The time a planet takes to revolve around the Sun is given by Kepler's Third Law, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Mathematically, this can be expressed as T^2 ∝ r^3, where T is the orbital period and r is the distance from the Sun.

Given that the original orbital period T1 is 200 days and the original distance r1 is reduced to one fourth, the new distance r2 is r1/4.

We want to find the new orbital period T2.

From Kepler's Third Law, we have the relationship:

(T1/T2)^2 = (r1/r2)^3

Substituting the given values, we get:

(200/T2)^2 = (r1/(r1/4))^3

Solving this equation gives:

(200/T2)^2 = 4^3

Taking the square root of both sides gives:

200/T2 = 4*2

Solving for T2 gives:

T2 = 200/(4*2) = 25 days

So, if the distance of the planet from the Sun is reduced to one fourth of the original distance, it will take 25 days to complete one revolution.

Question

The time a planet takes to revolve around the Sun is given by Kepler's Third Law, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Mathematically, this can be expressed as T^2 ∝ r^3, where T is the orbital period and r is the distance from the Sun.

Given that the original orbital period T1 is 200 days and the original distance r1 is reduced to one fourth, the new distance r2 is r1/4.

We want to find the new orbital period T2.

From Kepler's Third Law, we have the relationship:

(T1/T2)^2 = (r1/r2)^3

Substituting the given values, we get:

(200/T2)^2 = (r1/(r1/4))^3

Solving this equation gives:

(200/T2)^2 = 4^3

Taking the square root of both sides gives:

200/T2 = 4*2

Solving for T2 gives:

T2 = 200/(4*2) = 25 days

So, if the distance of the planet from the Sun is reduced to one fourth of the original distance, it will take 25 days to complete one revolution.

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