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.
Thanks!
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.
a) To calculate the Sun's luminosity, we can use the Stefan-Boltzmann law, which states that the luminosity of a blackbody is directly proportional to the fourth power of its temperature. The formula is given as:
L = 4πR^2σT^4,
where L is the luminosity, R is the radius, σ is the Stefan-Boltzmann constant, and T is the temperature.
Substituting the given values:
R = 7.0×10^5 km = 7.0×10^8 m,
T = 6000 K,
σ = 5.67 × 10^-8 W m^-2 K^-4,
we can calculate the luminosity:
L = 4π(7.0×10^8)^2 × 5.67 × 10^-8 × (6000^4).
b) The brightness or solar flux at Mars' distance can be calculated using the inverse square law. The solar flux is inversely proportional to the square of the distance between the Sun and Mars. The formula is given as:
Flux = L / (4πd^2),
where Flux is the solar flux, L is the Sun's luminosity, and d is the distance between the Sun and Mars.
Substituting the given values:
d = 1.5 AU = 1.5 × 149.6 × 10^9 m,
L (from part a),
we can calculate the solar flux:
Flux = (L) / (4π(1.5 × 149.6 × 10^9)^2).
c) To calculate the electrical power available to the MRO transmitter, we can use the formula:
Power = Conversion Efficiency × Solar Flux × Area,
where Power is the electrical power, Conversion Efficiency is the efficiency of the solar panel conversion, Solar Flux is the solar flux at Mars' distance (from part b), and Area is the area of the solar panels.
Substituting the given values:
Conversion Efficiency = 30% = 0.3,
Solar Flux (from part b),
Area = 100 ft^2 × (1 m / 3 ft)^2,
we can calculate the electrical power available:
Power = 0.3 × Flux × Area.
d) To calculate the power available to the MRO transmitter at Saturn, we can use the same formula as in part c. However, we need to use the new solar flux at Saturn's distance.
First, we calculate the new solar flux at Saturn's distance using the inverse square law, similar to part b:
Flux_Saturn = L / (4πd^2),
where Flux_Saturn is the solar flux at Saturn's distance, L is the Sun's luminosity, and d is the distance between the Sun and Saturn.
Substituting the given values:
d = 9.6 AU = 9.6 × 149.6 × 10^9 m,
L (from part a),
we can calculate the new solar flux:
Flux_Saturn = (L) / (4π(9.6 × 149.6 × 10^9)^2).
Then, we can calculate the power available to the MRO transmitter at Saturn:
Power_Saturn = 0.3 × Flux_Saturn × Area.
To calculate the ratio of MRO transmitter power at Saturn to a 60-watt light bulb, we divide the power at Saturn by 60 watts:
Ratio = Power_Saturn / 60 watts.