If all the linear dimensions of the solar system were reduced in some proportion, then what will be the length of Earth’s year?

###### Solution.

The gravitational force between the Earth and the Sun is in the radial direction and its magnitude is given by the Newton’s equation

F = G*M*m/r^2

where G is the gravitational constant, M and m are the masses of the Sun and the Earth respectively and r is the radius of the orbit.

In case of the circular motion the net force equals mass times acceleration (Newton’s Second Law of Motion), where acceleration could be calculated by ω^2*r or (2π/T)^2*r , where ω – the angular velocity and T – the period of one rotation.

Thereby,

F = m*a = m*ω^2*r = m*(2π/T)^2*r

from the last formula and the Newton’s Gravitational Equation we could get

G*M*m/r^2 = m*(2π/T)^2*r

or

G*M = (2π/T)^2 * r^3

Represent the mass of the Sun as a function of R and ρ, where R – the radius of the Sun and ρ – the density of the Sun.

M = ρ * (4/3*π) * R^3

Hence,

G * ρ * (4/3*π) * R^3 = (2π/T)^2 * r^3

or

T^2 = (3π/(Gρ))* (r/R)^3

From the last equation, it is clear that the length of Earth’s year (period of one rotation) depends from the density of the Sun and from the ratio of the radius of the orbit to the radius of the Sun. According to the problem, all the linear dimensions of the solar system were reduced then the length of Earth’s year remains the same. Of course, if the density of the Sun will remain the same.

Image Credit: NASA