Problem 82 - Master Resource Book in JEE Main Physics

Question

When the planet Jupiter is at a distance of \(824.7\) million \(\mathrm{km}\) from the earth, its angular diameter is measured to be \(35.72^{\prime \prime}\) of arc. The diameter of Jupiter can be calculated as (a) \(1329 \times 10^{7} \mathrm{~km}\) (b) \(1429 \times 10^{5} \mathrm{~km}\) (c) \(929 \times 10^{5} \mathrm{~km}\) (d) \(1829 \times 10^{5} \mathrm{~km}\)

Step-by-Step Solution

Answer

The diameter of Jupiter is approximately \(1429 \times 10^5\) km, option (b).

Step 1: Understand the Problem

The problem involves calculating the linear diameter of Jupiter using its angular diameter and the distance from Earth. The angular diameter is given in arcseconds, and the linear diameter can be found using trigonometric relations.

Step 2: Formula for Diameter Calculation

To calculate the linear diameter \( D \) of a planet from its angular diameter \( \theta \) and distance \( d \), use the formula: \[ D = 2d \tan\left(\frac{\theta}{2}\right) \] Given \( \theta = 35.72^{\prime\prime} \) and \( d = 824.7 \) million km, first convert \( \theta \) to radians.

Step 3: Convert Angular Diameter

Convert the angular diameter from arcseconds to radians. We know \[ 1^{\prime\prime} = \frac{1}{3600} \text{ degrees} = \frac{\pi}{648000} \text{ radians} \] Thus, \[ \theta = 35.72 \times \frac{\pi}{648000} \text{ radians} \approx 1.736 \times 10^{-4} \text{ radians} \]

Step 4: Calculate Linear Diameter

Substitute the values into the formula: \[ D = 2 \times 824.7 \times 10^6 \times \tan\left(\frac{1.736 \times 10^{-4}}{2}\right) \] Simplify using the small angle approximation, where \( \tan(x) \approx x \): \[ D \approx 2 \times 824.7 \times 10^6 \times \left(\frac{1.736 \times 10^{-4}}{2}\right) \] \[ D \approx 2 \times 824.7 \times 10^6 \times 8.68 \times 10^{-5} \approx 143075 \times 10^5 \text{ km} \]

Step 5: Choose the Closest Option

The closest option to \( 143075 \times 10^5 \) km is option (b), \(1429 \times 10^5 \) km.

Key Concepts

Angular Diameter

The concept of angular diameter is quite fascinating in astronomy. It refers to the apparent size of an object as seen from a particular point, often Earth. When you look at any planet or star, what you actually perceive is the angular diameter. It is how wide the object appears to your eye or telescope.

The measurement is typically made in arcseconds, where one arcsecond is a tiny fraction of a degree - specifically, 1/3600 of a degree. This precision is crucial when observing celestial bodies that are extremely far away. To give you an idea: when astronomers say Jupiter's angular diameter is 35.72 arcseconds, they're discussing how large Jupiter looks from Earth, not its actual size. This measurement allows us to start calculating how big Jupiter actually is.

Trigonometric Relations

Trigonometric relations help us bridge the gap between angular measurements and actual sizes. When we have an angular diameter and a distance, trigonometry provides a way to determine the actual size of an object in space.

The main equation used for such calculations is: \[ D = 2d \tan\left(\frac{\theta}{2}\right) \] Here, \(D\) is the linear diameter of the object (Jupiter, in this case), \(d\) is the distance to the object, and \(\theta\) is the angular diameter.

• This formula is derived from basic trigonometric laws, where the tangent of any angle in a right triangle is the opposite side divided by the adjacent side.
• In astronomy, this equation is adapted such that the opposite side represents the planet's linear diameter, and the adjacent side is the distance from Earth.

Knowing these relations helps in understanding how astronomers calculate real sizes of planets and stars from angular data.

Arcseconds to Radians

To find the actual physical diameter of Jupiter, we need to work with radians rather than arcseconds or degrees. This is because radians are the standard unit of angular measurement in mathematics and physics. Every arcsecond is a small slice of a degree, and any degree is a part of a full circle, which measures \(2\pi\) radians. Thus, to convert arcseconds to radians, use the relationship: \[ 1^{\prime\prime} = \frac{\pi}{648000} \text{ radians} \] This conversion step is crucial:

• Calculating with radians helps you accurately determine Jupiter's diameter using the trigonometric formula mentioned earlier.
• The given angular diameter, \(35.72^{\prime\prime}\), becomes \(35.72 \times \frac{\pi}{648000}\), approximately 1.736 x 10^{-4} radians when converted.

This seemingly small number is pivotal in all subsequent calculations.

Small Angle Approximation

In our exercise involving Jupiter, the small angle approximation is an essential concept that simplifies our calculations. When the angle \(\theta\) is very small, measured in radians, \(\tan(\theta)\) can be closely approximated by \(\theta\) itself. This helps simplify the calculation because:\[ \tan\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2} \] This approximation works because for very tiny angles, the curve of the tangent function is almost a straight line. While this might seem negligible, it greatly simplifies computation, especially when using a calculator.

• This leads to straightforward arithmetic rather than needing complex trigonometric function evaluations.
• In Jupiter's case, after simplifying, the computation effectively becomes an algebra problem, making it faster to find the actual diameter.

This approximation is used due to its effectiveness in delivering quick results without significant loss of accuracy for small angles.