Question

How many feet is the earth from the sun?

Answer 1

Hint: The size of the objects across in the universe vary over a wide range. They may be vary from the size of the order of \[{10^{ - 14}}\]m of the tiny nucleus of an atom to the size of the order of \[{10^{26}}\]m of the extent of the observable universe. The mean distance between moon and the sun is called the astronomical unit(A.U).
\[1A.U. = 1.5 \times 10^{11}\]

Complete answer:
Since, \[1{\text{ metre = 3}}{\text{.2 feet}}\]
Hence, in 1 A.U, we have
\[1.5 \times 10^{11} = 3.2 \times 1.5 \times {10}^{11} ft\]
\[\therefore 1 A.U. = 4.8 \times {10}^{11}ft\]
Hence, the distance between earth and the sun in feet is \[4.8 \times {10}^{11}\].

Note: Measurement of an A.U:- Those planets which are closer to the sun than the earth are called inferior planets. The orbits of these inferior planets are smaller than the earth.
The inferior planet P, sun S and the earth E are shown in the figure below.
The angle between the earth-sun direction and the planet-earth
direction and the planet earth direction is the planet’s elongation
and denoted by θ .
Clearly, the θ is maximum when the angle
between \[{r_{ps}}\]and \[{r_{pe}}\]is \[{90^0}\], i.e. angle subtended by the sun
and the earth is the planet is \[{90^0}\].
So, from the above figure, we have
\[\sin \theta = \dfrac{{{r_{ps}}}}{{{r_{es}}}} = \dfrac{{{r_{ps}}}}{{1{\text{A}}{\text{.U}}}}\]
\[ \Rightarrow {r_{ps}} = \sin \theta ({\text{A}}{\text{.U)}}\]
or \[\cos \theta = \dfrac{{{r_{pe}}}}{{{r_{pe}}}} = \dfrac{{{r_{ps}}}}{{1{\text{A}}{\text{.U}}}}\]
\[ \Rightarrow 1{\text{A}}{\text{.U = }}\dfrac{{{r_{pe}}}}{{\cos \theta }}\]
To find the \[{r_{pe}}\]radio signals are sent towards the planet from radar and after reflection are received back on the earth. if t time taken by the signal for its journey from earth to planet and back to the earth.
\[2{r_{pe}} = ct\]
\[ \Rightarrow {r_{pe}} = \dfrac{{ct}}{2}\]
Where the speed c = speed of the radio waves.
It is found that \[{\text{1 A}}{\text{.U}}{\text{. = (4}}{\text{.8 $\times$ 1}}{{\text{0}}^{{\text{11}}}})ft\].