Question

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth's rotation). Calculate the radius of such an orbit based on the data for the moon

Answer 1

Answer:

r = 4.24x10⁴ km.  

     

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                                                        

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$                            

$r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}$      

$r_{1} = 4.24 \cdot 10^{4} km$      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

Answer 2

The radius of orbit of Earth based on the data for the moon is $4.24 \times 10^{4} \;\rm km$.

Given data:

The orbital period of Earth Satellite is, T = 1 day.

We can use the Kepler's law to find the radius of orbit. Then applying the mathematical formula of Kepler's Law as,

$\dfrac{T^{2}}{T'^{2}}=\dfrac{r^{2}}{r'^{2}}$

Here,

T' is the orbital period of moon. (T' = 0.07481 y)

r is the radius of Earth orbit.

r' is the radius of moon.

Solve for r as,

$r = r' \sqrt{\dfrac{T^{2}}{T'^{2}}}\\\\r = 3.85 \times 10^{5} \times \sqrt{\dfrac{1^{2}}{({0.07481 \times 365)^{2}}}} \\\\\\r = 4.24 \times 10^{4} \;\rm km$

Thus, we can conclude that the radius of orbit of Earth based on the data for the moon is $4.24 \times 10^{4} \;\rm km$.

Learn more about the Kepler's law here:

https://brainly.com/question/1086445