# Kepler’s 3rd Law Formula and Numericals on it

## Kepler’s 3rd Law Formula

Statement: “The square of its revolution around the sun is directly proportional to cube of the distance of planet from sun”.

From fig, S = Sun

P = Planet

F = Force

If r is the arrange distance of the planet from the sun and T is period of revolution, then from the above statement.

T2 ∝ r3

T2 = Kr3

K is const of proportionality

I.e. T2/r3 = constant = K

So, T2 = Kr3 —- (1)

This is required equation.

### Numericals on Kepler’s 3rd Law:

The time period at Jupiter 11.86 years to complete the one revolution around sun. What is its distance from sun? Distance of the earth from the sun is 1.5×1011 m.

Given:  TJ = 11.86 years

re = 1.5×1011m

Te = 1 year

RJ =?

We know the Kepler’s third law

T2 ∝ R3

Te2 ∝ r3e —– (i)

TJ2 ∝ rJ3 —— (ii)

(Te/TJ)2 = (re/rJ)3

(Te/TJ)2/3 = re/rJ

rJ = re (TJ/Te)2/3

= 1.5×1011 (11.86/1)2/3

RJ = 7.90 × 1011 m

Updated: August 19, 2021 — 12:54 pm
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}