If the radius of an asteroid's uniform circular motion around the sun is four times the radius of the earth's revolution, then the cycle of the asteroid's motion is A4 B.6 c.8 d.9

If the radius of an asteroid's uniform circular motion around the sun is four times the radius of the earth's revolution, then the cycle of the asteroid's motion is A4 B.6 c.8 d.9

The answer is C, 8 years··
The rotational speed of planets in the solar system is basically the same·
It is the length of orbit that affects the revolution period
Perimeter = 2 * radius * pi
So~~
The orbit length of small stars is: 2 * (radius * 4) * PI
That is: eight times the length of the earth's orbit,
So the answer is - eight years

There is a planet much smaller than the earth that revolves around the sun with a period of 288 years. If we regard it and the earth's orbit around the sun as circles, ask him the distance from the sun
How many times is the distance between the earth and the sun

G is the constant of gravitation, M is the mass of the sun, M is the mass of the planet, R is the distance from the planet to the sun. From (GMM / R ^ 2) = (4 π ^ 2 * r * m / T ^ 2), R ^ 3 = (GMT ^ 2 / 4 π ^ 2) r = triple root sign (GMT ^ 2 / 4 π ^ 2), let the sun earth distance be r, and the sun planet distance be r; if the earth revolution period is t, and the planet revolution period is t, then R / r = three

The distance between a planet and the sun is 8 times of the average distance between the earth and the sun
Kepler's third law (periodic law): the ratio of the third power of the semimajor axis of the orbits of all planets to the second power of the revolution period is equal.
The formula is: A ^ 3 / T ^ 2 = K
A = semi major axis of planetary orbit
T = planetary revolution period
a^3/T^2＝（8a）^3/x^2
624 years
The revolution of the earth takes one year.
I don't understand this passage
a^3/T^2＝（8a）^3/x^2

It means that the square of the earth's cycle equals to the third power of the distance between the sun and the earth