Calculate the length of a parsec if you are on mars

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SUMMARY

The length of a parsec when calculated from Mars, using the mean distance of Mars from the sun (2.3 x 1013 cm) and the tangent of 1 arcsecond, results in approximately 4.97 light-years (LY). The calculation involves converting units from centimeters to light-years through a series of conversions. The method discussed is valid and confirms that the distance of 1 parsec can also be derived by adjusting for the ratio of Earth's orbital radius to Mars' orbital radius.

PREREQUISITES
  • Understanding of the CGS (centimeter-gram-second) system
  • Knowledge of angular measurements, specifically arcseconds
  • Familiarity with trigonometric functions, particularly tangent
  • Ability to perform unit conversions between centimeters, meters, kilometers, and light-years
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  • Research the concept of parsecs and their significance in astronomy
  • Study the relationship between distance and angular measurements in celestial mechanics
  • Learn about the methods of converting between different unit systems, especially CGS to SI
  • Explore the implications of orbital mechanics on distance measurements in the solar system
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Astronomy students, physics enthusiasts, and anyone interested in celestial measurements and unit conversions in astrophysics.

warfreak131
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Homework Statement



In class, we use the CGS system, so I do all my work in cgs units and then convert at the end.

The mean distance of Mars from the sun is 2.3 * 10^13cm

1 arcsecond = (1/60) of (1/60) of 1 degree, = 1/3600th degree

Let d = distance of 1 parsec

The Attempt at a Solution



\tan(1 sec) = \frac{2.3 \cdot 10^{13} cm}{d}

d=\frac{2.3{\cdot}10^{13} cm}{Tan(\frac{1}{3600})}

d=4.7{\cdot}10^{18} cm{\cdot}\frac{1 m}{100 cm}{\cdot}\frac{1 km}{1000 m}{\cdot}\frac{1 LY}{9.46{\cdot}10^{12} km}
d=4.7{\cdot}10^{18}{\cdot}\frac{1 LY}{9.46{\cdot}10^{12}}
d=4.97 LY

Is this correct?
 
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warfreak131 said:

Homework Statement



In class, we use the CGS system, so I do all my work in cgs units and then convert at the end.

The mean distance of Mars from the sun is 2.3 * 10^13cm

1 arcsecond = (1/60) of (1/60) of 1 degree, = 1/3600th degree

Let d = distance of 1 parsec

The Attempt at a Solution



Tan(1")=\frac{2.3{\cdot}10^{13} cm}{d}

d=\frac{2.3{\cdot}10^{13} cm}{Tan(\frac{1}{3600})}

d=4.7{\cdot}10^{18} cm{\cdot}\frac{1 m}{100 cm}{\cdot}\frac{1 km}{1000 m}{\cdot}\frac{1 LY}{9.46{\cdot}10^{12} km}
d=4.7{\cdot}10^{18}{\cdot}\frac{1 LY}{9.46{\cdot}10^{12}}
d=4.97 LY

Is this correct?

Can you solve this by just taking the standard parsec and multiplying this by the appropriate ratio of the Earth's orbital radius to that of the orbital radius of Mars?
 

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