Question

The radius of the Earth's circular orbit round the Sun is 149×10⁶ km. The Earth takes 365 days to orbit the Sun. The tangential velocity of the Earth is_____ km/hour. (π = 3.14) (Round off to one decimal place)

Answer 1

Correct Answer: 106817

$\text{1. Define Variables and Formula}$

The tangential velocity ($v$) of an object moving in a circular path is the distance traveled (circumference of the orbit, $C$) divided by the time taken ($T$, the orbital period).

$$\text{Velocity } (v) = \frac{\text{Distance}}{\text{Time}} = \frac{2\pi R}{T}$$

$\text{Variables}$

Orbital Radius ($R$): $149 \times 10^6 \text{ km}$

Orbital Period ($T$): $365 \text{ days}$

Constant ($\pi$): $3.14$

$\text{Required Unit Conversion}$

The final answer must be in $\mathbf{\text{km/hour}}$. We must convert the orbital period $T$ from days to hours.

$$T = 365 \text{ days} \times \frac{24 \text{ hours}}{1 \text{ day}} = 8760 \text{ hours}$$

$\text{2. Calculate the Tangential Velocity}$

Substitute the converted values into the velocity formula:

$$v = \frac{2 \pi R}{T}$$

$$v = \frac{2 \times 3.14 \times (149 \times 10^6 \text{ km})}{8760 \text{ hours}}$$

$$v = \frac{6.28 \times 149 \times 10^6}{8760} \text{ km/hour}$$

$$v = \frac{935.72 \times 10^6}{8760} \text{ km/hour}$$

$\text{Calculation}$

$$v \approx 106817.3516 \text{ km/hour}$$

$\text{3. Rounding}$

Rounding the result to one decimal place:

$$v \approx 106817.4 \text{ km/hour}$$