Physics, asked by cutiepie441, 10 months ago

Formula of kepler first law

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Answered by vinayakaj78
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Answered by amanraj143
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Explanation:

Hey mate

here is your answer

the formula of Kepler first law is

Mathematically, an ellipse can be represented by the formula:

{\displaystyle r={\frac {p}{1+\varepsilon \,\cos \theta }},} {\displaystyle r={\frac {p}{1+\varepsilon \,\cos \theta }},}

where {\displaystyle p} p is the semi-latus rectum, ε is the eccentricity of the ellipse, r is the distance from the Sun to the planet, and θ is the angle to the planet's current position from its closest approach, as seen from the Sun. So (r, θ) are polar coordinates.

For an ellipse 0 < ε < 1 ; in the limiting case ε = 0, the orbit is a circle with the Sun at the centre (i.e. where there is zero eccentricity).

At θ = 0°, perihelion, the distance is minimum

{\displaystyle r_{\min }={\frac {p}{1+\varepsilon }}} {\displaystyle r_{\min }={\frac {p}{1+\varepsilon }}}

At θ = 90° and at θ = 270° the distance is equal to {\displaystyle p} p.

At θ = 180°, aphelion, the distance is maximum (by definition, aphelion is – invariably – perihelion plus 180°)

{\displaystyle r_{\max }={\frac {p}{1-\varepsilon }}} {\displaystyle r_{\max }={\frac {p}{1-\varepsilon }}}

The semi-major axis a is the arithmetic mean between rmin and rmax:

{\displaystyle {\begin{aligned}r_{\max }-a&=a-r_{\min }\\[3pt]a&={\frac {p}{1-\varepsilon ^{2}}}\end{aligned}}} {\displaystyle {\begin{aligned}r_{\max }-a&=a-r_{\min }\\[3pt]a&={\frac {p}{1-\varepsilon ^{2}}}\end{aligned}}}

The semi-minor axis b is the geometric mean between rmin and rmax:

{\displaystyle {\begin{aligned}{\frac {r_{\max }}{b}}&={\frac {b}{r_{\min }}}\\[3pt]b&={\frac {p}{\sqrt {1-\varepsilon ^{2}}}}\end{aligned}}} {\displaystyle {\begin{aligned}{\frac {r_{\max }}{b}}&={\frac {b}{r_{\min }}}\\[3pt]b&={\frac {p}{\sqrt {1-\varepsilon ^{2}}}}\end{aligned}}}

The semi-latus rectum p is the harmonic mean between rmin and rmax:

{\displaystyle {\begin{aligned}{\frac {1}{r_{\min }}}-{\frac {1}{p}}&={\frac {1}{p}}-{\frac {1}{r_{\max }}}\\[3pt]pa&=r_{\max }r_{\min }=b^{2}\,\end{aligned}}} {\displaystyle {\begin{aligned}{\frac {1}{r_{\min }}}-{\frac {1}{p}}&={\frac {1}{p}}-{\frac {1}{r_{\max }}}\\[3pt]pa&=r_{\max }r_{\min }=b^{2}\,\end{aligned}}}

The eccentricity ε is the coefficient of variation between rmin and rmax:

{\displaystyle \varepsilon ={\frac {r_{\max }-r_{\min }}{r_{\max }+r_{\min }}}.} {\displaystyle \varepsilon ={\frac {r_{\max }-r_{\min }}{r_{\max }+r_{\min }}}.}

The area of the ellipse is

{\displaystyle A=\pi ab\,.} {\displaystyle A=\pi ab\,.}

The special case of a circle is ε = 0, resulting in r = p = rmin = rmax = a = b and A = πr2.

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