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Sagot :
Solution :
The semi-major axis of any terrestrial body may be defined as equal to mean distance of the body, so that the cube of its mean distance of the planet is directly proportional to the square of the sidereal period.
It is given by Kepler's third law,
[tex]$p^2 = a^3$[/tex]
where p is the orbital period
a is the semi major axis
Given :
a = [tex]$5 \times 10^{11}$[/tex] m
Therefore, a = [tex]$5 \times 10^{11}$[/tex] m
= 3.3 AU
So, [tex]$p^2=3.3^{3}$[/tex]
[tex]$p^2=35.93$[/tex]
p = 6 years (approximately)
or p = [tex]$1.8 \times 10^8$[/tex] seconds ( since, 1 year = [tex]$3.1 \times 10^7$[/tex] seconds )
The orbital period of an asteroid will be approximately 6 years when the factors given in the statement are taken into consideration, however the distance of the asteroid from the planet is not given.
The asteroid will have semi major axis of 5.0E11 in meters and the calculation is done by taking distance of such asteroid from the planet and this will be done by applying Kepler's third law.
- By keeping in mind the Kepler's third law we know that the orbital period squared is the cube of the semimajor axis. This can be represented by the following values.
[tex]p^{2} = a^{3}[/tex]
- The final output will be obtained as that the orbital period is 6 years after we put the values like a= 5.0E11 meters in the formula above which is the Kepler's third law.
- The values obtained will be halved because we want to achieve the semi major axis and the axis is exactly the double of the semi major axis we have obtained.
Hence, the orbital period axis of an asteroid with semi major axis of the given value will be approximately close to six years and some seconds.
To know more about Asteroids, refer to the links below.
https://brainly.com/question/9676376
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