Question

The mean distance of Earth from the Sun is \(9.29 \times 10^{7}\) miles. Assuming that the orbit of Earth around the Sun is circular and that 1 revolution takes 365 days, find the linear speed of Earth. Express your answer in miles per hour.

Short answer

The linear speed of Earth is approximately 66,700 miles per hour.

Step-by-step solution

- Calculate the circumference of the orbit

The circumference of Earth's orbit can be found using the formula for the circumference of a circle: \[ C = 2 \times \pi \times r \] where \( r \) is the radius (mean distance from the Sun). Here, \[ r = 9.29 \times 10^{7} \text{miles} \] so the circumference \( C \) is: \[ C = 2 \times \pi \times 9.29 \times 10^{7} \approx 5.84 \times 10^{8} \text{miles} \]

- Calculate the total time in hours

The given time for one revolution is 365 days. First, convert this time into hours:\[ 365 \text{days} \times 24 \text{hours/day} = 8760 \text{hours} \]

- Calculate the linear speed

The linear speed \( v \) is the total distance traveled divided by the total time. Using the circumference from Step 1 and the total time in hours from Step 2, the linear speed is: \[ v = \frac{5.84 \times 10^{8} \text{miles}}{8760 \text{hours}} = 6.67 \times 10^{4} \text{miles/hour} \]

Key concepts

circumference of a circle

To find the total distance Earth travels in its orbit around the Sun, we need to calculate the circumference of the orbit. The orbit is assumed to be a perfect circle for simplicity. The formula to calculate the circumference of a circle is: \[ C = 2 \times \pi \times r \] Where \( C \) is the circumference, and \( r \) is the radius (mean distance from the Sun). Here, the given mean distance is \( 9.29 \times 10^{7} \) miles. When we plug this value into the formula, we get: \[ C = 2 \times \pi \times 9.29 \times 10^{7} \approx 5.84 \times 10^{8} \text{miles} \] This computation gives us the total distance Earth travels in one complete orbit around the Sun.

unit conversion

In order to find the linear speed, we need to convert time units appropriately. The problem states that one complete revolution of Earth around the Sun takes 365 days. To calculate speed in miles per hour, we have to express this time in hours. We know there are 24 hours in a day, so we convert days into hours: \[ 365 \text{ days} \times 24 \text{ hours/day} = 8760 \text{ hours} \] Now we have the total time required for one revolution in hours. This helps us to find the linear speed accurately.

linear speed formula

Linear speed is the measure of how fast an object is moving along a path. The formula to calculate linear speed \( v \) is: \[ v = \frac{\text{Total Distance}}{\text{Total Time}} \] In our scenario, the total distance is the circumference of Earth's orbit, which we've calculated as approximately \( 5.84 \times 10^{8} \) miles. The total time taken for one revolution is 8760 hours. Plugging these values into our formula gives: \[ v = \frac{5.84 \times 10^{8} \text{ miles}}{8760 \text{ hours}} \approx 6.67 \times 10^{4} \text{ miles/hour} \] Thus, the linear speed of Earth in its orbit is approximately \( 6.67 \times 10^{4} \) miles per hour.

mean distance from the Sun

The mean distance from the Sun is essentially the average distance over the course of the Earth's orbit. For simplicity in calculations, we consider it to be a constant value. In this problem, the mean distance is provided as \( 9.29 \times 10^{7} \) miles. This distance is crucial for determining the circumference of the orbital path using the circle circumference formula. Understanding this average distance helps in modeling the Earth's orbit as a circle for simplifying our calculations.

orbital period of Earth

The orbital period of Earth is the time it takes for Earth to complete one full revolution around the Sun. This period is given as 365 days in this problem. Converting this into more convenient units like hours (for our speed calculation) involves simple unit conversion steps. Knowing the orbital period allows us to determine how much time it takes for Earth to travel the distance of its orbit, which in turn is essential for calculating the orbital speed.