Answer
384.6k+ views
Hint: To solve this problem first we will convert the given angle in degree than in radians to calculate the required distance, because for small angles , angle can be written as arc length divided by radius. In the above problem arc is radius of earth and radius is distance of the moon from earth.
Formula used:
$ \theta =\dfrac{arc}{radius}$
Complete step by step solution:
Given, moon subtends an angle of 57 minute at the base-line equal to radius of earth so, by converting it into degrees we have,
$\theta =57'$
$\Rightarrow \theta =\dfrac{57}{60}=\dfrac{19}{20}={{0.95}^{\circ }}$
$\Rightarrow \theta ={{0.95}^{\circ }}$
Now, to change angle theta into radian form we need to multiply the above value of theta with $\dfrac{\pi }{180}$so,
$\Rightarrow \theta ={{0.95}^{\circ }}\times \dfrac{\pi }{180}$
Now, we know that the above angle is very small so we can write angle theta by,
$\Rightarrow \theta =\dfrac{arc}{radius}$
Here, arc (Radius of earth R) and radius (Distance of the moon from earth D).
$\Rightarrow \theta =\dfrac{R}{D}$
$\Rightarrow D=\dfrac{R}{\theta }$
By putting the value of R from the problem and of theta calculated above we have,
$\Rightarrow D=\dfrac{6.4\times {{10}^{6}}}{{{0.95}^{\circ }}\times \dfrac{\pi }{180}}m$
$\Rightarrow D=\dfrac{11.52\times {{10}^{8}}}{{{0.95}^{\circ }}\times \pi }m$
$\Rightarrow D=3.86\times {{10}^{8}}m$
$\therefore $ The distance of the moon from the earth is $3.86\times {{10}^{8}}m$ so option (B) is correct.
Additional information:
In mathematics both radians and degrees represent angles in different notations, in which radian value can be defined using arc of circle and we know that $\pi $ radians is equal to ${{180}^{\circ }}$ so we can easily calculate the value of 1 degree which is given by,
$\Rightarrow {{1}^{\circ }}=\dfrac{\pi }{180}$
Note:
Earlier the distance between moon and earth is calculated by lunar distance which involved measuring the angle between the moon and reference points selected on earth, in above case that was the opposite point of radius of earth. In the late 1950s or 60s the methods were changed and experiments using radar or lasers with computer processing and modeling were used to calculate this distance.
Formula used:
$ \theta =\dfrac{arc}{radius}$
Complete step by step solution:
Given, moon subtends an angle of 57 minute at the base-line equal to radius of earth so, by converting it into degrees we have,
$\theta =57'$
$\Rightarrow \theta =\dfrac{57}{60}=\dfrac{19}{20}={{0.95}^{\circ }}$
$\Rightarrow \theta ={{0.95}^{\circ }}$
Now, to change angle theta into radian form we need to multiply the above value of theta with $\dfrac{\pi }{180}$so,
$\Rightarrow \theta ={{0.95}^{\circ }}\times \dfrac{\pi }{180}$
Now, we know that the above angle is very small so we can write angle theta by,
$\Rightarrow \theta =\dfrac{arc}{radius}$
Here, arc (Radius of earth R) and radius (Distance of the moon from earth D).
$\Rightarrow \theta =\dfrac{R}{D}$
$\Rightarrow D=\dfrac{R}{\theta }$
By putting the value of R from the problem and of theta calculated above we have,
$\Rightarrow D=\dfrac{6.4\times {{10}^{6}}}{{{0.95}^{\circ }}\times \dfrac{\pi }{180}}m$
$\Rightarrow D=\dfrac{11.52\times {{10}^{8}}}{{{0.95}^{\circ }}\times \pi }m$
$\Rightarrow D=3.86\times {{10}^{8}}m$
$\therefore $ The distance of the moon from the earth is $3.86\times {{10}^{8}}m$ so option (B) is correct.
Additional information:
In mathematics both radians and degrees represent angles in different notations, in which radian value can be defined using arc of circle and we know that $\pi $ radians is equal to ${{180}^{\circ }}$ so we can easily calculate the value of 1 degree which is given by,
$\Rightarrow {{1}^{\circ }}=\dfrac{\pi }{180}$
Note:
Earlier the distance between moon and earth is calculated by lunar distance which involved measuring the angle between the moon and reference points selected on earth, in above case that was the opposite point of radius of earth. In the late 1950s or 60s the methods were changed and experiments using radar or lasers with computer processing and modeling were used to calculate this distance.
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