Hi, this requires some basic knowledge of astrophysics. I missed last lecture that covered this, so though I'd prefer thorough solutions to these problems, I'd appreciate it even if you could tell me which formulas to use to solve each part of the problem.
NASA’s Mars Reconnaissance Orbiter (MRO) is currently in orbit around Mars. The space-craft carries six instruments to study the Martian atmosphere and surface, including a ground-penetrating radar to search for water beneath the surface. MRO communicates with Earth using a 10-foot diameter dish antenna and a transmitter powered by 100 square feet of solar panels. A very good modern conversion eﬃciency for solar panels is 30%, i.e. the panel converts 30% of the sunlight incident upon it into electrical power - the other 70% is lost.
a) Take the Sun to be a blackbody with a surface temperature of 6000 K. The Sun’s radius is 7.0×10^5 km. Calculate the Sun’s luminosity, in watts(Joules/second). (The luminosity you have calculated is called the “bolometric luminosity” because it sums the power emitted by the Sun at all wavelengths.)
b) Mars is 1.5 AU from the Sun. Calculate the brightness of the Sun at Mars’ distance (i.e., the solar ﬂux on Mars’ surface) expressed in watts per square meter.
c) Using the approximation (good to about 10%) that 1 meter = 3 feet, calculate how much electrical power will be available to the MRO transmitter, assuming that the
MRO solar panels are facing the Sun directly, and that they have a conversion eﬃciency of 30%. Express your result in watts.
d) Suppose now that a spacecraft identical to MRO were launched (presumably from a bigger rocket!) to observe Saturn’s moon Titan. How much power (in watts) would the MRO at Saturn have available to its transmitter? Saturn orbits at 9.6 AU from the Sun. For comparison, a typical light bulb in your home has a total power output of about 60 watts. What would be the ratio of MRO transmitter power at Saturn to
that of a 60 watt light bulb? This is why missions to the outer Solar System so often rely on power generated from the radioactive decay of a plutonium isotope, rather than from solar panels.
Physics - solar power in space - SraJMcGin, Wednesday, March 4, 2009 at 9:48am
Sorry, not my field, but hopefully one of the tutorials in this GOOGLE Search will help you: