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(a) The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?

(b) Moon is seen to be of $\left ( \frac{1}{2} \right )^{o}$ diameter from the earth. What must be the relative size compared to the earth?

(c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

Answers (1)

(a) Here we know that, $\theta =$ arc/radius

$= \frac{R_{e}}{60R_{e}}$

$= \frac{1}{60}rad$

Now, the angle from the moon to the diameter of the earth will be-

$2\theta = 2\left ( \frac{1}{60} \right )\left ( \frac{180}{\pi } \right )$

$= \frac{6^{o}}{\pi }$

$\cong \frac{6}{3.14}\cong 2^{o}$

(b) If the moon is seen from the earth diametrically, the angle will be $\frac{1}{2^{o}}$

& if the earth is seen from the moon, the angle will be $2^{o}$

Thus, the size of the moon/ size of the earth = $\frac{\frac{1}{2^{o}}}{2^{o}}= \frac{1}{4}$

Therefore, as compared to the earth, the size of the moon is $\frac{1}{4}$ the size(diameter) of the earth.

(c)Let r_em = x m be the distance between the earth and the moon

Now, the distance between the sun and the earth (r_se) = 400xm

On a solar eclipse, the sun is completely covered by the moon; also the angle formed by the diameter of the sun, moon and the earth are equal.

Thus, $\theta _{m}= \theta _{s}$

......... ( the angle from the earth to the moon is $\theta {_{m}}^{o}$ & angle from sun to earth is $\theta {_{s}}^{o}$

Now since $\theta _{m}= \theta _{s}$

$\frac{D_{m}}{r_{em}}= \frac{D_{s}}{r_{x}}$

$\frac{D_{m}}{x}= \frac{D_{s}}{400 x}$

$\frac{D_{s}}{D_{M}}= 400$

Thus, $4D_{m}= D_{e}or\; D_{m}= \frac{D_{e}}{4}$

Thus, $\frac{D_{s}}{\frac{D_{e}}{4}}= 400$

$\frac{4D_{s}}{D_{e}}= 400$

$\frac{D_{s}}{D_{e}}= \frac{400}{4}= 100$

Therefore, $D_{s}= 100\; D_{e}$

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