Question

8. A certain star has a mass four times that of our Sun. A planet in orbit around that star has a semimajor axis of 0.45 AU and an eccentricity of 0.32. What is the rate of precession of this planet's "perihelion"?

Answer 1

According to Kepler's orbital period of a planet, T, with a semi major axis a, around a star of mass M, is,

$T = 2\pi\sqrt{\frac{a^3}{GM}} = 2\pi\sqrt{\frac{(0.45*1.49*10^{11})^3}{4*6.67*10^{-11}*1.98*10^{30}}}\\\\ = 4.74*10^6s = 0.15~\textup{year}$G is gravitational constant, all quantities converted into SI units

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The apsidal precession due to relativistic effects is during one period of revolution is,

$\varepsilon =24\pi ^{3}{\frac {\alpha ^{2}}{T^{2}c^{2}\left(1-e^{2}\right)}}$,

Where,

• the semi major-axis of its orbit being α,
• the eccentricity of the orbit e
• Speed of light is c
• and the period of revolution T, then

$\varepsilon =24\pi ^{3}{\frac {(0.45*1.496*10^{11})^{2}}{(4.74*10^6)^{2}(3*10^8)^{2}\left(1-0.32^{2}\right)}}=1.85*10^{-6}~\textup{radian}$ - per 0.15year

so, in rad/s unit the rate of precession is,

$\frac{1.85*10^{-6}}{4.74*10^{6}} = 3.9*10^{-13} rad/s$