The acceleration of an object in circular motion is given by the formula a = v^2 / r, where v is the velocity and r is the radius. However, in this case, we are not given the velocity.

But we know that the Earth completes one orbit around the Sun in one year, and the distance it travels is the circumference of the orbit, which is 2πr. So we can calculate the velocity as the total distance divided by the time it takes.

First, let's calculate the velocity:

v = 2πr / T

where:
r = 1.496 x 10^11 m (the radius)
T = 1 year = 365.25 days = 365.25 * 24 * 60 * 60 seconds (the time for one orbit)

v = 2π * 1.496 x 10^11 m / (365.25 * 24 * 60 * 60 s)
v ≈ 29,785 m/s

Now, we can calculate the acceleration:

a = v^2 / r
a = (29,785 m/s)^2 / 1.496 x 10^11 m
a ≈ 0.00593 m/s^2

So, the acceleration of the Earth with respect to the Sun is approximately 0.00593 m/s^2, rounded to three significant digits.

Question

The acceleration of an object in circular motion is given by the formula a = v^2 / r, where v is the velocity and r is the radius. However, in this case, we are not given the velocity.

But we know that the Earth completes one orbit around the Sun in one year, and the distance it travels is the circumference of the orbit, which is 2πr. So we can calculate the velocity as the total distance divided by the time it takes.

First, let's calculate the velocity:

v = 2πr / T

where:
r = 1.496 x 10^11 m (the radius)
T = 1 year = 365.25 days = 365.25 * 24 * 60 * 60 seconds (the time for one orbit)

v = 2π * 1.496 x 10^11 m / (365.25 * 24 * 60 * 60 s)
v ≈ 29,785 m/s

Now, we can calculate the acceleration:

a = v^2 / r
a = (29,785 m/s)^2 / 1.496 x 10^11 m
a ≈ 0.00593 m/s^2

So, the acceleration of the Earth with respect to the Sun is approximately 0.00593 m/s^2, rounded to three significant digits.

Knowee AI · Accepted Answer

The acceleration of an object in circular motion is given by the formula a = v^2 / r, where v is the velocity and r is the radius. However, in this case, we are not given the velocity.

But we know that the Earth completes one orbit around the Sun in one year, and the distance it travels is the circumference of the orbit, which is 2πr. So we can calculate the velocity as the total distance divided by the time it takes.

First, let's calculate the velocity:

v = 2πr / T

where:
r = 1.496 x 10^11 m (the radius)
T = 1 year = 365.25 days = 365.25 * 24 * 60 * 60 seconds (the time for one orbit)

v = 2π * 1.496 x 10^11 m / (365.25 * 24 * 60 * 60 s)
v ≈ 29,785 m/s

Now, we can calculate the acceleration:

a = v^2 / r
a = (29,785 m/s)^2 / 1.496 x 10^11 m
a ≈ 0.00593 m/s^2

So, the acceleration of the Earth with respect to the Sun is approximately 0.00593 m/s^2, 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. The acceleration of the earth with respect to the sun is ______ m/s/s. Round to 3 significant digits.

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