Assume you want to deduce the radius of a planet in our Solar System as it occults a background star when the relative velocity between the planet and Earth is \(30 \mathrm{km} / \mathrm{s}\). If the star crosses through the middle of the planet and disappears for a total of 26 minutes, what is the planet's radius? a. \(3,000 \mathrm{km}\) b. \(23,000 \mathrm{km}\) c. \(15,000 \mathrm{km}\) d. \(5,000 \mathrm{km}\)

Short Answer

Expert verified
b. 23,000 km

Step by step solution

01

Calculate the Duration in Seconds

First, convert the duration from minutes to seconds. Since 1 minute equals 60 seconds, multiply the number of minutes by 60:\[26 \text{ minutes} \times 60 \text{ seconds/minute} = 1560 \text{ seconds}\]
02

Understand the Relative Motion

The relative velocity between the planet and Earth is given as 30 km/s. This is the speed at which a background star crosses the diameter of the planet during the occultation.
03

Relate Diameter to Time and Velocity

Use the formula for distance, which is the product of velocity and time, to find the diameter of the planet:\[ \text{Distance} = \text{Velocity} \times \text{Time} = 30 \text{ km/s} \times 1560 \text{ seconds} = 46800 \text{ km}\]This distance is the diameter of the planet.
04

Solve for the Radius

The radius of the planet is half of its diameter. So, divide the diameter by 2:\[ \text{Radius} = \frac{46800 \text{ km}}{2} = 23400 \text{ km}\]
05

Choose the Closest Answer

Among the given options, the closest value to the calculated radius of 23400 km is: b. \(23,000 \text{ km}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Relative Velocity
Relative velocity is the speed of one object as observed from another moving object. In this case, it's the speed at which the planet and Earth are moving towards or away from each other. The relative velocity between the planet and Earth is given as 30 km/s. This means that a background star would appear to cross the planet at this speed due to the combined motion of both the Earth and the planet.
The concept of relative velocity is essential when dealing with celestial bodies because their motion is not isolated. They are part of a dynamic system where multiple objects are simultaneously in motion. Calculating relative velocity helps us understand these interactions.
What is Occultation?
Occultation is an astronomical event that occurs when one object passes in front of another, blocking the latter from view. In our exercise, it refers to the planet obscuring a background star when viewed from Earth. The duration of this event helps in determining the size of the planet.
During the occultation, the background star disappears for a total of 26 minutes, which is crucial information to calculate the planet's radius. By observing how long the star is hidden, astronomers can deduce the size of the planet that is causing the obstruction. This duration, combined with the relative velocity, allows for precise calculations of planetary diameters and radii.
Calculating Distance and Speed
To find the radius of the planet, we need to understand the relationship between distance, speed, and time. The formula for distance is: \[ \text{Distance} = \text{Speed} \times \text{Time} \]In the problem, the planet's diameter can be found by multiplying the relative velocity (30 km/s) by the duration of the occultation in seconds (1560 seconds). This yields the diameter of the planet:
\[ \text{Diameter} = 30 \text{ km/s} \times 1560 \text{ seconds} = 46800 \text{ km} \]
To find the radius, we simply divide the diameter by 2:
\[ \text{Radius} = \frac{46800 \text{ km}}{2} = 23400 \text{ km} \]
The distance and speed measurements allow us to convert the observed occultation time into the physical dimensions of the planet, helping us to accurately deduce its radius. This is why understanding the concepts of distance, speed, and time is so critical in planetary science.

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Most popular questions from this chapter

Deep in the interiors of the giant planets, water is still a liquid even though the temperatures are tens of thousands of degrees above the boiling point of water. This can happen because a. the density inside the giant planets is so high. b. the pressure inside the giant planets is so high. c. the outer Solar System is so cold. d. space has very low pressure.

Jupiter is an oblate planet with an average radius of \(69,900 \mathrm{km},\) compared to Earth's average radius of \(6,370 \mathrm{km}\). a. Given that volume is proportional to the cube of the radius, how many Earth volumes could fit inside Jupiter? b. Jupiter is 318 times as massive as Earth. Show that Jupiter's average density is about one-fourth that of Earth's.

The Great Red Spot on Jupiter is a. a surface feature. b. a storm that has been raging for more than 300 years. c. caused by the interaction between the magnetosphere and Io. d. about the size of North America.

Jupiter's equatorial radius \(\left(R_{\text {Jup }}\right)\) is \(71,500 \mathrm{km}\), and its oblateness is \(0.065 .\) What is Jupiter's polar radius \(\left(R_{\text {Polar }}\right) ?\) (Oblateness is given by \(\left[R_{\text {Jup }}-R_{\text {Polar }}\right] / R_{\text {Jup }}\).)

Uranus occults a star at a time when the relative motion between Uranus and Earth is \(23.0 \mathrm{km} / \mathrm{s}\). An observer on Earth sees the star disappear for 37 minutes 2 seconds and notes that the center of Uranus passed directly in front of the star. a. On the basis of these observations, what value would the observer calculate for the diameter of Uranus? b. What could you conclude about the planet's diameter if its center did not pass directly in front of the star?

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