Question

the earth.
Example 4 : The parallax of a far off planet as measured from the two diametric extremes on the equator
the earth is 2.0 minute. If the radius of the earth is 6400 km, find the distance of the planet from​

Answer 1

Radius of the Earth (or Basis) = 6400 km .= 6.4 × 10⁶ km. 

Time = 2 minutes. 

This means that the angle sustained by the Earth on the Heavenly bodies is 2'

  

Thus, converting it into radians, 

∵ 1' = 2.91 × 10⁻⁴ radians. 

∴ 2' = 5.8 × 10⁻⁴ radians. 

Hence, Parallax angle (θ) = 5.8 × 10⁻⁴ rad

Now, Using the Formula, 

 θ = Basis/Distance

∴ Distance = (6.4 × 10⁶)/(5.8 × 10⁻⁴) rad

   = 1.10 × 10¹⁰ m. 

   = 1.1 × 10⁷ km. 

Hence, the distance between the Earth and the Planet is 1.1 × 10⁷ km. 

If it helps. Then mark me as brainliest please

Answer 2

SoluTion:

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We know that,

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Angle of parallax $\sf{\theta = 2\:minute}$

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$\longrightarrow$ $\sf{\theta = 2 \times \bigg( \dfrac{1}{60} \bigg)^{\circ}}$

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$\longrightarrow$ $\sf{\theta = 2 \times \bigg( \dfrac{1}{60} \bigg) \times \dfrac{\pi}{180}\:rad}$ $\sf{\bigg[ \because \: 1^{\circ} = \dfrac{\pi}{180} \:rad\:\bigg]}$

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$\longrightarrow$ $\sf{\theta\:=\:\dfrac{\pi}{5400}\:rad}$

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From parallax method,

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$\boxed{\red{\sf{BP\:=\:\dfrac{AB}{\theta}}}}$

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$\longrightarrow$ $\sf{BP\:=\:\dfrac{diameter\:of\:earth}{\theta}}$

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$\longrightarrow$ $\sf{BP\:=\:\dfrac{2 \times 6400}{\bigg(\dfrac{\pi}{5400} \bigg)}\:km}$

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$\longrightarrow$ $\sf{BP\:=\:2.2 \times 10^{7}\:km}$

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$\longrightarrow$ $\sf{\blue{BP\:=\:2.2 \times 10^{10}\:m}}$

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Hence, Distance of the planet from earth will be of 2.2 × 10¹⁰ m.