Enter the required parameters and the calculator will estimate the orbital period instantly.
“The time that an item or satellite tv for pc takes in final touch of 1 whole rotation around some other large item is called the orbital period”
The system this is used for the low earth orbit is as follows:
\(T = \sqrt(\dfrac{3\pi}{G*P})\)
To calculate the orbital length of a binary star machine, use the subsequent method::
\(T_b=2\pi \sqrt(\dfrac{a^{3}}{G(M_1 + M_2})\)
Go through the subsequent steps to calculate the orbital length::
Suppose the central body density is \(5.51 \, \text{g/cm}^3\), how do we find the orbital period for a low Earth orbit?
Solution:
Given:
Convert density into \(\text{kg/m}^3\):
\[ P = 5.51 \times 1000 \times \left(\frac{1}{100}\right)^3 = 5510 \, \text{kg/m}^3 \]
Gravitational constant (\(G\)):
\[ G = 6.67 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \]
Orbital period equation:
\[ T = \sqrt{\frac{3 \pi}{G \cdot P}} \]
Substitute the values:
\[ T = \sqrt{\frac{3 \times 3.14159}{(6.67 \times 10^{-11}) \cdot 5510}} \]
Calculate step-by-step:
Convert seconds to hours:
\[ T = \frac{5063.16}{3600} \approx 1.406 \, \text{hours} \]
Answer: The orbital period is approximately \(1.406 \, \text{hours}\).