# How to calculate the time the earth takes to go around the sun, using Newton’s Universal Law of Gravitation?

Last updated on January 5th, 2022 at 01:41 pm

**Here we will see how to calculate the time the Earth takes to go around the Sun once. **

We will use

*to calculate this.*

**Newton’s Universal Law of Gravitation**## **Calculating the time taken by the earth to go around the Sun Using Newton’s Universal Law of Gravitation**

Applying Newton’s universal law of Gravitation, we will find the value of the gravitational force experienced by the Earth due to the presence of the Sun.

This formula is given by: *F =* *Gm _{1}m_{2}/r^{2}*

Where *m*1 = mass of the Sun = 1.99 x 10^{30} kg*m*2 = mass of the Earth = 5.97 x 10^{24 }kg*r *= distance between the Sun and Earth = 1.49 x 10^{11} m

After calculation we get the value of this force as follows:

**So F = 3.56 x 10^{22} N**………………… (1)

We know that the Earth travels in an elliptical orbit around the Sun, but we can take this to be a circular orbit for the purposes of this calculation.

From our knowledge of circular motion, we know that the force acting on the Earth towards the center of the circle is the centripetal force given by the equation ** F = mv^{2} / r** ……………….. (2)

*And, here this centripetal force is supplied by the gravitational force.* *Hence the force in equations 1 and 2 are equal.*

Using equations (1) and (2):**So the velocity of the earth, v = (F r/m)^{1/2} = [( 3.56 x 10^{22} **N)(

**1.49 x 10**

^{11}m)/( 5.97 x 10

^{24 }kg)]

^{1/2}

**= 29846 m/s**

**The circumference of the orbit = 2π**)=

*r =*(2)(22/7)(**1.49 x 10**^{11}**9.38 x 10**

^{11}m**Time taken for 1 orbit = circumference/velocity = 9.38 x 10 ^{11}/29 846= 3.14 x 10^{7} s**

**This is equal to 1 year.**

**Summary:**

Thus, we can **calculate the time taken by the earth to go around the Sun Using Newton’s Universal Law of Gravitation**.

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