# 0.5 4.6 kepler’s laws  (Page 3/3)

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${F}_{\text{net}}={\text{ma}}_{\text{c}}=m\frac{{v}^{2}}{r}\text{.}$

The net external force on mass $m$ is gravity, and so we substitute the force of gravity for ${F}_{\text{net}}$ :

$G\frac{\text{mM}}{{r}^{2}}=m\frac{{v}^{2}}{r}\text{.}$

The mass $m$ cancels, yielding

$G\frac{M}{r}={v}^{2}\text{.}$

The fact that $m$ cancels out is another aspect of the oft-noted fact that at a given location all masses fall with the same acceleration. Here we see that at a given orbital radius $r$ , all masses orbit at the same speed. (This was implied by the result of the preceding worked example.) Now, to get at Kepler’s third law, we must get the period $T$ into the equation. By definition, period $T$ is the time for one complete orbit. Now the average speed $v$ is the circumference divided by the period—that is,

$v=\frac{2\pi r}{T}\text{.}$

Substituting this into the previous equation gives

$G\frac{\text{M}}{r}=\frac{{\mathrm{4\pi }}^{2}{r}^{2}}{{T}^{2}}\text{.}$

Solving for ${T}^{2}$ yields

${T}^{2}=\frac{{4\pi }^{2}}{\text{GM}}{r}^{3}\text{.}$

Using subscripts 1 and 2 to denote two different satellites, and taking the ratio of the last equation for satellite 1 to satellite 2 yields

This is Kepler’s third law. Note that Kepler’s third law is valid only for comparing satellites of the same parent body, because only then does the mass of the parent body $M$ cancel.

Now consider what we get if we solve ${T}^{2}=\frac{{4\pi }^{2}}{\text{GM}}{r}^{3}$ for the ratio ${r}^{3}/{T}^{2}$ . We obtain a relationship that can be used to determine the mass $M$ of a parent body from the orbits of its satellites:

$\frac{{r}^{3}}{{T}^{2}}=\frac{G}{{4\pi }^{2}}M\text{.}$

If $r$ and $T$ are known for a satellite, then the mass $M$ of the parent can be calculated. This principle has been used extensively to find the masses of heavenly bodies that have satellites. Furthermore, the ratio ${r}^{3}/{T}^{2}$ should be a constant for all satellites of the same parent body (because ${r}^{3}/{T}^{2}=\text{GM}/{4\pi }^{2}$ ). (See [link] ).

It is clear from [link] that the ratio of ${r}^{3}/{T}^{2}$ is constant, at least to the third digit, for all listed satellites of the Sun, and for those of Jupiter. Small variations in that ratio have two causes—uncertainties in the $r$ and $T$ data, and perturbations of the orbits due to other bodies. Interestingly, those perturbations can be—and have been—used to predict the location of new planets and moons. This is another verification of Newton’s universal law of gravitation.

## Section summary

• Kepler’s laws are stated for a small mass $m$ orbiting a larger mass $M$ in near-isolation. Kepler’s laws of planetary motion are then as follows:

Kepler’s first law

The orbit of each planet about the Sun is an ellipse with the Sun at one focus.

Kepler’s second law

Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times.

Kepler’s third law

The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun:

where $T$ is the period (time for one orbit) and $r$ is the average radius of the orbit.

• The period and radius of a satellite’s orbit about a larger body $M$ are related by
${T}^{2}=\frac{{4\pi }^{2}}{\text{GM}}{r}^{3}$

or

$\frac{{r}^{3}}{{T}^{2}}=\frac{G}{{4\pi }^{2}}M\text{.}$

## Problem exercises

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation).

Unreasonable Results

(a) Based on Kepler’s laws and information on the orbital characteristics of the Moon, calculate the orbital radius for an Earth satellite having a period of 1.00 h. (b) What is unreasonable about this result? (c) What is unreasonable or inconsistent about the premise of a 1.00 h orbit?

a) $5\text{.}\text{08}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{km}$

b) This radius is unreasonable because it is less than the radius of earth.

c) The premise of a one-hour orbit is inconsistent with the known radius of the earth.

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