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

Short Answer

Expert verified
Answer: The acceleration of the Earth in its orbit around the Sun is approximately 5.95 x 10^-3 m/s².

Step by step solution

01

Identify known values

The values we know: - Radius of Earth's orbit (r): Approximately 150 million kilometers (1.5 x 10^11 meters) - Time taken to complete one orbit (T): Approximately 1 year or 365.25 days
02

Calculate angular velocity (ω)

The angular velocity (ω) can be found using the formula: ω = 2π/T. Since the period (T) is given in days, we should first convert the period from days to seconds. 1 day = 24 hours/day × 60 min/hour × 60 sec/min = 86400 seconds T = 365.25 days × 86400 seconds/day = 3.15576 × 10^7 seconds Now, we can find the angular velocity ω: ω = 2π/T = 2π / (3.15576 × 10^7 s) ≈ 1.991 × 10^-7 rad/s
03

Calculate centripetal acceleration (a)

Now we have the angular velocity ω and the radius r, we can calculate the centripetal acceleration (a) using the formula: a = ω²r a = (1.991 × 10^-7 rad/s)² * (1.5 × 10^11 m) a ≈ (3.9686321 × 10^-14) * (1.5 × 10^11 m) a ≈ 5.9529 × 10^-3 m/s² Hence, the acceleration of the Earth in its orbit is approximately 5.95 x 10^-3 m/s².

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Key Concepts

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

Centripetal Acceleration
Understanding centripetal acceleration is crucial as it pertains to the force required to keep an object moving in a circle at constant speed. This inward force ensures an object's circular path, counteracting the inertia that would otherwise cause the object to move in a straight line. For objects in circular motion, such as the Earth in its orbit around the Sun, this acceleration is always directed towards the center of the circle.

In the context of Earth's orbit, centripetal acceleration can be comprehended by applying the formula:
\( a_c = \frac{v^2}{r} \), where \( a_c \) is the centripetal acceleration, \( v \) is the linear velocity, and \( r \) is the radius of the circular path.

However, since we often start with angular velocity, another form is used, expressed as:\( a_c = \omega^2 \times r \), where \( \omega \) is the angular velocity. This derivation shows how centripetal acceleration depends on the square of the angular velocity and the radius of the motion's path. As seen in the example of Earth's orbit, the actual computation involves squaring the angular velocity and multiplying by the radius to find the relatively small acceleration that keeps Earth in its annual journey around the Sun.
Angular Velocity

Connecting Angular Motion to Time

Angular velocity is a measure of how quickly an object rotates or revolves around a point or axis. Specifically, in the Earth's case, it details the speed at which our planet rotates around the Sun. The angular velocity, \( \omega \), is tied directly to the time it takes to complete one full orbit, known as the orbital period.

The formula to calculate angular velocity is:\( \omega = \frac{2\pi}{T} \), where \( T \) is the period of motion. For Earth, this period is one year, a known value that can be precisely converted into seconds to fit the formula's requirements. With angular velocity, we gain a fundamental understanding of the dynamic relationship between time and motion in circular paths and it is key to connecting the concept of time with physical motion in a cyclical context.
Circular Motion
Circular motion is the movement of an object along the circumference of a circle or rotation along a circular path. It's a ubiquitous concept, from the orbits of celestial bodies to the electrons spinning around the nucleus of an atom. This motion requires a constant force directed toward the center of the circle, also known as a 'centripetal' force, which keeps the object in its curving path instead of moving off in a straight line due to inertia.

When we observe the Earth's orbit, we're witnessing circular motion on a grand scale. Calculating the forces and accelerations associated with this motion can provide insights into foundational physics principles. The Earth's nearly circular orbit offers a prime example of how these concepts manifest in our universe, demonstrating the beautiful symphony of physics in celestial mechanics.

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Most popular questions from this chapter

Consider a \(53.0-\mathrm{cm}\) -long lawn mower blade rotating about its center at \(3400 . \mathrm{rpm}\) a) Calculate the linear speed of the tip of the blade. b) If safety regulations require that the blade be stoppable within \(3.00 \mathrm{~s}\), what minimum angular acceleration will accomplish this? Assume that the angular acceleration is constant.

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