft's MRI. What is the closest the impactor spacecraft can be and still "see" the entire comet nucleus? The camera has an array that can effectively image 1008 x 1008 pixels.

*****What is an arcsecond? An arc-second is one 60^{th} (1/60) of an arcminute. An arcminute is one 60^{th} (1/60) of a degree. Thus there are 3600 arc-seconds in one degree.

**A: Each of these questions requires the use of the small angle formula, but in different ways:**

NOTE: Answers are presented unrounded first, then with correct significant digits in red.

*1. What is the angular diameter, in arcseconds, of a HRI pixel?*

Here we just need to plug in the size of the pixel given for "diameter" and "distance", then rearrange and solve for the angular diameter, a.

a = | diameter | × 206265 arcseconds |

distance | ||

a = | 1.4 m | × 206265 arcseconds = 0.41253 = 0.41 arcseconds |

700,000 m |

*2. What is the angular diameter, in arcseconds, of a MRI pixel?*

This one is solved just like number 1...

a = | diameter | × 206265 arcseconds |

distance | ||

a = | 7 m | × 206265 arcseconds = 0.41253 = 2 arcseconds |

700,000 m |

*3. Given that the diameter of Tempel 1 is approximately 6 kilometers, how close (in kilometers) does the spacecraft have to be before the nucleus of the comet is bigger than 1 pixel on the HRI? How about for the MRI?*

Here we use the small angle formula to solve for "distance" (rearranging the formula using cross-multiplication). The "diameter" will be the size of the comet, and a will be the angular diameter of a pixel on the instrument.

For the HRI:

distance = | 206265 arcseconds | × diameter |

a | ||

distance = | 206265 arcseconds | × 6 km = 2,996,586 = 3,000,000 km |

0.41253 arcseconds |

For the MRI:

distance = | 206265 arcseconds | × diameter |

a | ||

distance = | 206265 arcseconds | × 6 km = 599,898 = 600,000 km |

2.06265 arcseconds |

*4. How close will the comet have to be for its nucleus to cover 3 pixels (the minimum number to be considered more than random noise) on the HRI? How about the MRI?*

We can solve this one the same way we solved number 3, but instead we'll use an a equal to 3 times the size of one pixel.

distance = | 206265 arcseconds | × diameter |

a | ||

distance = | 206265 arcseconds | × 6 km = 998,861 = 1,000,000 km |

0.41253 arcseconds × 3 |

For the MRI:

distance = | 206265 arcseconds | × diameter |

a | ||

distance = | 206265 arcseconds | × 6 km = 199,966 = 200,000 km |

2.06265 arcseconds × 3 |

*5. At the time of impact, the flyby spacecraft will be 234,641 km away from the comet (as we figured out in last month's Challenge). How many pixels across will the nucleus of the comet be on the HRI? How about the MRI?*

Here the "diameter" is the same 6 km we've been using for the comet's nucleus. The "distance" is given as 234,641 km. We're solving for a. Then we just divide this by the size of one pixel, and that gives us the total number of pixels it would fill.

a = | diameter | × 206265 arcseconds |

distance | ||

a = | 6 km | × 206265 arcseconds = 5.274 arcseconds |

234,641 km |

For the HRI:

5.274 arcseconds × | 1 pixel | = 13 pixels |

0.41253 arcseconds |

For the MRI:

5.274 arcseconds × | 1 pixel | = 3 pixels |

2.06265 arcseconds |

*6. If the flyby spacecraft were not in shield mode at the time of its closest approach, when it's only around 508 km away from the comet (see last month's Challenge), how many pixels across would the nucleus of the comet be on the HRI? How about the MRI?*

This one gets solved the same way we did number 5...

a = | diameter | × 206265 arcseconds |

distance | ||

a = | 6 km | × 206265 arcseconds = 2436.201 arcseconds |

508 km |

For the HRI:

2436.201 arcseconds × | 1 pixel | = 5899 = 5900 pixels |

0.41253 arcseconds |

For the MRI:

2436.201 arcseconds × | 1 pixel | = 1181 = 1000 pixels |

2.06265 arcseconds |

Which, by the way, is more pixels than these instruments have across, so the instrument would not be able to "see" the whole comet at one time. The level of detail of the comet's surface, however, would be phenomenal. Unfortunately, the instruments would not be safe this close, as there is too much dust and debris in the comet's nucleus.

*7. The impactor spacecraft has a camera identical to the one in the flyby spacecraft's MRI. What is the closest the impactor spacecraft can be and still "see" the entire comet nucleus? The camera has an array that can effectively image 1008 × 1008 pixels.*

We can solve this one the same way we solved number 3, but instead we'll use an a equal to 1008 times the size of one pixel for the MRI.

distance = | 206265 arcseconds | × diameter |

a | ||

distance = | 206265 arcseconds | × 6 km = 595 = 600 km |

(2.06265 arcseconds) × 1008 |