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Discovery Zone - Mission Challenge

See National Math Standards for this Challenge.

Image of the Moon

Q: What was the distance from the Deep Impact spacecraft to the moon on January 16, 2005?

On January 16, 2005, the Deep Impact spacecraft turned its cameras back towards Earth and captured some beautiful images of our Moon. To aid in calibrating the spacecraft instruments, astronomers would like to know precisely how far Deep Impact was from the moon when this picture was taken.

To understand this problem better, we need to know how astronomers locate objects in the sky. Much like navigators who rely on latitude and longitude to pinpoint their location on Earth, astronomers use a pair of numbers to uniquely identify every point on the celestial sphere. These numbers are called Right Ascension and Declination.

Declination is very similar to latitude measurements on Earth. We start by defining a celestial equator, which is simply an extension of the Earth's equator into the sky. This marks zero degrees declination. The range of declination values is from -90° to +90° with +90° situated at the north celestial pole, and -90° situated at the south celestial pole. Each degree is divided into 60 arcminutes, and each arcminute is divided into 60 arcseconds.

Right ascension is measured along the east-west direction, much like longitude on Earth, except it is measured in "hours", rather than degrees. The range for right ascension (or R.A.) is from 0h to 23h, incrementing as one looks east across the sky. Each hour is divided into 60 minutes, and each minute is further divided into 60 seconds.

As an example of how this is used, let's return to the beginning of the problem. At the time the moon picture was taken, Deep Impact was located at 13 hours, 12 minutes, 31.27 seconds right ascension (also written as 13h 12m 31.27s R.A.) and -3 degrees, 26 arcminutes, 38.2 arcseconds declination (also written as -3° 26' 38.2" Dec.). We also know that the spacecraft was 0.0084401211 AU from Earth (1 au = 1.5 x 108 km). An astronomical unit (AU) is a unit of distance that is approximately equal to the mean distance between the Earth and the Sun.

We can look up the same information for the moon at a particular time. The data can be retrieved via the JPL Horizon's webpage located at the following URL:

The celestial coordinates for both Deep Impact and the moon are summarized in this table:

January 16, 2005 Deep Impact Moon
R.A. 13h 12m 31.27s 1h 22m 47.56s
Dec. -3° 26' 38.2" 7° 18' 12.5"
Distance (AU) from the Earth 0.00844 0.00257

These data are sometimes referred to as the "ephemeris" of the objects. Now it's your turn to solve a problem of real interest to astronomers. Given the ephemeris data for Deep Impact spacecraft and the moon, plus some geometry and what you've just learned about Right Ascension and Declination, can you figure out how far the spacecraft was from the moon on January 16th?

Setting up the problem:

We can describe the location of an object in space (call it P) with the following vector notation:

Figure1: Graph of space P.

Vector from sigma to point P (representing any point in space) is defined by:
r = magnitude
phi = angle with respect to xy-plane
theta = angle with respect to positive x-axis

These coordinates, with the Earth located at the origin (sigma) relate to the ephemeris data as follows:
r = the distance from the Earth to the object we're interested in
phi = declination of the object
theta = right ascension of the object

In this problem, we are interested in finding the distance between two points in space, which we can call P1 and P2. These will represent the location of the Deep Impact spacecraft and the moon, respectively.

Figure 2: Graph showing distance between vectors P1 and P2.

To find the distance between any two points in space, we can use the distance formula:

d = sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)

where (x1, y1, z1) and (x2, y2, z2) are the Cartesian coordinates of the two points.

To use this, we need to first convert our (r, phi, theta) coordinates to (x, y, z) coordinates.

Figure 3: Graph depicting projections onto xy-plane. rz = projection of r onto z-axis = r sin phi
rxy = projection of r onto xy-plane = r cos phi
Figure 4: Graph depicting projections onto xy-plane. rx = projection of rxy onto x-axis = rxy cos theta
ry = projection of rxy onto y-axis = rxy sin theta

rx, ry, rz are going to be the (x, y, z) coordinates we need.

rxy = r cos phi
rx = rxy cos theta = (r cos phi) cos theta
ry = rxy sin theta = (r cos phi) sin theta

x = r cos phi cos theta

y = r cos phi sin theta

z = r sin phi
These three equations will give us the (x, y, z) coordinates if we have the distance, right ascension, and declination!

Before we use these transformations, we need to convert right ascension and declination to degrees (or radians).

R.A. recorded in hours, minutes, seconds
Dec. recorded in degrees, arcminutes, arcseconds

For right ascension: 360° = 24h = 1440m = 86,400s or,

1h = 360°/24h = 15°

1m = 360°/1440m = (1/4)°

1s = 360°/86400s = (1/240)°

For declination: 1° = 60' = 3600" or,

1' = (1/60)°

1" = (1/3600)°

We now have all the conversion and transformations we need to solve the problem!

Why did the science team need to make this calculation?

In addition to calibrating the instruments, we wanted to give the image of the Moon some perspective for the viewer. We were writing figure captions for release to the public. Ordinarily, the distance is calculated by the data processing pipeline, a series of computer programs that operates on the returned data as well as information from the spacecraft about its condition and location in space. At the time when the caption was being written, we did not have all of the information we needed in the data pipeline. It was still being set up. So we had to make the calculations ourselves. We were glad that we had studied trigonometry and pre-calculus so that we could do the calculation. Then we were also able to check that the computer code, when it was available, had been correctly programmed.

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Last Updated: 28 Jun 2010