To find the linear speed of the Earth in its orbit, we need to use the formula for the speed of an object moving in a circular path, which is v = 2πr/T, where:

- v is the linear speed
- r is the radius of the circle (the average orbital radius for the Earth)
- T is the period of the motion (the time it takes for the Earth to complete one orbit around the Sun)

We know that r = 1.496 x 10^11 m and T = 1 year = 365.25 days = 31557600 seconds (taking into account leap years).

Substituting these values into the formula, we get:

v = 2π(1.496 x 10^11 m) / 31557600 s

Now, we just need to calculate the value:

v ≈ 2 * 3.14159 * (1.496 x 10^11 m) / 31557600 s ≈ 29800 m/s

So, the Earth's linear speed in its orbit is approximately 29800 m/s, rounded to three significant digits.

Question

To find the linear speed of the Earth in its orbit, we need to use the formula for the speed of an object moving in a circular path, which is v = 2πr/T, where:

- v is the linear speed
- r is the radius of the circle (the average orbital radius for the Earth)
- T is the period of the motion (the time it takes for the Earth to complete one orbit around the Sun)

We know that r = 1.496 x 10^11 m and T = 1 year = 365.25 days = 31557600 seconds (taking into account leap years).

Substituting these values into the formula, we get:

v = 2π(1.496 x 10^11 m) / 31557600 s

Now, we just need to calculate the value:

v ≈ 2 * 3.14159 * (1.496 x 10^11 m) / 31557600 s ≈ 29800 m/s

So, the Earth's linear speed in its orbit is approximately 29800 m/s, rounded to three significant digits.

Knowee AI · Accepted Answer

To find the linear speed of the Earth in its orbit, we need to use the formula for the speed of an object moving in a circular path, which is v = 2πr/T, where:

- v is the linear speed
- r is the radius of the circle (the average orbital radius for the Earth)
- T is the period of the motion (the time it takes for the Earth to complete one orbit around the Sun)

We know that r = 1.496 x 10^11 m and T = 1 year = 365.25 days = 31557600 seconds (taking into account leap years).

Substituting these values into the formula, we get:

v = 2π(1.496 x 10^11 m) / 31557600 s

Now, we just need to calculate the value:

v ≈ 2 * 3.14159 * (1.496 x 10^11 m) / 31557600 s ≈ 29800 m/s

So, the Earth's linear speed in its orbit is approximately 29800 m/s, rounded to three significant digits.

The earth’s orbit about the sun is almost a circle. The average orbital radius for the earth is 1.496 x 10^11 m. Its linear speed in its orbit is _____ m/s. Round to three significant digits.

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