The Mars Polar Lander spacecraft was launched on January 3, 1999. On December 3,

Question

The Mars Polar Lander spacecraft was launched on January 3, 1999. On December 3, 1999, the day that Mars Polar Lander touched down on the Martian surface, the positions of the earth and Mars were given by these coordinates:

x y z
Earth 0.3182 AU 0.9329 AU 0.0000 AU
Mars 1.3087 AU -0.4423 AU -0.0414 AU

In these coordinates, the sun is at the origin and the plane of the earth's orbit is the xy-plane. The earth passes through the +x-axis once a year on the autumnal equinox, the first day of autumn in the northern hemisphere. One AU is equal to 1.496 * 10^8 km, the average distance from the earth to the sun.

a) In a diagram, show the positions of the sun, the earth, and Mars on December 3, 1999.

b) Find the following distances in AU on December 3,1999 :

i) from the sun to the Earth


ii) from the sun to Mars


iii) from the earth to Mars



c) As seen from the earth, what was the angle between the direction to the sun and the direction to Mars on December 3, 1999?

Explanation / Answer

First, let us define two vectors: the vector from the sun to Earth, SE; the vector from the sun to Mars SM. For notation, (x,y,z) is a point, whereas is a vector. Assuming the sun is at point (0,0,0), we find our vectors by subtracting the ending point from the starting point: SE=== SM== A. The distance is equal to the magnitude of the vector SE, or sqrt(x^2+y^2+z^2). The distance is 1.4745*10^8 km. Multiply by 1 AU/(1.496*10^8 km), the distance is 0.985 AU. B. Do the same as with A, but with vector SM. C. Create a new vector, ME, by subtracting the points of Earth from Mars. Take the magnitude of this vector and convert into AU. D. Take the dot product of SM and SE. Set this equal to |SE|*|SM|*cos(angle), where |SE| is the magnitude found in part A, and |SM| is the magnitude found in part B. E. Mars will *likely* not be visible. Given Earth's y position on the coordinate plane, an observer would only be able to see stars with positions of larger x and y values than Earth has. Mars has a smaller y-value, putting it likely out of the line of sight.

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