The parallax of a heavenly body …
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- The parallax of a heavenly body measured from two point diametrically opp. On the equator of the earth is 1.0minute.if the radius of the earth is 6400km, find the distance of the heavenly body from the centre of the earth in AU.?
Posted by Lochan Soni 6 years, 11 months ago
- 2 answers
Arun Soni 6 years, 9 months ago
{tex} \eqalign{ & Here,\theta = 1\min = \frac{1}{{60}} \times \frac{\pi }{{180}} \cr & \Rightarrow r = \frac{l}{\theta } \cr & l = 2 \times 6400 \times {10^3}m \cr & \Rightarrow r = \frac{{2 \times 6400 \times {{10}^3} \times 60 \times 180}}{\pi } \approx 4.4 \times {10^{10}}m \cr & 1A.U = 1.5 \times {10^{11}}m \cr &\Rightarrow r = \frac{{4.4 \times {{10}^{10}}}}{{1.5 \times {{10}^{11}}}} \approx 0.29AU \cr}{/tex}
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Arun Soni 6 years, 9 months ago
{tex} \eqalign{ & Here,\theta = 1\min = \frac{1}{{60}} \times \frac{\pi }{{180}} \cr & \ r = \frac{l}{\theta } \cr & l = 2 \times 6400 \times {10^3}m \cr &\Rightarrow r = \frac{{2 \times 6400 \times {{10}^3} \times 60 \times 180}}{\pi } \approx 4.4 \times {10^{10}}m \cr & 1A.U = 1.5 \times {10^{11}}m \cr &\Rightarrow r = \frac{{4.4 \times {{10}^{10}}}}{{1.5 \times {{10}^{11}}}} \approx 0.29AU \cr}{/tex}
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