During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.

Answers (1)

Thus, \Omega _{s}= \Omega _{m}

Thus, \frac{\pi (\frac{D_{s}}{2})^{2}}{r{_{s}}^{2}}= \frac{\pi (\frac{D_{m}}{2})^{2}}{r{_{m}}^{2}}

i.e., \frac{D_{s}^{2}}{4r_{s}^{2}}= \frac{D_{m}^{2}}{4r_{m}^{2}}

Thus, \frac{4R_{s}^{2}}{4r_{s}^{2}}= \frac{4R_{m}^{2}}{4r_{m}^{2}}

Taking square roots,

\frac{R_{s}}{r_{s}}=\frac{R_{m}}{r_{m}}

Or,

\frac{R_{s}}{R_{m}}=\frac{r_{s}}{r_{m}}

Therefore, the ratio of the size of the sun to the moon is equal to the distances from the sun to the moon from earth.

Most Viewed Questions

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions