Question

How can the orbital semi-major axis, eccentricity, and period be calculated if perihelion and aphelion are known?

Answer 1

Hint: Ellipse is the locus of all the points in the plane. Ellipse has its centre point, major axis, minoraxis. The major axis is the axis which is the longest diameter of the ellipse. The minor axis is called the shortest diameter of the ellipse. The eccentricity defines the elongation of the ellipse. The eccentricity, e lies in between \[0\] and \[1\] .

Complete step by step answer:

To find semi-major axis:
• We know that the longest diameter of the ellipse is called the major-axis. The half of the major axis is called the semi-major axis.
• We can find the semi-major axis by knowing the perihelion and aphelion distance.
• In the ellipse, the shortest distance from the centre of the ellipse is called perihelion. The farthest distance from the centre is called aphelion. In Greek, ‘Peri’ means close and ‘Apo’ means far. And ‘Helio’ means sun. Therefore here the ellipse is the orbit of the Earth around the sun.
• The sum of perihelion and the aphelion gives the value of the major axis as the sum defines the largest diameter. By just dividing the major axis, we can find semi-major axes.
• Lets take \[R\] is the average distance of the Earth from the sun. The perihelion value is taken as \[{R_p}\] and the aphelion value is taken as \[{R_a}\]
Therefore,
\[R = \dfrac{{{R_p} + {R_a}}}{2}\]
This \[R\]value gives the value of the semi-major axis.

To find the eccentricity:
• The value of eccentricity defines how much the ellipse is elongated. The value of eccentricity is denoted by ‘e’. The value of ‘e’ lies in between \[0\] and \[1\].
• The perihelion value is taken as \[{R_p}\] and the aphelion value is taken as \[{R_a}\]
\[ \Rightarrow {R_a} = 1 + e\]
\[ \Rightarrow {R_p} = 1 - e\]
Dividing these two values give,
\[\dfrac{{{R_a}}}{{{R_p}}} = \dfrac{{1 + e}}{{1 - e}}\]
\[ \Rightarrow {R_a}(1 - e) = {R_p}(1 + e)\]
\[ \Rightarrow {R_a} - {R_p} = \left( {{R_p} + {R_a}} \right)e\]
\[\therefore e = \dfrac{{{R_a} - {R_p}}}{{{R_p} + {R_a}}}\]
Hence the value of eccentricity can be found by perihelion and aphelion.

To find period:
• According to Kepler’s 3rd law, the square of the orbital time period is directly proportional to the cube of the average distance from the sun i.e. \[{T^2}\alpha {R^3}\]
\[\therefore \dfrac{{{T^2}}}{{{R^3}}} = C\]
\[ \Rightarrow T = \sqrt {C{R^3}} \]
• We already found the value of \[R\] by using perihelion and aphelion values. Now we can find the orbital time period by using the value of \[R\].

Note:
• Kepler’s 3rd law applies to orbiting bodies like the sun. The value of C only differs with respect to the average distance of the orbit.
• Earth’s average distance \[R\] is equal to 1AU.
• The Earth approaches its nearest point ‘perihelion’ with the sun in the month of January. It approaches its farthest point ‘aphelion’ in the month of July.