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The star Betelgeuse, in the constellation of Orion, is approximately $3.36 \times {10^{15}}$ miles from Earth. This is approximately $1.24 \times {10^6}$ times as far as Pluto’s minimum distance from Earth. What is Pluto’s approximate minimum distance from Earth? Write your answer in scientific notation.

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Last updated date: 20th Jun 2024
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Answer
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Hint: Let us assume that Pluto's approximate minimum distance from Earth be $x$ miles. Formulate an expression according to the condition given in the question that the distance of Betelgeuse from Earth is $1.24 \times {10^6}$ times Pluto's minimum distance from the Earth. Substitute the value of distance of Betelgeuse from the Earth to find Pluto’s distance.

Complete step by step answer
Let us assume that the approximate minimum distance of Pluto from the Earth is $x$ miles.
We are given the question that the distance of the Betelgeuse, in the constellation of Orion from the Earth is approximately $1.24 \times {10^6}$ times Pluto’s minimum distance from the Earth.
We can say, therefore, that the distance of Betelgeuse from the Earth is equal to the $1.24 \times {10^6}$ multiplied by the distance of Pluto from the Earth.
We have considered Pluto's distance from the Earth be $x$ miles.
Therefore,
\[{\text{Distance of Betelgeuse from Earth }} = 1.24 \times {10^6} \times x\]
We are given that the approximate distance of the Betelgeuse from the Earth is $3.36 \times {10^{15}}$ miles.
Substituting the value $3.36 \times {10^{15}}$ for the Distance of Betelgeuse from Earth in the equation \[{\text{Distance of Betelgeuse from Earth }} = 1.24 \times {10^6} \times x\], we get
\[3.36 \times {10^{15}} = 1.24 \times {10^6} \times x\]
Dividing the equation throughout by \[1.24 \times {10^6}\], we can solve for the value of $x$.
\[\dfrac{{3.36 \times {{10}^{15}}}}{{1.24 \times {{10}^6}}} = \dfrac{{1.24 \times {{10}^6}}}{{1.24 \times {{10}^6}}} \times x\]
$
  \dfrac{{3.36}}{{1.24}} \times {10^9} = x \\
  x \approx 2.7 \times {10^9} \\
$
Thus, the approximate minimum distance of Pluto from the Earth is \[2.7 \times {10^9}\]miles.

Note: The unit of the distance is miles throughout the solution. The solution is represented in scientific notation, as the distance is represented in the power of 10.