The Russian Znamya 2 was a 20-meter wide (flat) mirror, launched in 1992. It was
designed to test the feasibility of illuminating parts of the Earth’s surface by reflecting
sunlight from orbiting mirrors. Znamya 2 was deployed (for a few hours only) near
the space station MIR, orbiting at an altitude of
∼ 390 km.
(a) Calculate what size spot on the Earths surface was illuminated by Znamya 2.
(b) Assuming the mirror was optimally positioned so that it was
∼ perpendicular to
light from the Sun, how much solar energy did it receive per second ?
2007-10-16 02:23:27 · 1 answers · asked by multiplayertim 2 in Science & Mathematics ➔ Astronomy & Space
The Sun is about 1/2 degree wide in the sky, so the cone of light cast by an infinitely small mirror would be 1/2 degree wide or 390km * Tan 0.5 degree = 3.4 kilometer wide spot on the ground. So the 20 meter wide mirror make a spot 3.4 km + 20 meters wide, or 3.42 km wide. The area would be about 9,186,331 square meters, or almost 30,000 times larger in area than the 314 square meters area of the 20 meter mirror (assuming it is round). So the sunlight would be 30,000 times less intense than direct sunlight.
2007-10-16 03:02:04 · answer #1 · answered by campbelp2002 7 · 1⤊ 0⤋