/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 What is the acceleration of the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the acceleration of the Earth in its orbit? (Assume the orbit is circular.)

Short Answer

Expert verified
Answer: The approximate acceleration of Earth in its orbit around the Sun is 5.932 × 10^-3 m/s².

Step by step solution

01

Determine the radius of the Earth's orbit

The average distance from the Earth to the Sun, which represents the radius of the Earth's orbit, is approximately 93 million miles or 1 Astronomical Unit (AU). To use the centripetal acceleration formula, we should convert this distance to meters: 1 AU ≈ 1.496 × 10^11 meters
02

Find the Earth's orbital speed

The Earth takes about 365.25 days to complete one orbit around the Sun. To find the speed (v), we can use the formula: v = (2πr) / T, where r is the radius of the Earth's orbit and T is the period of revolution (time taken for one complete orbit). First, convert the time to seconds: 365.25 days × 24 hours/day × 60 minutes/hour × 60 seconds/minute ≈ 3.156 × 10^7 seconds Now, plug in the values to find the speed: v = (2π × 1.496 × 10^11 meters) / (3.156 × 10^7 seconds) ≈ 2.978 × 10^4 m/s
03

Calculate the centripetal acceleration

Now that we have the radius of the Earth's orbit (r) and its orbital speed (v), we can use the centripetal acceleration formula: a = v^2 / r Plug in the values to find the acceleration: a = (2.978 × 10^4 m/s)^2 / (1.496 × 10^11 meters) ≈ 5.932 × 10^-3 m/s^2
04

Express the result

The acceleration of the Earth in its orbit, assuming it is circular, is approximately 5.932 × 10^-3 m/s².

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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

A carousel at a carnival has a diameter of \(6.00 \mathrm{~m}\). The ride starts from rest and accelerates at a constant angular acceleration to an angular speed of 0.600 rev/s in \(8.00 \mathrm{~s}\). a) What is the value of the angular acceleration? b) What are the centripetal and angular accelerations of a seat on the carousel that is \(2.75 \mathrm{~m}\) from the rotation axis? c) What is the total acceleration, magnitude and direction, \(8.00 \mathrm{~s}\) after the angular acceleration starts?

The motor of a fan turns a small wheel of radius \(r_{\mathrm{m}}=\) \(2.00 \mathrm{~cm} .\) This wheel turns a belt, which is attached to a wheel of radius \(r_{f}=3.00 \mathrm{~cm}\) that is mounted to the axle of the fan blades. Measured from the center of this axle, the tip of the fan blades are at a distance \(r_{\mathrm{b}}=15.0 \mathrm{~cm} .\) When the fan is in operation, the motor spins at an angular speed of \(\omega=1200\). rpm. What is the tangential speed of the tips of the fan blades?

A point on a Blu-ray disc is a distance \(R / 4\) from the axis of rotation. How far from the axis of rotation is a second point that has, at any instant, a linear velocity twice that of the first point? a) \(R / 16\) b) \(R / 8\) c) \(R / 2\) d) \(R\)

Suppose you are riding on a roller coaster, which moves through a vertical circular loop. Show that your apparent weight at the bottom of the loop is six times your weight when you experience weightlessness at the top, independent of the size of the loop. Assume that friction is negligible.

You are holding the axle of a bicycle wheel with radius \(35.0 \mathrm{~cm}\) and mass \(1.00 \mathrm{~kg}\). You get the wheel spinning at a rate of 75.0 rpm and then stop it by pressing the tire against the pavement. You notice that it takes \(1.20 \mathrm{~s}\) for the wheel to come to a complete stop. What is the angular acceleration of the wheel?

See all solutions

Recommended explanations on Physics Textbooks

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.