Angular diameter of Jupiter is $36^0$. If the distance of Jupiter from the Earth is $825$ million kilometer, find the diameter of Jupiter.
Answer
654k+ views
Hint
In order to find the diameter of the Jupiter first of all we convert the angular diameter into radian by multiplying with $4.874 \times {10^{ - 6}}rad$ , then we use the formula i.e. diameter of the Jupiter is the product of the angular diameter and the distance of the Jupiter from the earth.
Complete step by step solution
In this question we have to find out the diameter of Jupiter.
For this, let the diameter of Jupiter is $d$.
It is also given that,
Distance of Jupiter from earth is $D = 825 \times {10^6} \times {10^3}m = 8.25 \times {10^{11}}m$
As we know that $1\deg ree = \dfrac{\pi }{{180}}radian$
Angular diameter is $\theta = 36'' = \dfrac{{36}}{{60 \times 60}} \times \dfrac{\pi }{{180}}rad$
$ \Rightarrow \theta = 36 \times 4.874 \times {10^{ - 6}}rad = 1.75 \times {10^{ - 4}}rad$
Now, as we also know that $angle = \dfrac{{arc}}{{radius}}$
Therefore, in this case angle is θ, arc is the diameter of Jupiter and radius is the distance of Jupiter from the earth.
$ \Rightarrow \theta = \dfrac{d}{D}$
$ \Rightarrow d = \theta \times D$
Now, put the given values. We get
$ \Rightarrow d = 1.75 \times {10^{ - 4}} \times 8.25 \times {10^{11}}$
$ \Rightarrow d = 1.44 \times {10^8}m = 1.44 \times {10^5}km$
Hence, the diameter of the Jupiter is $1.44 \times {10^5}km$
Note
In these types of the questions we have to use the formula, the angle subtended by one object on the other is the ratio of the arc of the object and the distance between these objects i.e. $angle = \dfrac{{arc}}{{radius}}$.
Care must be taken to convert the units, when we convert seconds into radian we should remember that we have to convert first into minutes then in degree by dividing by 3600 after that we convert this degree into radian by multiplying it with $\dfrac{\pi }{{180}}$.
In order to find the diameter of the Jupiter first of all we convert the angular diameter into radian by multiplying with $4.874 \times {10^{ - 6}}rad$ , then we use the formula i.e. diameter of the Jupiter is the product of the angular diameter and the distance of the Jupiter from the earth.
Complete step by step solution
In this question we have to find out the diameter of Jupiter.
For this, let the diameter of Jupiter is $d$.
It is also given that,
Distance of Jupiter from earth is $D = 825 \times {10^6} \times {10^3}m = 8.25 \times {10^{11}}m$
As we know that $1\deg ree = \dfrac{\pi }{{180}}radian$
Angular diameter is $\theta = 36'' = \dfrac{{36}}{{60 \times 60}} \times \dfrac{\pi }{{180}}rad$
$ \Rightarrow \theta = 36 \times 4.874 \times {10^{ - 6}}rad = 1.75 \times {10^{ - 4}}rad$
Now, as we also know that $angle = \dfrac{{arc}}{{radius}}$
Therefore, in this case angle is θ, arc is the diameter of Jupiter and radius is the distance of Jupiter from the earth.
$ \Rightarrow \theta = \dfrac{d}{D}$
$ \Rightarrow d = \theta \times D$
Now, put the given values. We get
$ \Rightarrow d = 1.75 \times {10^{ - 4}} \times 8.25 \times {10^{11}}$
$ \Rightarrow d = 1.44 \times {10^8}m = 1.44 \times {10^5}km$
Hence, the diameter of the Jupiter is $1.44 \times {10^5}km$
Note
In these types of the questions we have to use the formula, the angle subtended by one object on the other is the ratio of the arc of the object and the distance between these objects i.e. $angle = \dfrac{{arc}}{{radius}}$.
Care must be taken to convert the units, when we convert seconds into radian we should remember that we have to convert first into minutes then in degree by dividing by 3600 after that we convert this degree into radian by multiplying it with $\dfrac{\pi }{{180}}$.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

