Orbital Velocity of the Earth

Classical Mechanics Level 2

The period of Earth's orbit around the sun is 365 days. Given that the mass of the sun is about $2 \times 10^{30} \text{ kg}$ , compute the velocity to the nearest 1000 meters per second of the Earth as it orbits the sun. Approximate the orbit of the Earth around the sun as circular.

Useful constant : Newton's gravitational constant is $6.67\times 10^{-11} \text{ N} \cdot \text{m}^2 / \text{kg}^2$ .

25000 1000 0 30000

1 solution

Matt DeCross
May 13, 2016

Using $T = 2 \pi \sqrt{\frac{r^3}{GM}}$ , from the period $T$ , $G$ , and $M$ as given one computes that the radius of orbit of the Earth about the sun is:

$r = 1.5 \times 10^{11} \text{ m}.$

Now using the fact that the orbital velocity is related to the period and radius of orbit by:

$v = \frac{2\pi r}{T},$

plugging in the given numbers one finds:

$v = 29886 \text{ m}/\text{s} \approx 30000 \text{ m}/\text{s}.$

very nice solution

Arun Kumar - 2 years, 4 months ago

Orbital Velocity of the Earth

1 solution

0 pending reports