Question

If the earth is one -fourth of its present distance from the sun what will be the duration of the year.

Answer 1

Given :-

$R_1 = \dfrac{R}{4}$

To find :-

The duration of the year.

Solution:-

Let the present distance between earth and sun be R.

And the original period of revolution be T.

After decreasing the distance it's period of Revolution be $T_1$

A/Q

If earth distance is 1/4 of its original distance.

$R_1= \dfrac{R}{4}$

• Let, Recall about Kepler's law.

Kepler's Law :-

According to kepler's law The square of the time period of revolution of earth around the sun is directly proportional to the cube of distance between the earth and sun.

• By using Kepler's law.

→$T^2\propto R^3$

→$T^2 = K R^3$

→$K = \dfrac{T^2}{R^3}----1)$

→$T_1^2 = K R_1^3$

→$K = \dfrac{T_1^2}{R_1^3}---2)$

• From kepler's equation.

→$\dfrac{T_1^2}{T^2} = \dfrac{R_1^3}{R^3}$

→$\dfrac{T_1^2}{T^2} = (\dfrac{\dfrac{R}{4}}{R})^3$

→$\dfrac{T_1^2 }{T^2}= (\dfrac{R}{4R})^3$

→$T_1^2 = (\dfrac{1}{4})^3\times T^2$

• T = 365 days for earth.

→$T_1^2 = \dfrac{1}{64}\times (365)^2$

→$T_1^2 = \dfrac{1}{64}\times 133,225$

→$T_1^2 =2,081.64$

→$T_1 = \sqrt{2,081.64}$

→$T_1 = 45.62 days$

hence,

The period of revolution is 45.62 days.