The orbital period can be calculated using Kepler's 3rd law, which states that the square of the orbital period is proportional to the cube of the average distance from ....
1 Answers 1 viewsObserving their position on a star map and wait till they return to that position again. and counting days elapsed..
1 Answers 1 viewsThe Earth's tilt axis , or obliquity, is changing continuously over a 41,000 year cycle. The angle of tilt varies between 22.1 and 24.5 degrees. It is currently at 23.44...
1 Answers 1 viewsThis is only an approximation to the period, using Kepler's third (empirical) law for approximating the period..
1 Answers 1 viewsUse $v=sqrtmu((2/r)-(1/a))$, perihelion = $a(1-e)$ and aphelion $= a(1+e)$. Here e = 0.25, approximately. Now, for any planet, the ratio of the speeds at perihelion and aphelion becomes $sqrt((1+e)/(1-e))$ and...
1 Answers 1 viewsIt states that the orbital period, T is related to the distance, r as: $T^2 = (G.M)/(4pi)r^3$ where G is the universal gravitational constant (a very small number for reasons...
1 Answers 1 viewsorbital period is 224.7 days. Date from Planetary fact Sheet NASA,
1 Answers 1 viewsIf a is the semi-major axis and e is the eccentricity of the orbit of the asteroid, Aphelion/perihelion = 3 = (a(1+e))/(a(1-e)) = (1+e)/(1-e). solving for e, e = $1/2$...
1 Answers 1 views$a$ - Semi-major distance; $\quad b$ - Semi-minor distance; $c$ - Centre-to-Focal Point distance; $\quad e$ - Eccentricity of the ellipse; $d_p$ - Perihelion distance; $\quad d_a $ - Aphelion...
1 Answers 1 viewsTo do this problem, you need the $R = 6371 km$ For the satellite to be in a stable orbit at a height, h, its centripetal acceleration $V^2/(R + h)$...
1 Answers 1 views