Question

Mars is approximately half of the earth's diameter. When it is closest to the earth it is at about $\frac{1}{2}$A.U. from the earth. Calculate what size it will appear when seen through the same tescope. (Diameter of sun is 100 times to the diameter of the earth).Sun appears to be ${\frac{1}{2}}^{\circ }$ from the earth. Due to atmospheric fluctuations , eye can't resolve objects smaller than 1 arc minute?

Answer 1

Given,
Diameter of mars is approximately half of the earth's diameter,

$\frac{D_{mars}}{D_{earth}}=\frac{1}{2}\dots \left(1\right)$

Diameter of sun is 100 times to the diameter of the earth,

$\frac{D_{earth}}{D_{sun}}=\frac{1}{100}\dots \left(2\right)$

By multiplying the equation(1) and (2),

$\frac{D_{mars}}{D_{sun}}=\frac{1}{200}$

At 1 AU sun is seen as ${\left(\frac{1}{2}\right)}^{\circ }$ in diameter, and mars will be seen as ${\left(\frac{1}{400}\right)}^{\circ }$ in diameter.

At $\frac{1}{2}$A.U, mars will be seen as $\left(\frac{1}{200}\right)$in diameter.


With 100 magnification mars will be seen as ${\left(\frac{1}{2}\right)}^{\circ }=\left(\frac{1}{2}\right)\times 60=30$arcmin.

This is larger than the resolution limit due to atmospheric fluctuations. So, it will be looked magnified.

Final answer: $30$ arcmin