The parallax of a heavenly body …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
- 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 8 years, 7 months ago
- 2 answers
Arun Soni 8 years, 5 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}
Test
Related Questions
Posted by Mohammed Javith 1 year, 5 months ago
- 0 answers
Posted by Mansi Class 9Th 1 year, 4 months ago
- 0 answers
Posted by Nekita Baraily 1 year, 4 months ago
- 2 answers
Posted by M D 1 year, 4 months ago
- 1 answers
Posted by Pankaj Tripathi 1 year, 4 months ago
- 1 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Arun Soni 8 years, 5 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}
0Thank You