Question

Moon is seen to be of ${\dfrac{1}{2}^o}$ diameter from the earth. What must be the relative size of the moon as compared to the earth?

Answer 1

Hint:The Moon is the sole natural satellite of the Earth. It is the biggest natural satellite in the Solar System compared to the size of its planet,[f] the fifth biggest satellite in the Solar System overall, and larger than any dwarf planet, with a diameter of nearly one-quarter the diameter of Earth (equivalent to the breadth of Australia).


Complete step-by-step solution:
In our night sky, the moon is the brightest object. It looks to be rather huge, yet this is just due to the fact that it is the nearest celestial body. The moon is somewhat bigger than one-fourth (27%) of Earth's size, a far bigger ratio (1:4) than any other planet and its moons. The moon of Earth is the sixth biggest in the solar system. The average radius of the moon is 1,079.6 miles (1,737.5 kilometers). To obtain its diameter, multiply those figures by two: 2,159.2 miles (3,475 kilometres) is less than a third of the Earth's width. The equatorial circumference of the moon is 6,783.5 miles (10,917 km). According to NASA, "if Earth were the size of a nickel, the moon would be roughly the size of a coffee bean."
Now we know that
Since the moon is seen from earth (diametrically) angle = ${\dfrac{1}{2}^o}$
And suppose If the earth is seen from moon then the angle is = ${2^o}$
Hence we can write that
$\dfrac{{{\text{ size of moon }}}}{{{\text{ size (diameter) of earth }}}} = \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^\circ }}}{{{2^0}}} = \dfrac{1}{4}$
Hence, Size of the moon is $\dfrac{1}{4}$ the size (diameter) of earth.

Note:The earth's gravity, or pulsing power, is six times more than the moon's. As a result, if a person is weighed on the moon, he will only weigh one-sixth of his weight on Earth. Similarly, if a guy tries to leap on the moon with the same amount of energy, he will jump six times higher than he would be able to leap on the earth. Despite this, the moon's gravity is strong enough to generate ocean tides.