Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 42

Ceres has a diameter of \(975 \mathrm{km}\) and a period of about 9 hours. What is the rotational speed of a point on the surface of this dwarf planet?

Short Answer

Expert verified
Approximately 340.37 km/hour.

Step by step solution

01

Determine the Circumference

First, calculate the circumference of Ceres using its diameter. The formula for the circumference of a circle is \(C = \pi d\), where \(d\) is the diameter. Given that Ceres has a diameter of \(975 \, \mathrm{km}\): \[C = \pi \times 975 \, \mathrm{km} \approx 3063.36 \, \mathrm{km}\].
02

Convert the Period to Hours

The period of rotation is already given as about 9 hours. This means that a point on the surface completes one full rotation (the circumference calculated in Step 1) in 9 hours.
03

Calculate Rotational Speed

To find the rotational speed, divide the circumference by the rotational period. This can be done using the formula \( v = \frac{C}{T}\), where \( v\) is the rotational speed, \(C\) is the circumference, and \(T\) is the period. Thus: \[v = \frac{3063.36 \, \mathrm{km}}{9 \, \mathrm{hours}} \approx 340.37 \, \mathrm{km/hour}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

dwarf planet
Ceres is classified as a dwarf planet. But what does that mean exactly? A dwarf planet is a celestial body that orbits the sun, similar to the major planets in our solar system.
However, dwarf planets are distinct because they are not large enough to clear their orbital path of other debris. They are also not moons attached to other planets. Ceres, located in the asteroid belt between Mars and Jupiter, is an excellent example.
A typical feature of many dwarf planets is their relatively small size compared to regular planets. Ceres, for instance, has a diameter of about 975 kilometers, which is quite small compared to Earth's diameter of about 12,742 kilometers.
rotational speed
The rotational speed of a celestial body, like a planet or dwarf planet, refers to how quickly it spins around its axis. For Ceres, a point on its surface completes one full rotation in approximately 9 hours.
To understand rotational speed, think of it as how fast something is moving over the surface of the rotating object. The faster a point on the surface moves, the higher the rotational speed.
In calculations, such speed is usually measured in kilometers per hour (km/h). For Ceres, we will calculate how fast a point on its surface moves within one full rotation.
circumference calculation
The circumference of a circle or a sphere is the total distance around it. For a dwarf planet like Ceres, we need to know the circumference to calculate its rotational speed. We use the formula:
\(C = \pi d\) where \(C\) is the circumference and \(d\) is the diameter. Given Ceres' diameter of 975 kilometers, the circumference calculation looks like this:
\[C = \pi \times 975 \approx 3063.36 \text{ km}\].
This value tells us how far a point on the equator of Ceres would travel in one full rotation.
period of rotation
The period of rotation is the time it takes for a celestial body to complete one full spin around its axis. For Ceres, this period is approximately 9 hours. This means that if you were standing on the surface of Ceres, you would experience a complete day in just 9 hours.
Understanding the period of rotation is crucial because it helps us determine how fast Ceres spins. The shorter the period, the faster the rotational speed.
Think of the period of rotation like a clock cycle; every 9 hours, a point on Ceres' surface returns to its starting position.
velocity formula
Finally, to find the rotational speed, we utilize the velocity formula:\[v = \frac{C}{T}\] where \(v\) represents the velocity or rotational speed, \(C\) is the circumference, and \(T\) is the period of rotation.
For Ceres, we calculated the circumference (\(C\)) as approximately 3063.36 kilometers and the period (\(T\)) as 9 hours. Plugging these values into the formula gives us:
\[v = \frac{3063.36 \text{ km}}{9 \text{ hours}} \approx 340.37 \text{ km/hour}\].
This means that a point on Ceres' equator moves at about 340.37 kilometers per hour, giving us a clear idea of its rotational speed.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is zodiacal light, and what is its source?

Meteorites can provide information about a. the early composition of the Solar System. b. the composition of asteroids. c. other planets. d. all of the above

Most meteors that you see began as meteoroids the size of a. tiny dust grains. b. a pencil. c. a car. d. a house.

The three types of meteorites come from different parts of their parent bodies. Stony-iron meteorites are rare because a. they are hard to find. b. the volume of a differentiated body that has both stone and iron is small. c. there is very little iron in the Solar System. d. the magnetic field of the Sun attracts the iron.

In 1910 , Earth passed directly through the tail of Halley's Comet. Among the various gases in the tail was hydrogen cyanide, deadly to humans. Yet nobody became ill from this event. Why?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Fluids

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Physical Quantities And Units

Read Explanation

Famous Physicists

Read Explanation

Engineering Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free