Respuesta :
The formula that will be used to solve this question is:
P = 2π sqrt[ a^3 / (GM) ] where:
GM = 3.986004418 x 10^-14 m^3 sec^-2
a = 6378000 + 300000 = 6678000 meters
Substitute in the equation to get the period as follows:
P = 2π sqrt [( 6678000 )^3 / (3.986004418 x 10^-14) ]
P = 5431.0 seconds
P = 2π sqrt[ a^3 / (GM) ] where:
GM = 3.986004418 x 10^-14 m^3 sec^-2
a = 6378000 + 300000 = 6678000 meters
Substitute in the equation to get the period as follows:
P = 2π sqrt [( 6678000 )^3 / (3.986004418 x 10^-14) ]
P = 5431.0 seconds
The orbital period of a spacecraft at an orbit of 300 kilometers above the Earth's surface is around one hour and a half.
Further explanation
Objects, due to their inherent mass, tend to atract each other due to the laws of gravitation. In short words, objects with a huge ammount of mass atract other objects which are less massive, that is why we are all attracted towards the Earth (which is an incredible massive body).
In general this phenomenon is better seen at the astrological level, like planets attracting their moons or asteroids (this is the case of our problem). Even though, planets attract other bodies which are nearby, this doesn't mean that they will ever come into contact, this is the case for object which orbit those planets.
An orbit is a path which a certain body follows around a more massive object, like the moon orbiting the Earth, or the Earth orbiting around the Sun. These orbits are periodic, meaning they happen continuously over time, over and over again, the same way all times. By this reason, they have a period (which is the duration of such orbit).
At this point, there is a background theory that is necessary to derive the equation we're going to use to compute the orbital period, but it escapes the scope of this problem, so we're just going to use the equation. The equation to compute the orbital period is:
[tex]T= 2 \cdot \pi \cdot \sqrt{\frac{a^3}{\mu}}[/tex]
Where a is the distance between the orbiting object (in this case, the spacecraft) and the center of the orbited object (in this case the Earth), and [tex]\mu[/tex] is a constant dependent on the orbited object, it's called Standard Gravitational Parameter, for the Earth it has a value of [tex]3.986 \cdot 10^{14} \cdot \frac{m^3}{s^2}[/tex].
Since the radius of the Earth is 6371 Km, then a would be 6671 Km. Plugging values on the formula, and making sure to apply the correct units (notice how a is expressed in Km and [tex]\mu[/tex] has units of meter cube), we get that the orbital period is 5422 seconds, which is around one hour and half.
Learn more
- Moon orbiting around the Earth: https://brainly.com/question/6502290
- Orbit of Neptune: https://brainly.com/question/9708010
- Kepler's Laws: https://brainly.com/question/929044
Keywords
Orbit, gravitation, Earth, attraction laws
