A satellite orbiting in an equatorial plane has a period from perigee-perigee of 12 hours. Given that the eccentricity is 0.002, calculate the semi-major axis. G = 6.672 x 10^-11 m... A satellite orbiting in an equatorial plane has a period from perigee-perigee of 12 hours. Given that the eccentricity is 0.002, calculate the semi-major axis. G = 6.672 x 10^-11 m^3/kg/s^2, M = 5974 x 10^24 kg, r = 6378.1414 km. Read more

Steps to Solve

1. Convert the period to seconds

We are given the period $T$ in hours, but we need it in seconds for the units to be consistent with the gravitational constant $G$. $$T = 12 \text{ hours} = 12 \text{ hours} \cdot 60 \frac{\text{minutes}}{\text{hour}} \cdot 60 \frac{\text{seconds}}{\text{minute}} = 43200 \text{ seconds}$$

2. Calculate the gravitational parameter

The gravitational parameter $\mu$ is the product of the gravitational constant $G$ and the mass of the central body $M$. $$\mu = GM = (6.672 \times 10^{-11} \text{ m}^3\text{/kg/s}^2) \cdot (5974 \times 10^{24} \text{ kg}) = 3.9859 \times 10^{14} \text{ m}^3\text{/s}^2$$

3. Apply Kepler's Third Law to find the semi-major axis

Kepler's Third Law relates the period $T$ of an orbit to its semi-major axis $a$ and the gravitational parameter $\mu$. $$T^2 = \frac{4\pi^2 a^3}{\mu}$$ Rearrange to solve for $a$: $$a^3 = \frac{T^2 \mu}{4\pi^2}$$ $$a = \sqrt[3]{\frac{T^2 \mu}{4\pi^2}}$$ Substitute the known values: $$a = \sqrt[3]{\frac{(43200 \text{ s})^2 \cdot (3.9859 \times 10^{14} \text{ m}^3\text{/s}^2)}{4\pi^2}}$$ $$a = \sqrt[3]{\frac{7.4499 \times 10^{23} \text{ m}^3}{39.4784}}$$ $$a = \sqrt[3]{1.8871 \times 10^{22} \text{ m}^3}$$ $$a = 2.6637 \times 10^{7} \text{ m}$$

4. Convert the semi-major axis to kilometers

Convert the semi-major axis from meters to kilometers.

$$ a = 2.6637 \times 10^{7} \text{ m} = 2.6637 \times 10^{7} \text{ m} \cdot \frac{1 \text{ km}}{1000 \text{ m}} = 26637 \text{ km} $$