(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 rem = x m be the distance between the earth and the moon

Now, the distance between the sun and the earth (rse) = 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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