Question

Kepler's third law relates the period of a planet
to its orbital radius r, the constant G in Newton's law of gravi-
tation (F = Gm1m2/r^2), and the mass of the sun M.. What com-
bination of these factors gives the correct dimensions for the
period of a planet?​

Answer 1

Answer:

√(r³ / (MG))

Explanation:

The dimensions of each variable are:

r = [m]

G = [m³/kg/s²]

M = [kg]

Multiplying M and G eliminates kilograms:

MG = [m³/s²]

The radius cubed divided by MG eliminates meters:

r³ / (MG) = [s²]

The square root gives us seconds:

√(r³ / (MG)) = [s]

Answer 2

Kepler's third law relates the time  period of a planet to its orbital radius r, the gravitational constant G  and the mass of the sun M, the combination of these factors gives the correct dimensions for the period of a planet would be √( 4πr³/GM)

What are Kepler's laws of planetary motion?

There are three laws of Kepler as follows

The orbit of the planet is elliptical and the sun is present at one of the two focuses of the elliptical path followed by the revolving planet.

A line segment joining a planet and the Sun covers equal areas during equal intervals of the time period.

The square of a planet's orbital period is proportional to the cube of the length of the semi-major axis of its elliptical orbit.

Think about a small mass M traveling in a circle around a large mass M. The centripetal force for mass m comes from gravity. Taking circular motion as an example and applying Newton's second law,

gravitational force balances the centripetal force

GMm/r²  = mv²/r

GM/r = v²/r

Let us take a Time period of the orbital is P

v = 2πr/P

GM/r = 4πr²/P²

P² = 4πr³/GM

P = √( 4πr³/GM)

Thus the correct dimension for the time period in terms of the orbital radius (r), gravitational constant (G ), and the mass of the sun (M) would be √( 4πr³/GM).

Learn more about Kepler's laws of planetary motion

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