
How can I use Kepler's law of harmonies to predict the time Mars takes to orbit the sun?
Answer
546.3k+ views
Hint: There are three kepler laws which are referred to as the law of orbits of kepler, the law of regions of kepler, and the law of periods of kepler. Kepler's Law states that in elliptical orbits the planets move around the sun with the sun at one focus, so we called this law the law of planetary motions of Kepler.
Complete solution:
Kepler formulated his first two laws in 1609 to summarise the remarkable astronomical observations over many years made by his mentor, Tycho Brahe. Ten years later, his third law, sometimes called the Law of Harmonies, was published. It is a wonderful instance of a power law between an orbiting body's period and its average distance to the body it orbits.
Kepler's law of harmonies (also known as Kepler's third law) states that the orbital time period $T$ is related to the mean distance $R$ from the sun by
$T^{2} \propto R^{3}$
For Mars $R_{\operatorname{mars}}=2.28 \times 10^{11} \mathrm{~m}$
while for earth, ${{R}_{\text{earth }}}=1.50\times {{10}^{11}}~\text{m}$
Thus,
$\left(\dfrac{T_{\text {mars }}}{T_{\text {earth }}}\right)^{2}=\left(\dfrac{R_{\text {mars }}}{R_{\text {earth }}}\right)^{3}=\left(\dfrac{2.28 \times 10^{11} \mathrm{~m}}{1.50 \times 10^{11} \mathrm{~m}}\right)^{3} \approx 3.51$
Hence,
$\dfrac{T_{\text {mars }}}{T_{\text {earth }}}=\sqrt{3.51} \approx 1.87$
The time Mars takes to orbit the sun is $\frac{T_{\text {mars }}}{T_{\text {earth }}}=\sqrt{3.51} \approx 1.87$
Note:
Another law here is the Kepler Law of Periods, which states that the square of the planet's time period is directly proportional to the cube of its orbit's semimajor axis. Which is represented as $T^{2} \propto a^{3}$ mathematically. There is a small derivation, where $T$ is time, to derive this expression. This is Kepler's First Law, with the additional provision that in the case of a small planet revolving around a large sun, the sun will be at one of the focal points of the planet's elliptical orbit. Most planets have elliptical orbits. Most orbits are actually quite close to being circles.
Complete solution:
Kepler formulated his first two laws in 1609 to summarise the remarkable astronomical observations over many years made by his mentor, Tycho Brahe. Ten years later, his third law, sometimes called the Law of Harmonies, was published. It is a wonderful instance of a power law between an orbiting body's period and its average distance to the body it orbits.
Kepler's law of harmonies (also known as Kepler's third law) states that the orbital time period $T$ is related to the mean distance $R$ from the sun by
$T^{2} \propto R^{3}$
For Mars $R_{\operatorname{mars}}=2.28 \times 10^{11} \mathrm{~m}$
while for earth, ${{R}_{\text{earth }}}=1.50\times {{10}^{11}}~\text{m}$
Thus,
$\left(\dfrac{T_{\text {mars }}}{T_{\text {earth }}}\right)^{2}=\left(\dfrac{R_{\text {mars }}}{R_{\text {earth }}}\right)^{3}=\left(\dfrac{2.28 \times 10^{11} \mathrm{~m}}{1.50 \times 10^{11} \mathrm{~m}}\right)^{3} \approx 3.51$
Hence,
$\dfrac{T_{\text {mars }}}{T_{\text {earth }}}=\sqrt{3.51} \approx 1.87$
The time Mars takes to orbit the sun is $\frac{T_{\text {mars }}}{T_{\text {earth }}}=\sqrt{3.51} \approx 1.87$
Note:
Another law here is the Kepler Law of Periods, which states that the square of the planet's time period is directly proportional to the cube of its orbit's semimajor axis. Which is represented as $T^{2} \propto a^{3}$ mathematically. There is a small derivation, where $T$ is time, to derive this expression. This is Kepler's First Law, with the additional provision that in the case of a small planet revolving around a large sun, the sun will be at one of the focal points of the planet's elliptical orbit. Most planets have elliptical orbits. Most orbits are actually quite close to being circles.
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