Question

When the planet Jupiter is at a distance of 824.7 mil-

lion kilometres from the Earth, its angular diameter is

measured to be 35.72″ of arc. Calculate the diameter of

Jupiter.​

Answer 1

Answer:

Hence, diameter of jupiter is 1.4×10^5 km.

Answer 2

Answer :-

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the concept of Parallax Method has been used. We know that, we can measure distances of the stars that are less than 100 light years away from earth. Here we are given the distance of Jupiter from Earth in million kilometer.And then we can change the value of 35.72'' arc to radian. Then we can apply the formula and find the answer.

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★ Equations Used :-

$\\\;\boxed{\sf{1\;arc\:min\;\;or\;\;1''\;=\;\bf{4.874\;\times\;10^{-6}\;rad}}}$

$\\\;\boxed{\sf{Linear\;Diameter,\;(d)\;=\;\bf{Distance_{(in\;Km)} ,\;[D]\;\times\;Angular\;Diameter_{(in\;rad)} ,\;[\theta]}}}$

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★ Solution :-

Given,

» Distance of Jupiter from Earth = D = 824.7 million Kilometer

» Angular Diameter of Jupiter = θ = 35.72" or 35.72 arc min

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~ For the standard value of D and θ ::

We know that,

$\\\;\;\;\sf{:\rightarrow\;\;\;1\;million\;=\;\bf{10^{6}}}$

Then,

$\\\;\;\;\sf{:\rightarrow\;\;\;1\;million\;Km\;=\;\bf{10^{6}\;\;Km}}$

$\\\;\;\;\bf{:\rightarrow\;\;\;824.7\;million\;Km\;=\;\underline{\underline{\bf{824.74\;\times\;10^{6}\;\;Km}}}}$

Also,

$\\\;\boxed{\sf{1\;arc\:min\;\;or\;\;1''\;=\;\bf{4.874\;\times\;10^{-6}\;rad}}}$

$\\\;\boxed{\sf{35.72''\;=\;\underline{\underline{\bf{35.72\;\times\;4.874\;\times\;10^{-6}\;rad}}}}}$

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~ For the Diameter of the Jupiter :-

Since, Linear Diameter = Diameter of Jupiter.

Then,

$\\\;\;\;\sf{:\Longrightarrow\;\;\;Linear\;Diameter,\;[d]\;=\;\bf{Distance_{(in\;Km)} ,\;[D]\;\times\;Angular\;Diameter_{(in\;rad)} ,\;[\theta]}}$

$\\\;\;\;\sf{:\Longrightarrow\;\;\;Linear\;Diameter,\;[d]\;=\;\bf{824.74\;\times\;10^{6}\;\times\;35.72\;\times\;4.874\;\times\;10^{-6}}}$

$\\\;\;\;\sf{:\Longrightarrow\;\;\;Linear\;Diameter,\;[d]\;=\;\bf{143521\;\;Km}}$

$\\\;\;\;\sf{:\Longrightarrow\;\;\;Linear\;Diameter,\;[d]\;=\;\bf{1.435\;\times\;10^{5}\;\;Km}}$

$\\\;\large{\underline{\underline{\rm{Diameter\;of\;Jupiter\;is\;\;\boxed{\bf{1.435\;\times\;10^{5}\;\;Km}}}}}}$

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★ More to Know :-

• Intensity Method - Its based on inverse square law that, the intensity of illumination at a point is inversely proportional to the square of the distance from the source of light.

$\\\;\sf{\leadsto\;\;\;\dfrac{l_{1}}{l_{2}}\;=\;\dfrac{r_{2} ^{2}}{r_{1} ^{2}}}$

• Reflection or Echo Method - It gives the relation that :

$\\\;\sf{\leadsto\;\;S\;=\;\dfrac{v\;\times\;t}{2}}$

• Triangulation Method - It gives the height if an inaccessible object as,

$\\\;\sf{\leadsto\;\;h\;=\;\dfrac{d}{\cot \theta _{2}\;-\;\cot \theta _{1}}}$

• Relation between rad and arc units :

$\\\;\sf{\leadsto\;\;1^{\circ}\;=\;1.745\;\times\:10^{-2}\;\;rad}$

$\\\;\sf{\leadsto\;\;1'\;=\;2.908\;\times\:10^{-4}\;\;rad}$