Questions: The Earth orbits the sun in 365 days. What is the tangential speed, in m / s, of the earth in orbit? The average sun-earth distance is 1.50 x 10^11 m.

Transcript text: The Earth orbits the sun in 365 days. What is the tangential speed, in $\mathrm{m} / \mathrm{s}$, of the earth in orbit? The average sun-earth distance is $1.50 \times 10^{11} \mathrm{~m}$.

Solution

Solution Steps

Step 1: Determine the Circumference of Earth's Orbit

The Earth orbits the Sun in a nearly circular path. The circumference $ C $ of a circle is given by the formula: \[ C = 2\pi r \] where $ r $ is the radius of the orbit. Given that the average Sun-Earth distance is $ 1.50 \times 10^{11} \, \text{m} $, we have: \[ C = 2\pi \times 1.50 \times 10^{11} \, \text{m} \]

Step 2: Calculate the Circumference

Substitute the given value into the formula: \[ C = 2\pi \times 1.50 \times 10^{11} = 3.00\pi \times 10^{11} \, \text{m} \]

Step 3: Determine the Time Period of Earth's Orbit

The Earth completes one orbit around the Sun in 365 days. To convert this period into seconds, use the conversion: \[ 1 \, \text{day} = 24 \times 60 \times 60 \, \text{seconds} \] Thus, the time period $ T $ in seconds is: \[ T = 365 \times 24 \times 60 \times 60 \, \text{s} \]

Step 4: Calculate the Tangential Speed

The tangential speed $ v $ is given by the formula: \[ v = \frac{C}{T} \] Substitute the values for $ C $ and $ T $: \[ v = \frac{3.00\pi \times 10^{11}}{365 \times 24 \times 60 \times 60} \, \text{m/s} \]

Step 5: Compute the Tangential Speed

Calculate the numerical value: \[ v \approx \frac{3.00 \times 3.1416 \times 10^{11}}{31,536,000} \, \text{m/s} \approx 29,865 \, \text{m/s} \]